# 14.7: Triple Integrals in Cylindrical and Spherical Coordinates - Mathematics

We have seen that sometimes double integrals are simplified by doing them in polar coordinates; not surprisingly, triple integrals are sometimes simpler in cylindrical coordinates or spherical coordinates. We need to do the same thing here, for three dimensional regions.

The cylindrical coordinate system is the simplest, since it is just the polar coordinate system plus a (z) coordinate. A typical small unit of volume is the shape shown below "fattened up'' in the (z) direction, so its volume is (rDelta rDelta hetaDelta z), or in the limit, (r,dr,d heta,dz). A polar coordinates "grid".

Example (PageIndex{1})

Find the volume under (z=sqrt{4-r^2}) above the quarter circle inside (x^2+y^2=4) in the first quadrant.

Solution

We could of course do this with a double integral, but we'll use a triple integral:

[int_0^{pi/2}int_0^2int_0^{sqrt{4-r^2}} r,dz,dr,d heta=
int_0^{pi/2}int_0^2 sqrt{4-r^2}; r,dr,d heta=
{4piover3}.]

Compare this to Example 15.2.1.

Example (PageIndex{2})

An object occupies the space inside both the cylinder (x^2+y^2=1) and the sphere (x^2+y^2+z^2=4), and has density (x^2) at ((x,y,z)). Find the total mass.

Solution

We set this up in cylindrical coordinates, recalling that (x=rcos heta):

[eqalign{
int_0^{2pi}int_0^1int_{-sqrt{4-r^2}}^{sqrt{4-r^2}}
r^3cos^2( heta),dz,dr,d heta
&=int_0^{2pi}int_0^1 2sqrt{4-r^2};r^3cos^2( heta),dr,d hetacr
&=int_0^{2pi}
left({128over15}-{22over5}sqrt3 ight)cos^2( heta),d hetacr
&=left({128over15}-{22over5}sqrt3 ight)picr
}]

Spherical coordinates are somewhat more difficult to understand. The small volume we want will be defined by (Delta ho), (Deltaphi), and (Delta heta), as pictured in Figure (PageIndex{1}).

The small volume is nearly box shaped, with 4 flat sides and two sides formed from bits of concentric spheres. When (Delta ho), (Deltaphi), and (Delta heta) are all very small, the volume of this little region will be nearly the volume we get by treating it as a box. One dimension of the box is simply (Delta ho), the change in distance from the origin. The other two dimensions are the lengths of small circular arcs, so they are (rDeltaalpha) for some suitable (r) and (alpha), just as in the polar coordinates case. Figure (PageIndex{1}): A small unit of volume for a spherical coordinates (AP)

The easiest of these to understand is the arc corresponding to a change in (phi), which is nearly identical to the derivation for polar coordinates, as shown in the left graph in Figure (PageIndex{2}). In that graph we are looking "face on'' at the side of the box we are interested in, so the small angle pictured is precisely (Deltaphi), the vertical axis really is the (z) axis, but the horizontal axis is not a real axis---it is just some line in the (x)-(y) plane. Because the other arc is governed by ( heta), we need to imagine looking straight down the (z) axis, so that the apparent angle we see is (Delta heta). In this view, the axes really are the (x) and (y) axes. In this graph, the apparent distance from the origin is not ( ho) but ( hosinphi), as indicated in the left graph. Figure (PageIndex{2}): Setting up integration in spherical coordinates.

The upshot is that the volume of the little box is approximately (Delta ho( hoDeltaphi)( hosinphiDelta heta) = ho^2sinphiDelta hoDeltaphiDelta heta), or in the limit ( ho^2sinphi,d ho,dphi,d heta).

Example (PageIndex{3})

Suppose the temperature at ((x,y,z)) is [T=dfrac{1}{1+x^2+y^2+z^2}. onumber] Find the average temperature in the unit sphere centered at the origin.

Solution

In two dimensions we add up the temperature at "each'' point and divide by the area; here we add up the temperatures and divide by the volume, ((4/3)pi):

[{3over4pi}int_{-1}^1int_{-sqrt{1-x^2}}^{sqrt{1-x^2}}
int_{-sqrt{1-x^2-y^2}}^{sqrt{1-x^2-y^2}}
{1over1+x^2+y^2+z^2},dz,dy,dx onumber
]

This looks quite messy; since everything in the problem is closely related to a sphere, we'll convert to spherical coordinates.

[{3over4pi}int_0^{2pi}int_0^pi
int_0^1
{1over1+ ho^2}, ho^2sinphi,d ho,dphi,d heta
={3over4pi}(4pi -pi^2)=3-{3piover4}. onumber
]

## APEX Calculus

Just as polar coordinates gave us a new way of describing curves in the plane, in this section we will see how cylindrical and spherical coordinates give us new ways of desribing surfaces and regions in space.

### Subsection 14.7.1 Cylindrical Coordinates

In short, cylindrical coordinates can be thought of as a combination of the polar and rectangular coordinate systems. One can identify a point ((x_0,y_0,z_0) ext<,>) given in rectangular coordinates, with the point ((r_0, heta_0,z_0) ext<,>) given in cylindrical coordinates, where the (z)-value in both systems is the same, and the point ((x_0,y_0)) in the (x)-(y) plane is identified with the polar point (P(r_0, heta_0) ext<>) see Figure 14.7.1. So that each point in space that does not lie on the (z)-axis is defined uniquely, we will restrict (rgeq 0) and (0leq hetaleq 2pi ext<.>)

We use the identity (z=z) along with the identities found in Key Idea 10.4.6 to convert between the rectangular coordinate ((x,y,z)) and the cylindrical coordinate ((r, heta,z) ext<,>) namely:

These identities, along with conversions related to spherical coordinates, are given later in Key Idea 14.7.11.

Our rectangular to polar conversion formulas used (r^2=x^2+y^2 ext<,>) allowing for negative (r) values. Since we now restrict (rgeq 0 ext<,>) we can use (r=sqrt ext<.>)

###### Example 14.7.2 . Converting between rectangular and cylindrical coordinates.

Convert the rectangular point ((2,-2,1)) to cylindrical coordinates, and convert the cylindrical point ((4,3pi/4,5)) to rectangular.

Following the identities given above (and, later in Key Idea 14.7.11), we have (r = sqrt <2^2+(-2)^2>= 2sqrt<2> ext<.>) Using ( an( heta) = y/x ext<,>) we find ( heta = an^<-1>(-2/2) =-pi/4 ext<.>) As we restrict ( heta) to being between (0) and (2pi ext<,>) we set ( heta = 7pi/4 ext<.>) Finally, (z = 1 ext<,>) giving the cylindrical point ((2sqrt2,7pi/4,1) ext<.>)

In converting the cylindrical point ((4,3pi/4,5)) to rectangular, we have (x = 4cosig(3pi/4ig) = -2sqrt<2> ext<,>) (y = 4sinig(3pi/4ig) = 2sqrt<2>) and (z=5 ext<,>) giving the rectangular point ((-2sqrt<2>,2sqrt<2>,5) ext<.>)

Setting each of (r ext<,>) ( heta) and (z) equal to a constant defines a surface in space, as illustrated in the following example.

###### Example 14.7.3 . Canonical surfaces in cylindrical coordinates.

Describe the surfaces (r=1 ext<,>) ( heta = pi/3) and (z=2 ext<,>) given in cylindrical coordinates.

The equation (r=1) describes all points in space that are 1 unit away from the (z)-axis. This surface is a “tube” or “cylinder” of radius 1, centered on the (z)-axis, as graphed in Figure 11.1.12 (which describes the cylinder (x^2+y^2=1) in space).

The equation ( heta=pi/3) describes the plane formed by extending the line ( heta=pi/3 ext<,>) as given by polar coordinates in the (x)-(y) plane, parallel to the (z)-axis.

The equation (z=2) describes the plane of all points in space that are 2 units above the (x)-(y) plane. This plane is the same as the plane described by (z=2) in rectangular coordinates.

All three surfaces are graphed in Figure 14.7.4. Note how their intersection uniquely defines the point (P=(1,pi/3,2) ext<.>)

Cylindrical coordinates are useful when describing certain domains in space, allowing us to evaluate triple integrals over these domains more easily than if we used rectangular coordinates.

Theorem 14.6.25 shows how to evaluate (iiint_Dh(x,y,z), dV) using rectangular coordinates. In that evaluation, we use (dV = dz,dy,dx) (or one of the other five orders of integration). Recall how, in this order of integration, the bounds on (y) are “curve to curve” and the bounds on (x) are “point to point”: these bounds describe a region (R) in the (x)-(y) plane. We could describe (R) using polar coordinates as done in Section 14.3. In that section, we saw how we used (dA = r,dr,d heta) instead of (dA = dy,dx ext<.>)

Considering the above thoughts, we have (dV = dzig(r,dr,d hetaig) = r,dz,dr,d heta ext<.>) We set bounds on (z) as “surface to surface” as done in the previous section, and then use “curve to curve” and “point to point” bounds on (r) and ( heta ext<,>) respectively. Finally, using the identities given above, we change the integrand (h(x,y,z)) to (h(r, heta,z) ext<.>)

This process should sound plausible the following theorem states it is truly a way of evaluating a triple integral.

###### Theorem 14.7.5 . Triple Integration in Cylindrical Coordinates.

Let (w=h(r, heta,z)) be a continuous function on a closed, bounded region (D) in space, bounded in cylindrical coordinates by (alpha leq heta leq eta ext<,>) (g_1( heta)leq r leq g_2( heta)) and (f_1(r, heta) leq z leq f_2(r, heta) ext<.>) Then

###### Example 14.7.6 . Evaluating a triple integral with cylindrical coordinates.

Find the mass of the solid represented by the region in space bounded by (z=0 ext<,>) (z=sqrt<4-x^2-y^2>+3) and the cylinder (x^2+y^2=4) (as shown in Figure 14.7.7), with density function (delta(x,y,z) = x^2+y^2+z+1 ext<,>) using a triple integral in cylindrical coordinates. Distances are measured in centimeters and density is measured in grams/cm(^3 ext<.>)

We begin by describing this region of space with cylindrical coordinates. The plane (z=0) is left unchanged with the identity (r=sqrt ext<,>) we convert the hemisphere of radius 2 to the equation (z=sqrt<4-r^2> ext<>) the cylinder (x^2+y^2=4) is converted to (r^2=4 ext<,>) or, more simply, (r=2 ext<.>) We also convert the density function: (delta(r, heta,z) = r^2+z+1 ext<.>)

To describe this solid with the bounds of a triple integral, we bound (z) with (0leq zleq sqrt<4-r^2>+3 ext<>) we bound (r) with (0 leq r leq 2 ext<>) we bound ( heta) with (0 leq heta leq 2pi ext<.>)

Using Definition 14.6.26 and Theorem 14.7.5, we have the mass of the solid is

where we leave the details of the remaining double integral to the reader.

###### Example 14.7.8 . Finding the center of mass using cylindrical coordinates.

Find the center of mass of the solid with constant density whose base can be described by the polar curve (r=cos(3 heta)) and whose top is defined by the plane (z=1-x+0.1y ext<,>) where distances are measured in feet, as seen in Figure 14.7.9. (The volume of this solid was found in Example 14.3.10.)

We convert the equation of the plane to use cylindrical coordinates: (z= 1-rcos( heta)+0.1rsin( heta) ext<.>) Thus the region is space is bounded by (0 leq z leq 1-rcos( heta) + 0.1rsin( heta) ext<,>) (0 leq r leq cos(3 heta) ext<,>) (0 leq heta leq pi) (recall that the rose curve (r=cos(3 heta)) is traced out once on ([0,pi] ext<.>)

Since density is constant, we set (delta = 1) and finding the mass is equivalent to finding the volume of the solid. We set up the triple integral to compute this but do not evaluate it we leave it to the reader to confirm it evaluates to the same result found in Example 14.3.10.

From Definition 14.6.26 we set up the triple integrals to compute the moments about the three coordinate planes. The computation of each is left to the reader (using technology is recommended):

The center of mass, in rectangular coordinates, is located at ((-0.147,0.015,0.467) ext<,>) which lies outside the bounds of the solid.

### Subsection 14.7.2 Spherical Coordinates

In short, spherical coordinates can be thought of as a “double application” of the polar coordinate system. In spherical coordinates, a point (P) is identified with (( ho, heta,varphi) ext<,>) where ( ho) is the distance from the origin to (P ext<,>) ( heta) is the same angle as would be used to describe (P) in the cylindrical coordinate system, and (varphi) is the angle between the positive (z)-axis and the ray from the origin to (P ext<>) see Figure 14.7.10. So that each point in space that does not lie on the (z)-axis is defined uniquely, we will restrict ( ho geq 0 ext<,>) (0 leq heta leq 2pi) and (0 leq varphi leq pi ext<.>)

The symbol ( ho) is the Greek letter “rho.” Traditionally it is used in the spherical coordinate system, while (r) is used in the polar and cylindrical coordinate systems.

The following Key Idea gives conversions to/from our three spatial coordinate systems.

###### Key Idea 14.7.11 . Converting Between Rectangular, Cylindrical and Spherical Coordinates.

The role of ( heta) and (varphi) in spherical coordinates differs between mathematicians and physicists. When reading about physics in spherical coordinates, be careful to note how that particular author uses these variables and recognize that these identities will may no longer be valid.

###### Example 14.7.12 . Converting between rectangular and spherical coordinates.

Convert the rectangular point ((2,-2,1)) to spherical coordinates, and convert the spherical point ((6,pi/3,pi/2)) to rectangular and cylindrical coordinates.

This rectangular point is the same as used in Example 14.7.2. Using Key Idea 14.7.11, we find ( ho = sqrt <2^2+(-1)^2+1^2>= 3 ext<.>) Using the same logic as in Example 14.7.2, we find ( heta = 7pi/4 ext<.>) Finally, (cos(varphi) = 1/3 ext<,>) giving (varphi = cos^<-1>(1/3) approx 1.23 ext<,>) or about (70.53^circ ext<.>) Thus the spherical coordinates are approximately ((3,7pi/4,1.23) ext<.>)

Converting the spherical point ((6,pi/3,pi/2)) to rectangular, we have (x = 6sin(pi/2)cos(pi/3) = 3 ext<,>) (y = 6sin(pi/2)sin(pi/3) = 3sqrt<3>) and (z = 6cos(pi/2) = 0 ext<.>) Thus the rectangular coordinates are ((3,3sqrt<3>,0) ext<.>)

To convert this spherical point to cylindrical, we have (r = 6sin(pi/2) = 6 ext<,>) ( heta = pi/3) and (z = 6cos(pi/2) =0 ext<,>) giving the cylindrical point ((6,pi/3,0) ext<.>)

###### Example 14.7.13 . Canonical surfaces in spherical coordinates.

Describe the surfaces ( ho=1 ext<,>) ( heta = pi/3) and (varphi = pi/6 ext<,>) given in spherical coordinates.

The equation ( ho = 1) describes all points in space that are 1 unit away from the origin: this is the sphere of radius 1, centered at the origin.

The equation ( heta = pi/3) describes the same surface in spherical coordinates as it does in cylindrical coordinates: beginning with the line ( heta = pi/3) in the (x)-(y) plane as given by polar coordinates, extend the line parallel to the (z)-axis, forming a plane.

The equation (varphi=pi/6) describes all points (P) in space where the ray from the origin to (P) makes an angle of (pi/6) with the positive (z)-axis. This describes a cone, with the positive (z)-axis its axis of symmetry, with point at the origin.

All three surfaces are graphed in Figure 14.7.14. Note how their intersection uniquely defines the point (P=(1,pi/3,pi/6) ext<.>)

Spherical coordinates are useful when describing certain domains in space, allowing us to evaluate triple integrals over these domains more easily than if we used rectangular coordinates or cylindrical coordinates. The crux of setting up a triple integral in spherical coordinates is appropriately describing the “small amount of volume,” (dV ext<,>) used in the integral.

Considering Figure 14.7.15, we can make a small “spherical wedge” by varying ( ho ext<,>) ( heta) and (varphi) each a small amount, (Delta ho ext<,>) (Delta heta) and (Deltavarphi ext<,>) respectively. This wedge is approximately a rectangular solid when the change in each coordinate is small, giving a volume of about

Given a region (D) in space, we can approximate the volume of (D) with many such wedges. As the size of each of (Delta ho ext<,>) (Delta heta) and (Deltavarphi) goes to zero, the number of wedges increases to infinity and the volume of (D) is more accurately approximated, giving

Again, this development of (dV) should sound reasonable, and the following theorem states it is the appropriate manner by which triple integrals are to be evaluated in spherical coordinates.

It is generally most intuitive to evaluate the triple integral in Theorem 14.7.16 by integrating with respect to ( ho) first it often does not matter whether we next integrate with respect to ( heta) or (varphi ext<.>) Different texts present different standard orders, some preferring (dvarphi, d heta) instead of (d heta, dvarphi ext<.>) As the bounds for these variables are usually constants in practice, it generally is a matter of preference.

###### Theorem 14.7.16 . Triple Integration in Spherical Coordinates.

Let (w=h( ho, heta,varphi)) be a continuous function on a closed, bounded region (D) in space, bounded in spherical coordinates by (alpha_1 leq varphi leq alpha_2 ext<,>) (eta_1 leq heta leq eta_2) and (f_1( heta,varphi) leq ho leq f_2( heta,varphi) ext<.>) Then

###### Example 14.7.17 . Establishing the volume of a sphere.

Let (D) be the region in space bounded by the sphere, centered at the origin, of radius (r ext<.>) Use a triple integral in spherical coordinates to find the volume (V) of (D ext<.>)

The sphere of radius (r ext<,>) centered at the origin, has equation ( ho = r ext<.>) To obtain the full sphere, the bounds on ( heta) and (varphi) are (0leq heta leq 2pi) and (0 leq varphi leq pi ext<.>) This leads us to:

the familiar formula for the volume of a sphere. Note how the integration steps were easy, not using square–roots nor integration steps such as Substitution.

###### Example 14.7.18 . Finding the center of mass using spherical coordinates.

Find the center of mass of the solid with constant density enclosed above by ( ho=4) and below by (varphi = pi/6 ext<,>) as illustrated in Figure 14.7.19.

We will set up the four triple integrals needed to find the center of mass (i.e., to compute (M ext<,>) (M_ ext<,>) (M_) and (M_)) and leave it to the reader to evaluate each integral. Because of symmetry, we expect the (x)- and (y)- coordinates of the center of mass to be 0.

While the surfaces describing the solid are given in the statement of the problem, to describe the full solid (D ext<,>) we use the following bounds: (0 leq ho leq 4 ext<,>) (0 leq heta leq 2pi) and (0 leq varphi leq pi/6 ext<.>) Since density (delta) is constant, we assume (delta =1 ext<.>)

To compute (M_ ext<,>) the integrand is (x ext<>) using Key Idea 14.7.11, we have (x = hosin(varphi)cos( heta) ext<.>) This gives:

which we expected as we expect (overline = 0 ext<.>)

To compute (M_ ext<,>) the integrand is (y ext<>) using Key Idea 14.7.11, we have (y = hosin(varphi)sin( heta) ext<.>) This gives:

which we also expected as we expect (overline = 0 ext<.>)

To compute (M_ ext<,>) the integrand is (z ext<>) using Key Idea 14.7.11, we have (z = hocos(varphi) ext<.>) This gives:

Thus the center of mass is ((0,0,M_/M) approx (0,0,2.799) ext<,>) as indicated in Figure 14.7.19.

This section has provided a brief introduction into two new coordinate systems useful for identifying points in space. Each can be used to define a variety of surfaces in space beyond the canonical surfaces graphed as each system was introduced.

However, the usefulness of these coordinate systems does not lie in the variety of surfaces that they can describe nor the regions in space these surfaces may enclose. Rather, cylindrical coordinates are mostly used to describe cylinders and spherical coordinates are mostly used to describe spheres. These shapes are of special interest in the sciences, especially in physics, and computations on/inside these shapes is difficult using rectangular coordinates. For instance, in the study of electricity and magnetism, one often studies the effects of an electrical current passing through a wire that wire is essentially a cylinder, described well by cylindrical coordinates.

This chapter investigated the natural follow–on to partial derivatives: iterated integration. We learned how to use the bounds of a double integral to describe a region in the plane using both rectangular and polar coordinates, then later expanded to use the bounds of a triple integral to describe a region in space. We used double integrals to find volumes under surfaces, surface area, and the center of mass of lamina we used triple integrals as an alternate method of finding volumes of space regions and also to find the center of mass of a region in space.

Integration does not stop here. We could continue to iterate our integrals, next investigating “quadruple integrals” whose bounds describe a region in 4–dimensional space (which are very hard to visualize). We can also look back to “regular” integration where we found the area under a curve in the plane. A natural analogue to this is finding the “area under a curve,” where the curve is in space, not in a plane. These are just two of many avenues to explore under the heading of “integration.”

### Exercises 14.7.3 Exercises

Explain the difference between the roles (r ext<,>) in cylindrical coordinates, and ( ho ext<,>) in spherical coordinates, play in determining the location of a point.

In cylindrical, (r) determines how far from the origin one goes in the (x)-(y) plane before considering the (z)-component. Equivalently, if on projects a point in cylindrical coordinates onto the (x)-(y) plane, (r) will be the distance of this projection from the origin.

In spherical, ( ho) is the distance from the origin to the point.

Why are points on the (z)-axis not determined uniquely when using cylindrical and spherical coordinates?

If (r=0) or ( ho=0 ext<,>) then the point in each coordinate system lies on the (z)-axis regardless of the value of ( heta ext<.>)

What surfaces are naturally defined using cylindrical coordinates?

Cylinders (tubes) centered at the origin, parallel to the (z)-axis planes parallel to the (z)-axis that intersect the (z)-axis planes parallel to the (x)-(y) plane.

What surfaces are naturally defined using spherical coordinates?

Spheres centered at the origin planes parallel to the (z)-axis that intersect the (z)-axis cones centered on the (z)-axis with point at the origin.

In the following exercises, points are given in either the rectangular, cylindrical or spherical coordinate systems. Find the coordinates of the points in the other systems.

Points in rectangular coordinates: ((2,2,1)) and ((-sqrt<3>,1,0))

Points in cylindrical coordinates: ((2,pi/4,2)) and ((3,3pi/2,-4))

Points in spherical coordinates: ((2,pi/4,pi/4)) and ((1,0,0))

Cylindrical: ((2sqrt 2,pi/4,1)) and ((2,5pi/6,0)) Spherical: ((3,pi/4,cos^<-1>(1/3))) and ((2,5pi/6,pi/2))

Rectangular: ((sqrt 2,sqrt 2,2)) and ((0,-3,-4)) Spherical: ((2sqrt 2,pi/4,pi/4)) and ((5,3pi/2,pi- an^<-1>(3/4)))

Rectangular: ((1,1,sqrt<2>)) and ((0,0,1)) Cylindrical: ((sqrt<2>,pi/4,sqrt<2>)) and ((0,0,1))

Points in rectangular coordinates: ((0,1,1)) and ((-1,0,1))

Points in cylindrical coordinates: ((0,pi,1)) and ((2,4pi/3,0))

Points in spherical coordinates: ((2,pi/6,pi/2)) and ((3,pi,pi))

Cylindrical: ((1,pi/2,1)) and ((1,pi,1)) Spherical: ((sqrt 2,pi/2,pi/4)) and ((sqrt<2>, pi, pi/4))

Rectangular: ((0,0,1)) and ((-1,-sqrt 3,0)) Spherical: ((1,pi,0)) and ((2,4pi/3,pi/2))

Rectangular: ((sqrt 3,1,0)) and ((0,0,-3)) Cylindrical: ((2,pi/6,0)) and ((0,pi,-3))

In the following exercises, describe the curve, surface or region in space determined by the given bounds.

Bounds in cylindrical coordinates:

(r=1 ext<,>) (0leq hetaleq 2pi ext<,>) (0leq zleq 1)

(1leq rleq 2 ext<,>) (0leq hetaleq pi ext<,>) (0leq zleq 1)

Bounds in spherical coordinates:

( ho=3 ext<,>) (0leq hetaleq2pi ext<,>) (0leqvarphileq pi/2)

(2leq holeq3 ext<,>) (0leq hetaleq2pi ext<,>) (0leqvarphileq pi)

A cylindrical surface or tube, centered along the (z)-axis of radius 1, extending from the (x)-(y) plane up to the plane (z=1) (i.e., the tube has a length of 1).

This is a region of space, being half of a tube with “thick” walls of inner radius 1 and outer radius 2, centered along the (z)-axis with a length of 1, where the half “below” the (x)-(z) plane is removed.

This is upper half of the sphere of radius 3 centered at the origin (i.e., the upper hemisphere).

This is a region of space, where the ball of radius 2, centered at the origin, is removed from the ball of radius 3, centered at the origin.

Bounds in cylindrical coordinates:

(1leq rleq 2 ext<,>) ( heta= pi/2 ext<,>) (0leq zleq 1)

(r= 2 ext<,>) (0leq hetaleq 2pi ext<,>) (z=5)

Bounds in spherical coordinates:

(0leq holeq2 ext<,>) (0leq hetaleqpi ext<,>) (varphi = pi/4)

( ho=2 ext<,>) (0leq hetaleq2pi ext<,>) (varphi = pi/6)

A square portion of the (y)-(z) plane with corners at ((0,1,0) ext<,>) ((0,1,1) ext<,>) ((0,2,1)) and ((0,2,0) ext<.>)

This is a curve, a circle of radius 2, centered at ((0,0,5) ext<,>) lying parallel to the (x)-(y) plane (i.e., in the plane (z=5)).

This is a region of space, a half of a solid cone with rounded top, where the rounded top is a portion of the ball of radius 2 centered at the origin and the sides of the cone make an angle of (pi/4) with the positive (z)-axis. The bounds on ( heta) mean only the portion “above” the (x)-(z) plane are retained.

This is a curve, a circle of radius 1 centered at ((0,0,sqrt 3) ext<,>) lying parallel to the (x)-(y) plane.

In the following exercises, standard regions in space, as defined by cylindrical and spherical coordinates, are shown. Set up the triple integral that integrates the given function over the graphed region.

## 14.7: Triple Integrals in Cylindrical and Spherical Coordinates - Mathematics

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1. Be comfortable setting up and computing triple integrals in cylindrical and spherical coordinates. 2. Understand the scaling factors for triple integrals in cylindrical and spherical coordinates, as well as where they come from. 3. Be comfortable picking between cylindrical and spherical coordinates.

### Cylindrical Coordinates

#### Recap Video

Here is a video highlights the main points of the section.

#### Example Video

Here is an example of setting up the bounds for a triple integral in cylindrical coordinates.

#### Problems

The integrand in cylindrical coordinates is The integral therefore becomes

This is the region under a paraboloid and inside a cylinder. The reason cylindrical coordinates would be a good coordinate system to pick is that the condition means we will probably go to polar later anyway, so we can just go there now with cylindrical coordinates.

The paraboloid’s equation in cylindrical coordinates (i.e. in terms of , , and ) is Thus, our bounds for will be Now that we have , we can look at the -plane for our polar bounds. The disc in polar is Therefore, the integral becomes

### Spherical Coordinates

Throughout this section, you will need to use “” to denote “.”

#### Recap Video

Here is a video highlights the main points of the section.

#### Example Video

Here is an example of setting up the bounds for a triple integral in spherical coordinates.

### Picking a Coordinate System

For the following problems, you should decide whether to use Cartesian, cylindrical, or spherical coordinates to evaluate the triple integral. It would be good practice for you to try all three and see which ones are realistic options for each problem.

You once did the next problem as a double integral. Now do it as a triple integral and convince yourself it is the same thing.

• In cylindrical coordinates, the integral would be
• In spherical, the integral would be (input “p” for phi):
• Evaluating either way gives an answer of: .

You could set this up in cylindrical or spherical. Notice the intersection is the circle . In cylindrical, the setup would be This is actually solvable because the square root will disappear after integrating with respect to .

In spherical, the plane has equation , or . The sphere has equation . These give the bounds. To get , notice that the region starts at , and it will go up to the intersection of the plane with the sphere, which is , or . This gives the bounds. The setup is therefore

Evaluate the triple integral in cylindrical coordinates.

Let’s start by converting the limits of integration from rectangular coordinates to cylindrical coordinates, starting with the innermost integral. These will be the limits of integration for . z. which means they need to be solved for . z. once we get them to cylindrical coordinates. The upper limit . 3. can stay the same since . z=z. when we go from rectangular to cylindrical coordinates, but the lower limit needs to be converted using the conversion formulas.

Using the trigonometric identity . sin^2+cos^2=1. we can simplify to

This means that the limits of integration with respect to . z. in cylindrical coordinates are . [r,3].

Next we’ll do the limits of integration for the middle integral. These will be the limits of integration for . x. which means they need to be solved for . r. once we get them to cylindrical coordinates.

The lower limit is given by

The upper limit is given by

It looks like the limits of integration for . r. in cylindrical coordinates will be given by . [-3,3]. However, remember that . r. represents the radius, or distance from the origin. It doesn’t make sense to say that we’re . -3. units away from the origin. Instead, we always say that the lower bound for . r. is . 0. such that . 0. is the closest we can be to the origin (right on the origin), and . 3. is the furthest we can be from the origin. So the limits of integration for . r. will be . [0,3].

Finally, we’ll do the limits of integration for the outer integral. These will be the limits of integration for . y. which means they need to be solved for . heta. once we get them to cylindrical coordinates. But since we’re going to . heta. we can just assume that the interval is . [0,2pi]. because that interval represents the full set of values for . heta. which is just the angle between any point and the positive direction of the . x. -axis.

Next we’ll use the conversion formulas to convert the function itself into cylindrical coordinates.

Putting all of this, plus . dV=r dz dr d heta. into the integral gives

. int^<2pi>_0int^3_<0>int^3_rrzcos< heta>left(r dz dr d heta ight).

. int^<2pi>_0int^3_<0>int^3_rr^2zcos< heta> dz dr d heta. we’ll need to convert the limits of integration, the function itself, and dV from rectangular coordinates to cylindrical coordinates.

We always integrate from the inside out, which means we’ll integrate first with respect to . z. treating all other variables as constants.

Now we’ll integrate with respect to . r. treating all other variables as constants.

Like cartesian (or rectangular) coordinates and polar coordinates, cylindrical coordinates are just another way to describe points in three-dimensional space.

Remember that cylindrical coordinates are exactly the same as polar coordinates, just in three-dimensional space instead of two-dimensional space. Since polar coordinates in 2D are given as . (r, heta). cylindrical coordinates just require us to add a value for . z. to account for 3D space, which means cylindrical coordinates are given as . (r, heta,z). Rectangular coordinates are given as . (x,y,z).

where . x. is the distance of . (x,y,z). from the origin along the . x. -axis

where . y. is the distance of . (x,y,z). from the origin along the . y. -axis

where . z. is the distance of . (x,y,z). from the origin along the . z. -axis

Cylindrical coordinates are given as . (r, heta,z).

where . r. is the distance of . (r, heta,z). from the origin

where . heta. is the angle between . r. (the line connecting . (r, heta,z). to the origin) and the positive direction of the . x. -axis

where . z. is the the distance of . (r, heta,z). from the origin along the . z. -axis

To convert between cylindrical coordinates and rectangular coordinates, we use the conversion formulas

## 01.Triple Integrals Cylindrical Coordinates

Spherical coordinates calculator converts between Cartesian and spherical coordinates in a 3D space. When converting from the rectangular to the spherical system, our spherical coordinate calculator assumes that the origins of both systems overlap. Flux is the total force you feel, the total number of bananas you see flying by your surface. Think of flux like weight. Flux Factors. ## 14.7: Triple Integrals in Cylindrical and Spherical Coordinates - Mathematics

### Lecture Description

This video lecture, part of the series Vector Calculus by Prof. Christopher Tisdell, does not currently have a detailed description and video lecture title. If you have watched this lecture and know what it is about, particularly what Mathematics topics are discussed, please help us by commenting on this video with your suggested description and title. Many thanks from,

- The CosmoLearning Team

### Course Index

1. Applications of Double integrals
2. Path Integrals: How to Integrate Over Curves
3. What is a Vector Field?
4. What is the Divergence?
5. What is the Curl?
6. What is a Line Integral?
7. Applications of Line Integrals
8. Fundamental Theorem of Line Integrals
9. What is Green's Theorem?
10. Green's Theorem
11. Parametrised Surfaces
12. What is a Surface Integral? (Part I)
13. More On Surface Integrals
14. Surface Integrals and Vector Fields
15. Divergence Theorem of Gauss
16. How to Solve PDEs via Separation of Variables and Fourier Series
17. Vector Revision
18. Intro to Curves and Vector Functions
19. Limits of Vector Functions
20. Calculus of Vector Functions: One Variable
21. Calculus of Vector Functions Tutorial
22. Vector Functions Tutorial
23. Intro to Functions of Two Variables
24. Limits of Functions of Two Variables
25. Partial Derivatives
26. Partial Derivatives and PDEs Tutorial
27. Multivariable Functions: Graphs and Limits
28. Multivariable Chain Rule and Differentiability
29. Chain Rule: Partial Derivative of $arctan (y/x)$ w.r.t. $x$
30. Chain Rule & Partial Derivatives
31. Chain Rule: Identity Involving Partial Derivatives
32. Multivariable Chain Rule
33. Leibniz' Rule: Integration via Differentiation Under Integral Sign
34. Evaluating Challenging Integrals via Differentiation: Leibniz Rule
37. Directional dDerivative of $f(x,y)$
38. Tangent Plane Approximation and Error Estimation
40. Partial Derivatives and Error Estimation
41. Multivariable Taylor Polynomials
42. Taylor Polynomials: Functions of Two Variables
43. Multivariable Calculus: Limits, Chain Rule and Arc Length
44. Critical Points of Functions
45. How to Find Critical Points of Functions
46. How to Find Critical Points of Functions
47. Second Derivative Test: Two Variables
48. Multivariable Calculus: Critical Points and Second Derivative Test
49. How to Find and Classify Critical Points of Functions
50. Lagrange Multipliers
51. Lagrange Multipliers: Two Constraints
52. Lagrange Multipliers: Extreme Values of a Function Subject to a Constraint
53. Lagrange Multipliers Example
54. Lagrange multiplier Example: Minimizing a Function Subject to a Constraint
55. Second Derivative Test, Max/Min and Lagrange Multipliers
56. Intro to Jacobian Matrix and Differentiability
57. Jacobian Chain Rule and Inverse Function Theorem
58. Intro to Double Integrals
59. Double Integrals Over General Regions
60. Double Integrals: Volume Between Two Surfaces
61. Double Integrals: Volume of a Tetrahedron
62. Double Integral
63. Double Integrals and Area
64. Double Integrals in Polar Co-ordinates
65. Reversing Order in Double Integrals
66. Double Integrals: Reversing the Order of Integration
67. Applications of Double Integrals
68. Double Integrals and Polar Co-ordinates
69. Double Integrals
70. Centroid and Double Integral
71. Center of Mass, Double Integrals and Polar Co-ordinates
72. Triple Integral
73. Triple integrals in Cylindrical and Spherical Coordinates
74. Triple integrals & Center of Mass
75. Change of Variables in Double Integrals
76. Path Integral (Scalar Line Integral) From Vector Calculus
77. Line Integral Example in 3D-Space
78. Line Integral From Vector Calculus Over a Closed Curve
79. Line Integral Example From Vector Calculus
80. Divergence of a Vector Field
81. Curl of a Vector Field (ex. no.1)
82. Curl of a Vector Field (ex. no.2)
83. Divergence Theorem of Gauss
84. Intro to Fourier Series and How to Calculate Them
85. How to Compute a Fourier Series: An Example
86. What are Fourier Series?
87. Fourier Series
88. Fourier Series and Differential Equations

### Course Description

In this course, Prof. Chris Tisdell gives 88 video lectures on Vector Calculus. This is a series of lectures for "Several Variable Calculus" and "Vector Calculus", which is a 2nd-year mathematics subject taught at UNSW, Sydney. This playlist provides a shapshot of some lectures presented in Session 1, 2009 and Session 1, 2011. These lectures focus on presenting vector calculus in an applied and engineering context, while maintaining mathematical rigour. Thus, this playlist may be useful to students of mathematics, but also to those of engineering, physics and the applied sciences. There is an emphasis on examples and also on proofs. Dr Chris Tisdell is Senior Lecturer in Applied Mathematics.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1.

It is more convenient to calculate this integral in cylindrical coordinates. Projection of the region of integration onto the (xy)-plane is the circle ( + le 1) or (0 le ho le 1) (Figure (3)).

Notice that the integrand can be written as

Then the integral becomes

[I = intlimits_0^ <2pi > intlimits_0^1 << ho ^4> ho d ho > intlimits_0^1 .]

The second integral contains the factor ( ho) which is the Jacobian of transformation of the Cartesian coordinates into cylindrical coordinates. All the three integrals over each of the variables do not depend on each other. As a result the triple integral is easy to calculate as

Use spherical coordinates to find the volume of the triple integral, where . B. is a sphere with center . (0,0,0). and radius . 4.

Using the conversion formula . ho^2=x^2+y^2+z^2. we can change the given function into spherical notation.

. intintint_Bx^2+y^2+z^2 dV=intintint_B ho^2 dV.

Then we’ll use . dV= ho^2sin d ho d heta dphi. to make a substitution for . dV.

. intintint_B ho^2left( ho^2sin d ho d heta dphi ight).

. intintint_B ho^4sin d ho d heta dphi.

Now we’ll find limits of integration. We already know the limits of integration for . phi. and . heta. since they are always the same if we’re dealing with a full sphere, so we get

. int_0^piint_0^ d ho d heta dphi.

Since . ho. defines the radius of the sphere, and we’re told that this sphere has its center at . (0,0,0). and radius . 4. . ho. is defined on . [0,4]. so

. int_0^piint_0^ d ho d heta dphi. We can use triple integrals and spherical coordinates to solve for the volume of a solid sphere.

We always integrate inside out, so we’ll integrate with respect to . ho. first, treating all other variables as constants.

Now we’ll integrate with respect to . heta. treating all other variables as constants.

Finally, we’ll integrate with respect to . phi.

This is the volume of the region bounded beneath the surface . x^2+y^2+z^2. and above the sphere defined by . B.

## Solutions for Chapter 14.7: Triple Integrals in Cylindrical and Spherical Coordinates

Solutions for Chapter 14.7: Triple Integrals in Cylindrical and Spherical Coordinates

• 14.7.1: In Exercises 16, evaluate the iterated integral.2020r cos dr d dz
• 14.7.2: In Exercises 16, evaluate the iterated integral.202r0 rz dz dr d
• 14.7.3: In Exercises 16, evaluate the iterated integral.
• 14.7.4: In Exercises 16, evaluate the iterated integral.20e32 d d d
• 14.7.5: In Exercises 16, evaluate the iterated integral.
• 14.7.6: In Exercises 16, evaluate the iterated integral.4040cos 02 sin cos .
• 14.7.7: In Exercises 7 and 8, use a computer algebra system to evaluate the.
• 14.7.8: In Exercises 7 and 8, use a computer algebra system to evaluate the.
• 14.7.9: In Exercises 912, sketch the solid region whose volume is given by .
• 14.7.10: In Exercises 912, sketch the solid region whose volume is given by .
• 14.7.11: In Exercises 912, sketch the solid region whose volume is given by .
• 14.7.12: In Exercises 912, sketch the solid region whose volume is given by .
• 14.7.13: In Exercises 1316, convert the integral from rectangular coordinate.
• 14.7.14: In Exercises 1316, convert the integral from rectangular coordinate.
• 14.7.15: In Exercises 1316, convert the integral from rectangular coordinate.
• 14.7.16: In Exercises 1316, convert the integral from rectangular coordinate.
• 14.7.17: Volume In Exercises 1720, use cylindrical coordinates to find the v.
• 14.7.18: Volume In Exercises 1720, use cylindrical coordinates to find the v.
• 14.7.19: Volume In Exercises 1720, use cylindrical coordinates to find the v.
• 14.7.20: Volume In Exercises 1720, use cylindrical coordinates to find the v.
• 14.7.21: Mass In Exercises 21 and 22, use cylindrical coordinates to find th.
• 14.7.22: Mass In Exercises 21 and 22, use cylindrical coordinates to find th.
• 14.7.23: In Exercises 2328, use cylindrical coordinates to find the indicate.
• 14.7.24: In Exercises 2328, use cylindrical coordinates to find the indicate.
• 14.7.25: In Exercises 2328, use cylindrical coordinates to find the indicate.
• 14.7.26: In Exercises 2328, use cylindrical coordinates to find the indicate.
• 14.7.27: In Exercises 2328, use cylindrical coordinates to find the indicate.
• 14.7.28: In Exercises 2328, use cylindrical coordinates to find the indicate.
• 14.7.29: Moment of Inertia In Exercises 29 and 30, use cylindrical coordinat.
• 14.7.30: Moment of Inertia In Exercises 29 and 30, use cylindrical coordinat.
• 14.7.31: Volume In Exercises 31 and 32, use spherical coordinates to find th.
• 14.7.32: Volume In Exercises 31 and 32, use spherical coordinates to find th.
• 14.7.33: Mass In Exercises 33 and 34, use spherical coordinates to find the .
• 14.7.34: Mass In Exercises 33 and 34, use spherical coordinates to find the .
• 14.7.35: Center of Mass In Exercises 35 and 36, use spherical coordinates to.
• 14.7.36: Center of Mass In Exercises 35 and 36, use spherical coordinates to.
• 14.7.37: Moment of Inertia In Exercises 37 and 38, use spherical coordinates.
• 14.7.38: Moment of Inertia In Exercises 37 and 38, use spherical coordinates.
• 14.7.39: Give the equations for the coordinate conversion from rectangular t.
• 14.7.40: Give the equations for the coordinate conversion from rectangular t.
• 14.7.41: Give the iterated form of the triple integral in cylindrical form.
• 14.7.42: Give the iterated form of the triple integral in spherical form.
• 14.7.43: Describe the surface whose equation is a coordinate equal to a cons.
• 14.7.44: When evaluating a triple integral with constant limits of integrati.
• 14.7.45: Find the volume of the four-dimensional spherex2 y 2 z2 w2 a2by eva.
• 14.7.46: Use spherical coordinates to show that x2 y2 z2 e x2y2z2 dx dy dz 2. 1
##### ISBN: 9780618606245

Calculus: Early Transcendental Functions was written by and is associated to the ISBN: 9780618606245. This expansive textbook survival guide covers the following chapters and their solutions. Since 46 problems in chapter 14.7: Triple Integrals in Cylindrical and Spherical Coordinates have been answered, more than 73596 students have viewed full step-by-step solutions from this chapter. This textbook survival guide was created for the textbook: Calculus: Early Transcendental Functions , edition: 4. Chapter 14.7: Triple Integrals in Cylindrical and Spherical Coordinates includes 46 full step-by-step solutions.

A rectangular graphical display of categorical data.

An angle whose vertex is the center of a circle

See Cartesian coordinate system.

The factor Ae-a in an equation such as y = Ae-at cos bt

A vector in the direction of a line in three-dimensional space

The length of the chord through the focus and perpendicular to the axis.

A method of solving a system of n linear equations in n unknowns.

See Polynomial function in x.

See Natural logarithmic regression

A model of population growth: ƒ1x2 = c 1 + a # bx or ƒ1x2 = c1 + ae-kx, where a, b, c, and k are positive with b < 1. c is the limit to growth

Any of the real numbers in a matrix

The process of expanding a fraction into a sum of fractions. The sum is called the partial fraction decomposition of the original fraction.

lim x:a- ƒ(x) = limx:a+ ƒ(x) but either the common limit is not equal ƒ(a) to ƒ(a) or is not defined

In the plane i = <1, 0> and j = <0,1> in space i = <1,0,0>, j = <0,1,0> k = <0,0,1>