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8.2: Area of a Surface of Revolution - Mathematics


The concepts we used to find the arc length of a curve can be extended to find the surface area of a surface of revolution. Surface area is the total area of the outer layer of an object. We wish to find the surface area of the surface of revolution created by revolving the graph of (y=f(x)) around the (x)-axis as shown in the following figure.

As we have done many times before, we are going to partition the interval ([a,b]) and approximate the surface area by calculating the surface area of simpler shapes. We start by using line segments to approximate the curve, as we did earlier in this section. For (i=0,1,2,…,n), let (P={x_i}) be a regular partition of ([a,b]). Then, for (i=1,2,…,n,) construct a line segment from the point ((x_{i−1},f(x_{i−1}))) to the point ((x_i,f(x_i))). Now, revolve these line segments around the (x)-axis to generate an approximation of the surface of revolution as shown in the following figure.

Notice that when each line segment is revolved around the axis, it produces a band. These bands are actually pieces of cones (think of an ice cream cone with the pointy end cut off). A piece of a cone like this is called a frustum of a cone.

To find the surface area of the band, we need to find the lateral surface area, (S), of the frustum (the area of just the slanted outside surface of the frustum, not including the areas of the top or bottom faces). Let (r_1) and (r_2) be the radii of the wide end and the narrow end of the frustum, respectively, and let (l) be the slant height of the frustum as shown in the following figure.

We know the lateral surface area of a cone is given by

[ ext{Lateral Surface Area } =πrs,]

where (r) is the radius of the base of the cone and (s) is the slant height (Figure (PageIndex{7})).

Since a frustum can be thought of as a piece of a cone, the lateral surface area of the frustum is given by the lateral surface area of the whole cone less the lateral surface area of the smaller cone (the pointy tip) that was cut off (Figure (PageIndex{8})).

The cross-sections of the small cone and the large cone are similar triangles, so we see that

[ dfrac{r_2}{r_1}=dfrac{s−l}{s}]

Solving for (s), we get =s−ls

[egin{align*} dfrac{r_2}{r_1} &=dfrac{s−l}{s} r_2s &=r_1(s−l) r_2s &=r_1s−r_1l r_1l &=r_1s−r_2s r_1l &=(r_1−r_2)s dfrac{r_1l}{r_1−r_2} =s end{align*}]

Then the lateral surface area (SA) of the frustum is

[egin{align*} S &= ext{(Lateral SA of large cone)}− ext{(Lateral SA of small cone)} [4pt] &=πr_1s−πr_2(s−l) [4pt] &=πr_1(dfrac{r_1l}{r_1−r_2})−πr_2(dfrac{r_1l}{r_1−r_2−l}) [4pt] &=dfrac{πr^2_1l}{r^1−r^2}−dfrac{πr_1r_2l}{r_1−r_2}+πr_2l [4pt] &=dfrac{πr^2_1l}{r_1−r_2}−dfrac{πr_1r2_l}{r_1−r_2}+dfrac{πr_2l(r_1−r_2)}{r_1−r_2} [4pt] &=dfrac{πr^2_1}{lr_1−r_2}−dfrac{πr_1r_2l}{r_1−r_2} + dfrac{πr_1r_2l}{r_1−r_2}−dfrac{πr^2_2l}{r_1−r_3} [4pt] &=dfrac{π(r^2_1−r^2_2)l}{r_1−r_2}=dfrac{π(r_1−r+2)(r1+r2)l}{r_1−r_2} [4pt] &= π(r_1+r_2)l. label{eq20} end{align*}]

Let’s now use this formula to calculate the surface area of each of the bands formed by revolving the line segments around the (x-axis). A representative band is shown in the following figure.

Note that the slant height of this frustum is just the length of the line segment used to generate it. So, applying the surface area formula, we have

[egin{align*} S &=π(r_1+r_2)l &=π(f(x_{i−1})+f(x_i))sqrt{Δx^2+(Δyi)^2} &=π(f(x_{i−1})+f(x_i))Δxsqrt{1+(dfrac{Δy_i}{Δx})^2} end{align*}]

Now, as we did in the development of the arc length formula, we apply the Mean Value Theorem to select (x^∗_i∈[x_{i−1},x_i]) such that (f′(x^∗_i)=(Δy_i)/Δx.) This gives us

[S=π(f(x_{i−1})+f(x_i))Δxsqrt{1+(f′(x^∗_i))^2} onumber]

Furthermore, since(f(x)) is continuous, by the Intermediate Value Theorem, there is a point (x^{**}_i∈[x_{i−1},x[i]) such that (f(x^{**}_i)=(1/2)[f(xi−1)+f(xi)],

so we get

[S=2πf(x^{**}_i)Δxsqrt{1+(f′(x^∗_i))^2}. onumber]

Then the approximate surface area of the whole surface of revolution is given by

[ ext{Surface Area} ≈sum_{i=1}^n2πf(x^{**}_i)Δxsqrt{1+(f′(x^∗_i))^2}. onumber]

This almost looks like a Riemann sum, except we have functions evaluated at two different points, (x^∗_i) and (x^{**}_{i}), over the interval ([x_{i−1},x_i]). Although we do not examine the details here, it turns out that because (f(x)) is smooth, if we let n(→∞), the limit works the same as a Riemann sum even with the two different evaluation points. This makes sense intuitively. Both (x^∗_i) and x^{**}_i) are in the interval ([x_{i−1},x_i]), so it makes sense that as (n→∞), both (x^∗_i) and (x^{**}_i) approach (x) Those of you who are interested in the details should consult an advanced calculus text.

Taking the limit as (n→∞,) we get

[ egin{align*} ext{Surface Area} &=lim_{n→∞}sum_{i=1}n^2πf(x^{**}_i)Δxsqrt{1+(f′(x^∗_i))^2} [4pt] &=∫^b_a(2πf(x)sqrt{1+(f′(x))^2}) end{align*}]

As with arc length, we can conduct a similar development for functions of (y) to get a formula for the surface area of surfaces of revolution about the (y-axis). These findings are summarized in the following theorem.

Surface Area of a Surface of Revolution

Let (f(x)) be a nonnegative smooth function over the interval ([a,b]). Then, the surface area of the surface of revolution formed by revolving the graph of (f(x)) around the x-axis is given by

[ ext{Surface Area}=∫^b_a(2πf(x)sqrt{1+(f′(x))^2})dx]

Similarly, let (g(y)) be a nonnegative smooth function over the interval ([c,d]). Then, the surface area of the surface of revolution formed by revolving the graph of (g(y)) around the (y-axis) is given by

[ ext{Surface Area}=∫^d_c(2πg(y)sqrt{1+(g′(y))^2}dy]

Example (PageIndex{4}): Calculating the Surface Area of a Surface of Revolution 1.

Let (f(x)=sqrt{x}) over the interval ([1,4]). Find the surface area of the surface generated by revolving the graph of (f(x)) around the (x)-axis. Round the answer to three decimal places.

Solution

The graph of (f(x)) and the surface of rotation are shown in Figure (PageIndex{10}).

We have (f(x)=sqrt{x}). Then, (f′(x)=1/(2sqrt{x})) and ((f′(x))^2=1/(4x).) Then,

[egin{align*} ext{Surface Area} &=∫^b_a(2πf(x)sqrt{1+(f′(x))^2}dx [4pt] &=∫^4_1(sqrt{2πsqrt{x}1+dfrac{1}{4x}})dx [4pt] &=∫^4_1(2πsqrt{x+14}dx. end{align*}]

Let (u=x+1/4.) Then, (du=dx). When (x=1, u=5/4), and when (x=4, u=17/4.) This gives us

[egin{align*} ∫^1_0(2πsqrt{x+dfrac{1}{4}})dx &= ∫^{17/4}_{5/4}2πsqrt{u}du [4pt] &= 2πleft[dfrac{2}{3}u^{3/2} ight]∣^{17/4}_{5/4} [4pt] &=dfrac{π}{6}[17sqrt{17}−5sqrt{5}]≈30.846 end{align*}]

Exercise (PageIndex{4})

Let ( f(x)=sqrt{1−x}) over the interval ( [0,1/2]). Find the surface area of the surface generated by revolving the graph of ( f(x)) around the (x)-axis. Round the answer to three decimal places.

Hint

Use the process from the previous example.

Answer

[ dfrac{π}{6}(5sqrt{5}−3sqrt{3})≈3.133]

Example ( PageIndex{5}): Calculating the Surface Area of a Surface of Revolution 2

Let ( f(x)=y=dfrac[3]{3x}). Consider the portion of the curve where ( 0≤y≤2). Find the surface area of the surface generated by revolving the graph of ( f(x)) around the ( y)-axis.

Solution

Notice that we are revolving the curve around the ( y)-axis, and the interval is in terms of ( y), so we want to rewrite the function as a function of ( y). We get ( x=g(y)=(1/3)y^3). The graph of ( g(y)) and the surface of rotation are shown in the following figure.

We have ( g(y)=(1/3)y^3), so ( g′(y)=y^2) and ( (g′(y))^2=y^4). Then

[egin{align*} ext{Surface Area} &=∫^d_c(2πg(y)sqrt{1+(g′(y))^2})dy [4pt] &=∫^2_0(2π(dfrac{1}{3}y^3)sqrt{1+y^4})dy [4pt] &=dfrac{2π}{3}∫^2_0(y^3sqrt{1+y^4})dy. end{align*}]

Let ( u=y^4+1.) Then ( du=4y^3dy). When ( y=0, u=1), and when ( y=2, u=17.) Then

[egin{align*} dfrac{2π}{3}∫^2_0(y^3sqrt{1+y^4})dy &=dfrac{2π}{3}∫^{17}_1dfrac{1}{4}sqrt{u}du [4pt] &=dfrac{π}{6}[dfrac{2}{3}u^{3/2}]∣^{17}_1=dfrac{π}{9}[(17)^{3/2}−1]≈24.118. end{align*}]

Exercise (PageIndex{5})

Let ( g(y)=sqrt{9−y^2}) over the interval ( y∈[0,2]). Find the surface area of the surface generated by revolving the graph of ( g(y)) around the ( y)-axis.

Hint

Use the process from the previous example.

Answer

( 12π)


Surface of revolution

Examples of surfaces of revolution generated by a straight line are cylindrical and conical surfaces depending on whether or not the line is parallel to the axis. A circle that is rotated around any diameter generates a sphere of which it is then a great circle, and if the circle is rotated around an axis that does not intersect the interior of a circle, then it generates a torus which does not intersect itself (a ring torus).


8.2: Area of a Surface of Revolution - Mathematics

In this section we are going to look once again at solids of revolution. We first looked at them back in Calculus I when we found the volume of the solid of revolution. In this section we want to find the surface area of this region.

So, for the purposes of the derivation of the formula, let’s look at rotating the continuous function (y = fleft( x ight)) in the interval (left[ ight]) about the (x)-axis. We’ll also need to assume that the derivative is continuous on (left[ ight]). Below is a sketch of a function and the solid of revolution we get by rotating the function about the (x)-axis.

We can derive a formula for the surface area much as we derived the formula for arc length. We’ll start by dividing the interval into (n) equal subintervals of width (Delta x). On each subinterval we will approximate the function with a straight line that agrees with the function at the endpoints of each interval. Here is a sketch of that for our representative function using (n = 4).

Now, rotate the approximations about the (x)-axis and we get the following solid.

The approximation on each interval gives a distinct portion of the solid and to make this clear each portion is colored differently. Each of these portions are called frustums and we know how to find the surface area of frustums.

The surface area of a frustum is given by,

and (l) is the length of the slant of the frustum.

For the frustum on the interval (left[ <<>>,> ight]) we have,

and we know from the previous section that,

Before writing down the formula for the surface area we are going to assume that (Delta x) is “small” and since (fleft( x ight)) is continuous we can then assume that,

So, the surface area of the frustum on the interval (left[ <<>>,> ight]) is approximately,

The surface area of the whole solid is then approximately,

and we can get the exact surface area by taking the limit as (n) goes to infinity.

If we wanted to we could also derive a similar formula for rotating (x = hleft( y ight)) on (left[ ight]) about the (y)-axis. This would give the following formula.

These are not the “standard” formulas however. Notice that the roots in both of these formulas are nothing more than the two (ds)’s we used in the previous section. Also, we will replace (fleft( x ight)) with (y) and (hleft( y ight)) with (x). Doing this gives the following two formulas for the surface area.

Surface Area Formulas

There are a couple of things to note about these formulas. First, notice that the variable in the integral itself is always the opposite variable from the one we’re rotating about. Second, we are allowed to use either (ds) in either formula. This means that there are, in some way, four formulas here. We will choose the (ds) based upon which is the most convenient for a given function and problem.

Now let’s work a couple of examples.

The formula that we’ll be using here is,

since we are rotating about the (x)-axis and we’ll use the first (ds) in this case because our function is in the correct form for that (ds) and we won’t gain anything by solving it for (x).

Let’s first get the derivative and the root taken care of.

Here’s the integral for the surface area,

There is a problem however. The (dx) means that we shouldn’t have any (y)’s in the integral. So, before evaluating the integral we’ll need to substitute in for (y) as well.

Previously we made the comment that we could use either (ds) in the surface area formulas. Let’s work an example in which using either (ds) won’t create integrals that are too difficult to evaluate and so we can check both (ds)’s.

Note that we’ve been given the function set up for the first (ds) and limits that work for the second (ds).

Solution 1
This solution will use the first (ds) listed above. We’ll start with the derivative and root.

We’ll also need to get new limits. That isn’t too bad however. All we need to do is plug in the given (y)’s into our equation and solve to get that the range of (x)’s is (1 le x le 8). The integral for the surface area is then,

Note that this time we didn’t need to substitute in for the (x) as we did in the previous example. In this case we picked up a (dx) from the (ds) and so we don’t need to do a substitution for the (x). In fact, if we had substituted for (x) we would have put (y)’s into the integral which would have caused problems.

Solution 2
This time we’ll use the second (ds). So, we’ll first need to solve the equation for (x). We’ll also go ahead and get the derivative and root while we’re at it.

We used the original (y) limits this time because we picked up a (dy) from the (ds). Also note that the presence of the (dy) means that this time, unlike the first solution, we’ll need to substitute in for the (x). Doing that gives,

Note that after the substitution the integral was identical to the first solution and so the work was skipped.

As this example has shown we can use either (ds) to get the surface area. It is important to point out as well that with one (ds) we had to do a substitution for the (x) and with the other we didn’t. This will always work out that way.

Note as well that in the case of the last example it was just as easy to use either (ds). That often won’t be the case. In many examples only one of the (ds) will be convenient to work with so we’ll always need to determine which (ds) is liable to be the easiest to work with before starting the problem.


Theory of Intense Beams of Charged Particles

5.13.1 The Problem Statement

We have seen that using the surface of revolution as a basic stream tube, based on the assumption that Vl, Vψ depend only on l, reduces the problem under consideration to the integration of an ordinary differential equation and, possibly, to the calculation of a quadrature for η. A further task may consist of abandoning the requirement V α = V α (ξ 1 ) and constructing the exact solutions based on studying the group properties of Eqs. (5.225) formulated for a basic surface that is not necessarily a surface of revolution. As such a surface, we can use, as example, any of the surfaces we came across in Section 2 while studying the exact solutions of beam equations (plane, circular cylinder, and cone, as well as helicoid) ( Syrovoy, 1989 ).


Surface of revolution

Examples of surfaces of revolution generated by a straight line are cylindrical and conical surfaces depending on whether or not the line is parallel to the axis. A circle that is rotated around any diameter generates a sphere of which it is then a great circle, and if the circle is rotated around an axis that does not intersect the interior of a circle, then it generates a torus which does not intersect itself (a ring torus).

A portion of the curve x = 2 + cos z rotated around the z-axis

The sections of the surface of revolution made by planes through the axis are called meridional sections. Any meridional section can be considered to be the generatrix in the plane determined by it and the axis.[2]

The sections of the surface of revolution made by planes that are perpendicular to the axis are circles.

Some special cases of hyperboloids (of either one or two sheets) and elliptic paraboloids are surfaces of revolution. These may be identified as those quadratic surfaces all of whose cross sections perpendicular to the axis are circular.
Area formula

If the curve is described by the parametric functions x(t), y(t), with t ranging over some interval [a,b], and the axis of revolution is the y-axis, then the area Ay is given by the integral

provided that x(t) is never negative between the endpoints a and b. This formula is the calculus equivalent of Pappus's centroid theorem.[3] The quantity

comes from the Pythagorean theorem and represents a small segment of the arc of the curve, as in the arc length formula. The quantity 2πx(t) is the path of (the centroid of) this small segment, as required by Pappus' theorem.

Likewise, when the axis of rotation is the x-axis and provided that y(t) is never negative, the area is given by[4]

If the continuous curve is described by the function y = f(x), a ≤ x ≤ b, then the integral becomes

for revolution around the x-axis, and

for revolution around the y-axis (provided a ≥ 0). These come from the above formula.[5]

For example, the spherical surface with unit radius is generated by the curve y(t) = sin(t), x(t) = cos(t), when t ranges over [0,π]. Its area is therefore

For the case of the spherical curve with radius r, y(x) = √r2 − x2 rotated about the x-axis

A minimal surface of revolution is the surface of revolution of the curve between two given points which minimizes surface area.[6] A basic problem in the calculus of variations is finding the curve between two points that produces this minimal surface of revolution.[6]

There are only two minimal surfaces of revolution (surfaces of revolution which are also minimal surfaces): the plane and the catenoid.[7]
Coordinate expressions

A surface of revolution given by rotating a curve described by y=f(x) around the x-axis may be most simply described in cylindrical coordinates by ( ). In Cartesian coordinates, this yields the parametrization in terms of z and ( heta ) as ( ). If instead we revolve the curve around the y-axis, then the curve is described in cylindrical coordinates by ( ), yielding the expression ( ) in terms of the parameters r and ( heta ).

If x and y are defined in terms of a parameter t, then we obtain a parametrization in terms of t and ( heta ). If x and y are functions of t, then the surface of revolution obtained by revolving the curve around the x-axis is described in cylindrical coordinates by the parametric equation ( ), and the surface of revolution obtained by revolving the curve around the y-axis is described by ( ). In Cartesian coordinates, these (respectively) become ( ) and ( ). The above formulae for surface area then follow by taking the surface integral of the constant function 1 over the surface using these parametrizations.

Geodesics on a surface of revolution

Meridians are always geodesics on a surface of revolution. Other geodesics are governed by Clairaut's relation.[8]
Toroids
Main article: Toroid
A toroid generated from a square

A surface of revolution with a hole in, where the axis of revolution does not intersect the surface, is called a toroid.[9] For example, when a rectangle is rotated around an axis parallel to one of its edges, then a hollow square-section ring is produced. If the revolved figure is a circle, then the object is called a torus.
Applications of surfaces of revolution

The use of surfaces of revolution is essential in many fields in physics and engineering. When certain objects are designed digitally, revolutions like these can be used to determine surface area without the use of measuring the length and radius of the object being designed.
See also

Channel surface, a generalisation of a surface of revolution
Gabriel's Horn
Lemon (geometry), surface of revolution of a circular arc
Liouville surface, another generalization of a surface of revolution
Solid of revolution
Surface integral
Generalized helicoid
Translation surface (differential geometry)

Middlemiss Marks Smart. "15-4. Surfaces of Revolution". Analytic Geometry (3rd ed.). p. 378. LCCN 68015472.
Wilson, W.A. Tracey, J.I. (1925), Analytic Geometry (Revised ed.), D.C. Heath and Co., p. 227
Thomas, George B. "6.7: Area of a Surface of Revolution 6.11: The Theorems of Pappus". Calculus (3rd ed.). pp. 206–209, 217–219. LCCN 69016407.
Singh, R.R. (1993). Engineering Mathematics (6 ed.). Tata McGraw-Hill. p. 6.90. ISBN 0-07-014615-2.
Swokowski, Earl W. (1983), Calculus with analytic geometry (Alternate ed.), Prindle, Weber & Schmidt, p. 617, ISBN 0-87150-341-7
Weisstein, Eric W. "Minimal Surface of Revolution". MathWorld.
Weisstein, Eric W. "Catenoid". MathWorld.
Pressley, Andrew. “Chapter 9 - Geodesics.” Elementary Differential Geometry, 2nd ed., Springer, London, 2012, pp. 227–230.

External links
Weisstein, Eric W. "Surface of Revolution". MathWorld.
"Surface de révolution". Encyclopédie des Formes Mathématiques Remarquables (in French).


8.2: Area of a Surface of Revolution - Mathematics

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Mathematical Expression Editor

We compute surface area of a frustrum then use the method of “Slice, Approximate, Integrate” to find areas of surface areas of revolution.

The area of a frustum

In order to perform the approximation step, we first need to discuss the surface area of a frustrum.

To compute the area of a surface of revolution, we approximate that this area is equal to the sum of areas of basic shapes that we can lay out flat. The argument for this goes way back to the great physicist and mathematician, Archimedes of Alexandria. To follow his argument, we have to begin by computing the area of a ‘lamp shade’ or frustum.

And of course, few things are more interesting than the area of a frustum:

  • denote the number of trapezoids,
  • denote the length of the top of each trapezoid,
  • denote the height of each trapezoid,
  • denote the length of the bottom of each trapezoid,

then from geometry, we have that each of the trapezoids, one of which is shown below:

  • is the circumference of the top circle,
  • is the slant height of the frustum as shown in the above figure, and
  • is the circumference of the bottom circle,

and by way of limit laws we find Now, letting

  • be the radius of the circle defining the top of the frustum,
  • be the slant height of the frustum, and
  • be the radius of the circle defining the base of the frustum,

The area of a surface of revolution

Let’s consider a function with a continuous derivative, and form a surface of revolution formed by this curve by rotating the portion of the curve from to about the -axis:

We can find a formula that gives the surface area of this surface of revolution using the procedure of “Slice, Approximate, Integrate”!

Step 1: Slice Since we have the curve to be revolved expressed as a function of , we choose to slice with respect to :

Step 2: Approximate We have seen how to find the surface area of a frustrum, so we should thus approximate each slice as a frustrum.

Thus the surface area, of this frustum is: Note that there is a value between and such that , so we write:

and can find the total approximate surface area by using frustra by adding together all of the surface areas:

Step 3: Integrate The formula above has good conceptual meaning, it does not readily pass to an integral quite yet! and have seen that we can express free in terms of either or , which allows us to express the infinitesimal by:

Note also that as the slice widths shrink, the value above approaches the distance that the corresponding slice is away from the axis of rotation.

To make sure that we emphasize this freedom in expressing as well as the inherent geometric results we used to build the surface area, we write:

where the radius is the distance from the axis of rotation to the slice and is the slant height of the slice.

To compute this surface area, we first choose to express either:

We then have to express the distance in terms of the variable of integration. This will always be a vertical or horizontal distance, which can be computed just as we have been doing in previous sections!

Using the remark, and letting and we can therefore write:

where the radius is the distance from the axis of rotation to the slice at .

An important concept to note is that the slice is located at a point on the curve. The choice of variable of integration may require that we express either or in terms of the other by using the equation that describes the curve. We will see this in the following examples.

We begin by considering looking at a picture,

As a brief aside, note that this slice gives rise to the following frustrum when revolved about the -axis:

Let’s first set up the integral with respect to . In order to do this, we choose:

Note that here is a vertical distance that must be expressed in terms of . Since the slice is located at a point on the curve :

Note that here is a vertical distance that must be expressed in terms of . Since the slice is located at a point on the curve :

Now, since we have , write with me: So

This integral can be computed using the substitution . Working out the details (which you should do on your own), gives:

As our final example, we will compute the surface area of the sphere.

Final thoughts

The key formulas in this section are:

We are free to choose the variable of integration here since we can express in terms of either or easily. Once this choice of variable has been determined, we need to express the radius of the infinitesimal frustrum and the limits for the integral in terms of the variable of integration.

This radius is the distance from the axis of rotation to the slice at , which is either a horizontal or vertical distance. We just need to make sure that we express it in terms of the variable of integration appropriately.

Many of the integrals that arise in the context of these problems can be difficult. Careful differentiation and algebra, as well as a good grasp of integration techniques can be vital when finding surface areas. As usual, this can be challenging and practice is the key here.

“Math is not a spectator sport. It’s not a body of knowledge. It’s not symbols on a page. It’s something you play with, something you do” - Keith Devlin


What's going on?

So how can it be that we have a finite volume, but an infinite surface area?

One way of looking at it is to recall how a volume of solid of revolution is actually calculated using integral calculus. We are finding the sum of the volumes of an infinite number of disks, radius y (for different values of x). The disks have area

and infinitesimal height dx. So each disk has volume

dx" />

The sum of an infinite number of such disks will converge, since this general sum converges:

The dispute

The paradox about an object having finite volume but infinite surface area caused a lot of dispute regarding the nature of infinity among mathematicians of the 17th century, including Galileo and Wallis. Such paradoxes are a great way to get us thinking!

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Posted in Mathematics category - 22 Mar 2018 [Permalink]

6 Comments on “The object with finite volume but infinite surface area”

I was wondering, though, with regard to the surface area of the solid of revolution, as

is

When the derivative is squared, it's

The end result would be the same, though, as above as x approaches infinity, with the expression in the integrand being,

Good evening
A very nice problem with a great answer.
I would like to add or I must say that I want comments on these lines particularly which I am going to write. If we see this problem practically then
Actually the volume we get after integrating is tending to that volume as length is infinitesimally large. So it will hardly make any difference in volume whereas it's surface area will keep on increasing.

@Christian: Thanks for pointing out the error, which I have fixed in the post. I didn't question myself as the end result was the same, as you said!

So what would happen if someone were to hold the Horn vertically and fill it up with π litres of paint? Would the inside surface of the Horn get covered in paint entirely?

@Jan: Excellent question! Counter-intuitive it may be, but I suspect not. Anyone else like to weigh in?

Great article about the object with finite volume but infinite surface area.


8.2: Area of a Surface of Revolution - Mathematics

Consider a plane y=f(x) in the x-y plane between ordinates x=a and x=b. If a certain portion of this curve is revolved about an axis, a solid of revolution is generated.

  1. Cartesian Form:
    • Area of solid formed by revolving the arc of curve about x-axis is-
    • Area of revolution by revolving the curve about y axis is-

  • About x-axis:
  • About y-axis:
  • About the x-axis: initial line


    Here replace r by f(θ)
  • About the y-axis:


    Here replace r by f(θ)
  • Limits for x: x = a to x = b

    Here PM is in terms of x.
  • Limits for y: y = c to y = d

    Here PM is in terms of y.

Example:
Find the area of the solid of revolution generated by revolving the parabola about the x-axis.
Explanation:
Now we are given with the Cartesian form of the equation of parabola and the parabola has been rotated about the x-axis. Hence we use the formula for revolving Cartesian form about x-axis which is:

Here . Now we need to calculate dy/dx

Differentiating w.r.t x we get:

Using

Now we are provided with limits of x as x=0 to x=3. Plugging our calculated values in the above formula we get:

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Lateral & Surface Areas, Volumes

The lateral area of a regular pyramid or right cone is similar to that of prisms, but since each face is a triangle (or triangle-like), there is a factor of one half. The lateral area is thus half the slant height times the perimeter. The slant height is the distance from the vertex to the edge of the base where it is halfway between the base's vertices. If the pyramid is irregular and certainly if the cone is oblique, the surface area might not be calculatable using elementary techniques (which is a fancy way to say you may need calculus). It depends on if you can obtain the altitude (slant height) of each triangular face.

Surface Area = Lateral Area + n × Bases
n = 2 for prisms/cylinders n = 1 for pyramids/cones n = 0 for spheres.

The surface area of a pyramid or cone is the lateral area plus the area of the single base.

The surface area of a sphere is equal to 4 r 2 . Analogous to the unit circle is the unit sphere. Similarly, just as there are 2 radians of angle in one revolution, there are 4 steradians of solid angle in all directions.

Example: Consider a right pyramid A-BCDE with vertex A and square base BCDE of length 20" on each side and a slant height of 26". What are its lateral and surface areas?

Answer: We don't need the height for this calculation, but we will calculate it anyway to stress the difference between slant height and height. The slant height is the hypotenuse of a right triangle where the height is one leg and 20"/2 = 10" is the other leg. Thus 10 2 + h 2 = 26 2 or 100 + h 2 = 676. Thus h 2 = 576 or h = 24". The lateral surfaces are all triangles with a base of 20" and a height (the slant height) of 26". There are four of them. Thus the lateral area is 4×½吐"吖" = 1040 in 2 . The base is 20" square or 400 in 2 . Thus the [total] surface area is 1440 in 2 .

Understanding surface area may be clearer if you refer back to the net associated with the object. At left is a net for a cube and at right a portion of a net for a sphere. Each of these portions of a sphere is called a gore .

Now is a good time to review something learned in algebra, namely ( x + y ) 2 = x 2 + xy + xy + y 2 = x 2 + 2 xy + y 2 . The diagram at the right should clarify this further, help you remember the FOIL method , as well as give a physical basis for this relationship. (Remember also, the square root of ( x 2 + y 2 ) does NOT equal x + y .) Consider extending the FOIL method first into trinomials: ( a + b + c )( d + e + f ) = ad + ae + af + bd + be + bf + cd + ce + cf . The distributive property is another way to consider this situation. Here the box method is useful.

  d e f
a ad ae af
b bd be bf
c cd ce cf

Now extend the method into three dimensions to find: V = ( a + b )( c + d )( e + f ) = ace + acf + ade + adf + bce + bcf + bde + bdf . This would be useful in finding the volume, which is why it is difficult to display in two dimensions.

  • Every polyhedral region has an unique volume , dependent only on your unit cube .
  • A box has a volume of length × width × height ( V = lwh ).
  • Congruent figures have equivalent volume.
  • Total volume is the sum of all nonoverlapping regions.

By knowing the volume, one can determine the dimensions of a polyhedron. Specifically for a cube with edge s and volume s 3 , given a cube with volume 1000 cubic centimeters (1 liter), you can take the cube root to determine each side had length 10 centimeters or about 3.937 inches. Since one gallon is 231 cubic inches, it is thus about 3.785 liters. Other unit conversions can be expected and are summarized in Numbers lesson 9. Cube roots and volume are at the heart of an ancient impossible geometric construction from antiquity, the Delian Cube Doubling problem. Another important concept is that if you double the dimensions of a cube, the volume goes up by a factor of 8=2 3 , just like area went up by a factor of 4=2 2 . This is a problem commonly encountered when converting from cubic feet into cubic yards!

Example: Suppose you wish to pour concrete 4" deep in your driveway which is 90' long and 9' wide.

Answer: You quickly discover that there are 90࡯঩=270 feet 3 . However, there are only 10 yards 3 since each yard is 3 feet and 3 3 =27.

Calculating in the "native unit" of yards: 30ࡩয can help prevent such an error. By "native" we mean here that the final results are expected in cubic yards. If the initial units are converted to yards fewer mistakes will be made. It is EXTREMELY common to erroneously divide by 3 or 9 and not 27 when converting cubic feet into cubic yards.

Example: Suppose you wish to find the volume of the square based right pyramid A-BCDE given in an earlier example with slant height 26" and base 20" on each side.

Answer: The height is 24" as calculated in the previous example. Thus the volume is (1/3)× B × h = (1/3) × 20" × 20" × 24" = 3200 in 3 .

Volume Formulae

Prism or cylinder: V = base area × height
Pyramid or cone: V = (1/3) × base area × height
Sphere: V = (4/3) × (radius) 3

Typically these formula are written as V = Bh (prism or cylinder), V =(1/3) Bh (pyramid or cone), or V =(4/3) r 3 (sphere). Note how a big B is used to signify that this is a two dimensional base or area and not the same (linear) b we use in triangles.

Oblique Prisms and cylinders have the same volume as a right prism or cylinder with the same height and base area. Think of a stack of paper whose top has been pushed to one side. The stack is no longer vertical. However, the volume of paper hasn't changed. In the formula for finding the volume of an oblique prism please note that the height is the perpendicular segment between the top and bottom bases. When you learn calculus you will discover the surface area of a sphere to be the derivative with respect to r of the sphere's volume formula. A similar thing happens between area of a circle and its circumference. This may be happenstance or there may be a deep reason which I'd like to know.

Example: A favorite volume/surface area problem is as follows. A swimming pool is 24' long, 20' wide, 3' deep at the shallow end, and 10' deep at the deep end. The floor slopes evenly. What is the inside surface of the swimming pool and what is the volume (in gallons)?

Answer: The swimming pool is a trapezoidal prism. The floor is 25 feet long since 10' - 3' = 7' and 7 2 +24 2 = 49 + 576 = 625 = 25 2 . The surface area is the sum of 5 surfaces: 2 congruent trapezoidal sides (½(3+10)㩌), 2 rectangular ends (3㩈 + 10㩈), and the bottom (20㩍). This is 2𤚤 + 60 + 200 + 500 = 1072 feet 2 . The volume is Base × height, where Base is the area of one side (½(3+10)㩌), and the height is the width of the pool (20). Thus the volume is 3120 feet 3 or 23339 gallons (multiply by 12 3 cubic inches per cubic foot and divide by 231 cubic inches per gallon).

A gedanken experiment (thought experiment) used to justify the volume formula for a sphere is as follows. First, remember the circle area activity where we cut the circle into 16 wedges, then rearranged the wedges into a r × r parallelogram. Along the same lines, cut a sphere into pyramids. The total area of the bases of these pyramids is 4 r 2 . The height of each is r . Hence the formula is derived. Along the same lines, some have suggested remembering the 1/3 in conal volume formulae by correlating it to the analoguous two dimensional trianglular area formula which has a 1/2 in it.

Given two solids included between parallel planes. If every plane cross section parallel to the given planes has the same area in both solids, then the volumes of the solids are equal. This is know as Cavalieri's Principle .

The Greek Archimedes is one of the three greatest mathematicians of all time. Among his important discoveries is the relationship between the volumes of the cone, sphere, and cylinder. In fact, this discovery was so much his favorite that he requested it to be inscribed on his tombstone. Specifically, consider a sphere of radius r , two cones each with the same radius and height ( r ), and a cylinder with the same radius and height (2 r ). The cylinder will contain either the two cones or the sphere. Their volumes can easily be seen to be (4/3) r 3 , 2(1/3) r 3 , and 2 r 3 . Thus the cones plus the sphere equals the cylinder exactly. (Actually, Archimedes is more commonly credited with showing the sphere's volume to be 2/3's that of the cylinder.) See the corresponding diagrams in the textbook related to the proof of Cavalieri's Principle.

Example: Question 10.2#24 in our text asked the students about cones made from circles (radius 4") with central angles of 45°, 60°, and 120° removed (which got taped to the board amid Madonna jokes). Perform the following. Find the volume of each cone. Find the central angle which maximizes volume. Find the central angle which maximized volume to surface ratio.

Answer: For bonus points hand in your solution by the time the chapter reviews are due.


CALC 2: Area of a Surface of Revolution: x=y+y^3 from 0 to 4

Hi, so I'm having trouble with one of my online calc homework..

Consider the following.
x=y+y^3 from 0 to 4
(a) Set up an integral for the area of the surface obtained by rotating the curve about the x-axis and the y-axis.
(i) the x-axis, the answer is S= 2piy(sqrt((3y^2+1)^2)+1)dy
(ii) the y-axis, the answer is S= 2pi(y^3+y)sqrt((3y^2+1)^2+1)dy

(b) Use the numerical integration capability of a calculator to evaluate the surface areas correct to four decimal places.
(i) the x-axis, the answer is 1258.6212
(ii) the y-axis, I'm really stuck on how to take the integral. I used wolfram to show how they did it, but I don't understand how they got it! I would really like to understand how to do this problem!

Subhotosh Khan

Super Moderator

Hi, so I'm having trouble with one of my online calc homework..

Consider the following.
x=y+y^3 from 0 to 4
(a) Set up an integral for the area of the surface obtained by rotating the curve about the x-axis and the y-axis.
(i) the x-axis, the answer is S= 2piy(sqrt((3y^2+1)^2)+1)dy
(ii) the y-axis, the answer is S= 2pi(y^3+y)sqrt((3y^2+1)^2+1)dy

(b) Use the numerical integration capability of a calculator to evaluate the surface areas correct to four decimal places.
(i) the x-axis, the answer is 1258.6212
(ii) the y-axis, I'm really stuck on how to take the integral. I used wolfram to show how they did it, but I don't understand how they got it! I would really like to understand how to do this problem!

Please share your work with us . even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

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Watch the video: Area of a Surface of Revolution (December 2021).