Volumes

Section 7.1 Volumes

An important application of integration is the calculation of volumes of three-dimensional objects. Maybe you wondered, sometime, where the various volume formulas come from: why does a sphere have volume \(\frac{4\pi r^3}{4}\text{,}\) or a cone \(\frac{\pi r^2 h}{3}\text{?}\) The oldest versions of these formlas were almost certainly the product of approximation and guessing, but now I claim to know them exactly. What is the proof? How are they calculated? Moveover, if there are new three-dimensional shapes described, techniques to calculate their volume are needed. This is useful in all sorts of designs, where designers need to know how much material the component will requre, or how heavy it will be (assuming density is known). Integrals can answer these questions.

This week, I’ll be introducing two methods for calculating volume. Both methods are based on the definition of the integral: as the limit of a sum of approximate pieces. For each method, I’ll develop a particular way of cutting up the object into 2-dimensional slices whose area is known. Then I’ll multiply by some width \(\Delta w\) to get a volume for the slice. The sum of a number of such slices will approximate the area. Then I take the limit. In the limit, the sum becomes an integral and the width \(\Delta w\) becomes an infinitesimal \(dw\text{.}\)

Subsection 7.1.1 Volumes by Cross-Sectional Slices

The first of the two methods use cross-sectional slices. These are slices made by some plane as the plane moves along the shape. The idea is to choose an orientation for the slices such that they are some familiar shape: rectangles, triangles, circles. Then their area is known and I can setup the approximation system and resulting integral I described above. There are three pieces to the integral. Let me assume the slices perpendicular to the \(x\) axis, so that each slice is identified by its \(x\) coordinates (though, of course, this will all work for any axis by simply changing the variable name). First I have the range in \(x\text{,}\) \(x \in [a,b]\text{,}\) that covers the object. This will become the bounds of integration. Second, I have the area of the slice, \(A(x)\text{,}\) as a function of the \(x\) position that identifies the slice. This will form the integrand. Finally, I have the width. In the limit that defined the integral, this becomes the \(dx\) which completes the integral. Then I can write the integral for volume by cross-sectional slices.

\begin{equation*} V = \int_a^b A(x) dx \end{equation*}

Example 7.1.1.

Consider a sphere of radius \(R\) is centered at the origin in \(\RR^3\text{.}\) Such an object is given by the equation \(x^2 + y^2 + z^2 = R^2\) as a locus in \(\RR^3\text{.}\) I will slice the sphere with vertical planes; the cross sections will be circles. At the front of the sphere, the slice is just a point. As the slices move down the sphere, they get larger. At the middle, the slice is a circle of radius \(R\text{.}\) Then the slices start shrinking down again until there is just a point at the back of the sphere. I’ll assume the slices are perpendicular to the \(x\) axis and are identified by their \(x\) position as \(x \in [-R,R]\text{.}\)

Then I need to know the radius of each disc, based on its \(x\) position. I could treat the maximum \(y\) values as the radius. If I set \(z=0\) in the locus, then \(x^2 + y^2 = R^2\text{,}\) so the maximum \(y\) value would be \(y = \sqrt{R^2 - x^2}\text{.}\) This is the radius of the slice at position \(x\)

Then I need to set up the integral. Implicitly, of course, the integral is a limit of approximations. However, it is usually possible to go directly to setting up the integral using \(dx\) (or whatever variable indicates the position of the slide) as the width term. I have a range for the \(x\) positions: \(x \in [-R,R]\text{,}\) which forms the bounds for the integral. I have a circle of radius \(\sqrt{R^2 - x^2}\) at each \(x\) position. The area of this circle is \(\pi (\sqrt(R^2 - x^2)^2 = \pi(R^2 - x^2)\text{.}\) That’s the area of the cross section, which is the function I will integrate. The width is \(dx\text{.}\) Those are all the pieces of the integral. The resulting integral is a reasonable polynomial integral.

\begin{align*} V \amp = \int_{-R}^R \pi (R^2 - x^2) dx\\ \amp = \pi \int_{-R}^R R^2 - x^2 dx = \left. \pi R^2 x - \pi \frac{x^3}{3} \right|_{-R}^R\\ \amp = \pi R^3 - (-\pi R^3) - \left( \frac{\pi R^3}{3} + \frac{\pi R^3}{3} \right) = 2\pi R^3 - \frac{2\pi R^3}{3} = \frac{4\pi R^3}{3} \end{align*}

The result is the well known formula for the volume of a sphere.

The next example will be an example of a surface of revolution. This is a very common construction for anything that has a circular cross section. The idea is to take the graph of a function \(y = f(x)\) and think of the graph sitting inside \(\RR^3\text{.}\) Then I split the graph around the \(x\) axis, like an object on a lathe. The result is an object with circular symmetry around the \(x\) axis. The radius of each slice is simply the function \(f(x)\text{.}\) This means that the integral to describe the volume of a surface of revoluation uses the function \(\pi (f(x))^2\) for the area of each circular slice.

\begin{equation*} V = \int_a^b \pi (f(x))^2 dx \end{equation*}

This construction can be done around any axis by simply changing the variable names in the construction.

Example 7.1.2.

The parabaloid with radius \(a\) at height \(h\) is described by the function \(y = h \left(\frac{x}{a}\right)^2\) rotated around the \(y\) axis. The slices here are along the \(y\) axis, not the \(x\) axis; this time \(x\) is the radius. If I solve for \(x\) in the function, I get \(x = a \sqrt{\frac{y}{h}}\text{.}\) If this is the radius, the area is \(\pi \left( a \sqrt{\frac{y}{h}} \right)^2 = \frac{\pi a^2 y}{h}\text{.}\) The range of \(y\) is \(y \in [0,h]\text{.}\) This lets me setup the integral for a surface of revolution using \(y\) as the variable. The integral is a straightforward polynomial integral.

\begin{align*} V \amp = \int_0^h \pi a^2 \frac{y}{h} dy = \frac{\pi a^2}{h} \int_0^h y dy\\ \amp = \frac{\pi a^2 h^2}{2h} = \frac{\pi a^2 h}{2} \end{align*}

I get a volume equation which is similar to a cone \(\left(\frac{1}{3} \pi r^2 h\right)\) but divided by \(2\) instead of \(3\text{.}\)

Example 7.1.3.

The interesting named Horn of Gabriel is a surface of revolution under the graph of \(f(x) = \frac{1}{x}\) for \(x \in [1,\infty)\text{.}\) It looks like an infinitely long trumpet. I can use an improper integral to calculate its volume, setting up the integral with circular cross-sectional slices oriented along the \(x\) axis. The radius is \(\frac{1}{x}\text{.}\)

\begin{equation*} V = \int_1^\infty \pi \frac{1}{x^2} dx = \lim_{a \rightarrow \infty} \int_1^a \frac{\pi}{x^2} = \lim_{a \rightarrow \infty} \left. \frac{-\pi}{x} \right|_1^a = \lim_{a \rightarrow \infty} \frac{-\pi}{a} + \pi = \pi \end{equation*}

The volume is finite! Even though the Horn of Gabriel extends to infinity, it narrows quickly enough that it only contains a finite volume. These kind of strange things can happen with limits and infinite: an infinitely long object with a finite volume.

Subsection 7.1.2 Volumes by Shells

In the previous section, I used cross-sectional slices, slicing with a plane to produce the cross-section with reasonable shapes and known areas. However, there are other ways to slice up a volume. In this section, I look at the general method of shells. Instead of thinking of the object as a collection of slices, I’ll think of it as a collection of nested shells. You could think of the (rough) spheres that form an onion, or stacking Matryoshka dolls. Only now, at least in the limit, there are infinitely many infinitely thin shells.

The setup is the same as for slices: the shells are identified by some variable \(x\) over some range \(x \in [a,b]\text{.}\) Each shell, identified by its position \(x\text{,}\) has an area \(A(x)\text{.}\) In the approximation, the shells also has a width \(\Delta x\) and the approximate volume of the shell is \(A(x) \Delta x\text{.}\) In the limit, the sum becomes and integral and the width becomes the infinitesimal \(dx\text{,}\) leading to the same kind of integral as for cross-sectional slices.

\begin{equation*} V = \int_a^b A(x) dx \end{equation*}

The slices were previously oriented along an axis. However, shells can be describes in a variety of ways. If the shells are radial, then perhaps \(r\) is the appropriate variable instead of \(x\text{.}\) Regardless of the variable name, the setup in the same.

Thought there could be a huge variety of the types of shells used, I’m going to restrict to cylindrical shells. These are useful because the surface areas is reasonable. A thin cylinder of radius \(r\) and height \(h\) could be cut and unrolled into a rectangle. The height of the rectangle is still \(h\text{,}\) but the width is the circumference \(2\pi r\text{.}\) Therefore, the area of the shell is \(2\pi Rh\text{.}\) Assuming that the height depends on the radius, I can write \(h = h(r)\text{.}\) Putting this into the integral gives the general setup for cylindrical shells.

\begin{equation*} V = \int_a^b 2 \pi r h(r) dr \end{equation*}

Let me move on to some examples using cylindrical shells.

Example 7.1.4.

Figure 7.1.5. Cross Section of a Bowl

Consider a pottery bowl whose cross section is roughly the area between \(x^2\) and \(x^4\) on the interval \([-1,1]\text{.}\) The cross section is show in Figure 7.1.5.

Sliced into cylindrial shells, each shell has height \(h = r^2 - r^4\text{,}\) based on the radius \(r\) from the centre of the bowl. I set up the integral with for cylindrical shells using the radius variable \(r\) and height \(r^2 - r^4\text{.}\) The integral is a straightforward polynomial integral.

\begin{align*} V \amp = \int_0^1 2\pi r (r^2 - r^4) dr\\ \amp = 2\pi \int_0^1 r^3 - r^5 dr\\ \amp = \left. 2\pi \left( \frac{r^4}{4} - \frac{r^6}{6} \right) \right|_0^1\\ \amp = \left. 2\pi \left( \frac{3r^4 - 2r^6}{12} \right) \right|_0^1 = \frac{2\pi}{12} = \frac{\pi}{6} \end{align*}

Example 7.1.6.

Figure 7.1.7. Cross Section of a Bell

Consider a bell which is the surface of revolution about the \(y\) axis of the function \(e^{-x^2}\) between \([-2,2]\text{.}\) A cross section of this bell is shown in Figure 7.1.7. I setup the cylindrical shells integral in the radius variable with height \(e^{-r^2}\text{.}\) The integral needs a substitution, which I’ve shown below; I change the bounds of the substitution as well, so that I don’t have to do the reverse substitution.

\begin{align*} V \amp = \int_0^2 2\pi re^{-r^2} dr\\ u \amp = r^2\\ du \amp = 2r dr\\ u(0) \amp = 0\\ u(2) \amp = 4 \\ V \amp = \int_0^4 \pi e^{-u} du = -\pi e^{-u} \Bigg|_0^4 = \pi \left( 1 - \frac{1}{e^4} \right) = \frac{\pi(e^4-1)}{e^4} \end{align*}

Example 7.1.8.

An interesting example is the calculation of the volume of a torus (colloquially, a doughnut). A torus is defined by a larger radius \(a\text{,}\) which is the distance from the centre of the torus to the centre of any cross-sectional circle, and a smaller radius \(b\text{,}\) which is the radius of any cross-sectional circle.

Figure 7.1.9. A Cross-Section of a Torus

I’m going to try to calculate the volume of the torus using cylindrical shells and the radius variable \(r\text{.}\) To set this up, I’ll look at the cross-section of the torus, in Figure 7.1.9.

First, the bounds of the cylinrdical shell will be \(r \in [b-a, b+a]\text{,}\) since \(a\) is the distance to the centre of the small circle.

Second, the circle on the right can be descirbe as the conventional locus of a circle but offset by \(a\) in the \(r\) axis (treating the horizontal axis here as the \(r\) axis.) If I call the vertical axis \(h\text{,}\) If I call the vertical axis \(h\text{,}\) this offset locus is \((r-a)^2 + h^2 = b^2\text{.}\) I can solve for \(h = \sqrt{b^2 - (r-a)^2}\text{.}\) This is only half the height of a cylindrical shell, since it goes both above and below, so I multiply by 2 to get the full height \(2 \sqrt{b^2 - (r-a)^2}\text{.}\) This lets me write the cylindrical shell integral for calculating the volume of the torus.

\begin{align*} V \amp = 2 \int_{a-b}^{a+b} 2\pi r \sqrt{b^2 - (r-a)^2} dr \end{align*}

A substitution helps here. I change the bounds as well, as usual. After the substitution, the integral will split into two by linearity.

\begin{align*} u \amp = r-a \implies r = u+a\\ du \amp = dr\\ u(a-b) \amp = -b\\ u(a+b) \amp = b\\ V \amp = 4\pi \int_{-b}^b (u+a) \sqrt{b^2 - u^2} du\\ \amp = 4\pi \left[ \int_{-b}^b u \sqrt{b^2-u^2} du + \int_{-b}^b a \sqrt{b^2-u^2} du \right] \end{align*}

I’ll deal with the two integrals separately. The first integral can also be solved with substitution, uving \(v = b^2 - u^2\text{.}\) I’ve not shown the details of this substitution; just the resulting antiderivative.

\begin{align*} \int_{-b}^b u \sqrt{b^2-u^2}du \amp = \left. \frac{1}{-2} \frac{2}{3} (b^2 -u^2)^{\frac{3}{2}} \right|_{-b}^b = 0 \end{align*}

This integral evaluates to zero, so it will not contribute to the volume at all. Now I move on to the second integral.

\begin{align*} \int_{-b}^b \sqrt{b^2 - u^2} \amp \end{align*}

This needs a trig substitution, Since the variable \(u\) is in the expression \(b^2 - u^2\) in the square root, the sine substitution is appropriate.

\begin{align*} u \amp = b \sin \theta\\ du \amp = b \cos \theta d \theta\\ u = b \amp \implies \theta = \frac{\pi}{2}\\ u = - b \amp \implies \theta = - \frac{\pi}{2}\\ \int_{-b}^b \sqrt{b^2 - u^2} \amp = \int_{-\pi/2}^{\pi/2} b\cos \theta b \cos \theta d\theta b^2 \int_{-\pi/2}^{\pi/2} \cos^2 \theta d\theta \end{align*}

After the trig substitution, I need a half-angle identity to deal with the even power of cosine. After the half-angle identity, I will get two reasonable integrals

\begin{align*} \amp = b^2 \int_{-\pi/2}^{\pi/2} \frac{1 + \cos 2\theta}{2} d \theta\\ \amp = b^2 \int_{-\pi/2}^{\pi/2} \frac{1}{2} d\theta+ b^2 \int_{-\pi/2}^{\pi/2} \frac{\cos 2\theta}{2} d\theta\\ \amp = \frac{b^2}{2} \left( \frac{\pi}{2} - \frac{-\pi}{2} \right) + 0\\ \amp = \frac{b^2\pi}{2} \end{align*}

Then I put this (and the zero for the first integral) back in the full expression to calculate the volume.

\begin{align*} V \amp = 4\pi \left[ 0 + \frac{b^2 \pi}{2} \right] = 2\pi^2 a^2 b \end{align*}

Therefore, the volume of a torus with major radius \(a\) and minor radius \(b\) is \(2\pi^2 a^2 b\text{.}\) As a curious obvservation, I can write this as \((2\pi a)(\pi b^2)\text{.}\) The \(2\pi a\) factor is the circumference of the large circle with radius \(a\text{,}\) which lies at the center of each cross-sectional circle. The \(\pi b^2\) factor is the radius of each cross-sectional circle. Therefore, the torus has the same volume of a cylinder with height equal to this circumference and radius \(b\text{.}\)

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