**Lesson overview**

In this lesson, we'll discuss how by using the

concept of a definite integral one can calculate the volume of something called an *oblate spheroid*. An oblate spheroid is essentially just a sphere which is compressed or stretched along one of its dimensions while leaving its other two dimensions unchanged. For example, the Earth is technically not a sphere—it is an oblate spheroid. To find the volume of an oblate spheroid, we'll start out by finding the volume of a *paraboloid .* (If you cut an oblate spheroid in half, the two left over pieces would be paraboloids.) To do this, we'll draw an \(n\) number of cylindrical shells inside of the paraboloid; by taking the Riemann sum of the volume of each cylindrical shell, we can obtain an estimate of the volume enclosed inside of the paraboloid. If we then take the limit of this sum as the number of cylindrical shells approaches infinity and their volumes approach zero, we'll obtain a definite integral which gives the exact volume inside of the paraboloid. After computing this definite integral, we'll multiply the result by two to get the volume of the oblate spheroid.

**Finding volume of an oblate spheroid**

In Figure 1, I have graphed the ellipse \(\frac{x^2}{9}+{y^2}{4}=1\) on the \(xy\)-plane. If we rotate the eclipse about either the \(x\)-axis or \(y\)-axis, the ellipse will trace out the closed surface illustrated in Figure 3. The volume of revolution which that surface encloses is called an oblate spheroid. In this lesson, we'll use the concept of a definite integral to calculate the volume of an oblate spheroid. To calculate this volume, we'll first approximate the volume by summing the volumes of an \(n\) number of cylindrical shells (see Figure 2) drawn within the oblate spheroid. After that, we'll take the limit of this sum as \(n→∞\).

But before we do that, let's discuss how to construct a cylindrical shell and how to calculate its volume. Let's subdivide the interval on the \(x\)-axis, \(Δx=3-0\), into an \(n\) number of equally spaced tick marks; let's label each tick mark with \(x_i\) where \(i=1,...,n\). In Figure 1, I have drawn a rectangle with with \(Δx=x_{i+1}-x_i\) and height \(f(x_i)\). If we rotate this rectangle about the \(y\)-axis, the rectangle will trace out the cylindrical shell illustrated in Figure 2. To calculate the volume of the cylindrical shell, we must take the product of the area of the cylindrical shell's base with its height. The ring \(QQ'RR'\) with width \(Δx_{i+1}-x_i\) in Figure 2 is the cylindrical shell's base. Let's subtract the area of the inner circle \(QQ'\) from the area of the outer circle \(RR'\) in Figure 2 to get the area of the cylindrical shell's base:

$$A=π(x_{i+1})^2-π(x_i)^2.\tag{1}$$

Using basic algebra, we can rewrite Equation (1) as

$$A=π\frac{x_i+x_{i+1}}{2}\biggl[2(x_{i+1}-x_i)\biggr].\tag{2}$$

The term \((x_i+x_{i+1}/2\) in Equation (2) is the average value of \(x_i\) and \(x_{i+1}\). In Figure 1 (click to enlarge), I have labeled the average of these two values as \(\bar{x}_i\) on the \(x\)-axis. Substituting \(\bar{x}_i\) into Equation (2), we have

$$A=2π\bar{x}_iΔx.\tag{3}$$

(You might be asking yourself why we went through the trouble of rewriting Equation (1) of the form expressed in Equation (3). The reason why we did this will become evident when we wish to express the limit of the sum of the volumes of each cylindrical shell as a definite integral. But we'll discuss this in more detail shortly.)

As you can see from Figure 2, the hieght of a cylindrical shell is \(f(x_i)\). The volume of the \(i^{th}\) cylindrical shell is therefore given by

$$ΔV_i=2π\bar{x}_if(x_i)Δx.\tag{4}$$

To estimate the volume of the paraboloid, let's sum the volumes of all the cylindrical shells to get

$$S_n=\sum_{i=1}^n2π\bar{x}_if(x_i)Δx.\tag{5}$$

When defining a definite integral, we always start with a sum of the form

$$S_m=\sum_{i=1}^mg(x_i)Δx;\tag{6}$$

then, we take the limit of such a sum as \(m→∞\) to get

$$\int_a^bg(x)dx=\lim_{m→∞}\sum_{i=1}^mg(x_i)Δx.$$

The problem with Equation (5) is that the term \(2π\bar{x}_if(\bar{x}_i)Δx\) isn't the same as the \(g(x_i)\) in Equation (6). We cannot define a function \(h(\bar{x}_i)\) or \(h(x_i)\) that we can set equal to \(2π\bar{x}_if(\bar{x}_i)Δx\). The term \(2π\bar{x}_if(\bar{x}_i)Δx\) requires two input values (namely, \(\bar{x}_i)\) and \(x_i\)) to specify its value whereas functions like \(g(x_i)\) in Equation (6) require only one input value (namely, \(x_i\)) to specify its value. Fortunately, there is a way around this problem. Recall that it does not matter whether we take a left-hand side Reimann sum (in which case, the height of the rectangle would be \(g(x_i)\)), a right-hand side Reimann sum (this is when the height of each rectangle is given by \(g(x_{i+1}\)), or a midpoint Reiman sum (when the height of a rectangle is given by \(g(\frac{x_i+x_{i+1}}{2})=g(\bar{x}_i)\)). (We shall not discuss the reasons why this is here; but if you do not understand why this is, I strongly encourage you to review the topic.) For similar reasons, we could replace the \(f(x_i)\) in Equation (5) with either \(f(x_{i+1}\) or \(f(\bar{x}_i)\); doing so will not change the limit of the sum. (Indeed, we could replace \(f(x_i)\) in Equation (5) with \(f(x_i*)\) (where \(x_i≤x_i*≤x_{i+1}\) and, although the Equation (5) would give a different approximation of the paraboloid, the limit of Equation (5) would remain the same. To understand why this is, it would be a good idea to review the concept of limits.) Swapping the \(f(x_i)\) in Equation (5) with \(f(\bar{x}_i)\), we get a different sum (which we'll specify by \(S_n'\)) given by

$$S_n'=\sum_{i=1}^n2π\bar{x}_if(\bar{x}_i)Δx.\tag{7}$$

What's nice about Equation (7) is that the term \(2π\bar{x}_if(\bar{x}_i)Δx\) is expressed entirely in terms of the single variable \bar{x}_i\). Thus, Equation (7) is of the same form as Equation (6). If \(n→∞\) (which is to say, if the number of cylindrical shells within the paraboloid approaches infinity), then the sum \(S'_n\) will get closer and closer to equaling the exact volume of the paraboloid. Thus

$$\lim_{n→∞}\sum_{i=1}^n2π\bar{x}_if(\bar{x}_i)Δx=\int_0^32πxf(x)dx.\tag{8}$$

To evaluate the integral in Equation (8), we need to find out what the function \(f(x)\) is. \(f(x)\) represents the height (which is to say, the \(y\)-value) associated with each rectangle on the interval \(Δx=3-0\). In other words, \(f(x)\) is the \(y\)-coordinate associated with each point along the quarter-ellipse in the first quadrant of the \(xy\)-plane illustrated in Figure 2. Recall that the equation \(\frac{x^2}{9}+\frac{y^2}{4}=1\) was used to graph each \((x,y)\) coordinate along the ellipse in Figure 1. If we restrict the domain of this function to values of \(x\) and \(y\) where \(0≤y≤2\), then the equation \(\frac{x^2}{9}+\frac{y^2}{4}=1\) could be used to graph the quarter-ellipse in the first quadrant of the \(xy\)-plane in Figure 1. Thus, for the aforementioned restrictions on the domain, the \(y\) in the equation, \(\frac{x^2}{9}+\frac{y^2}{4}=1\), specifies the \(y\)-coordinate of each point along the quarter-ellipse. It therefore also specifies the height of each rectangle under the quarter-ellipse. This means that \(f(x)=y(x)\). Using the equation \(\frac{x^2}{9}+\frac{y^2}{4}=1\), we can solve for \(f(x)=y(x)\):

$$\frac{x^2}{9}+\frac{(f(x))^2}{4}=1$$

$$\frac{(f(x))^2}{4}=1-\frac{x^2}{9}$$

$$f(x)=\sqrt{4-\frac{4}{9}x^2}.\tag{9}$$

Substituting Equation (9) into the integral in Equation (8), we have

$$\text{Volume of paraboloid}=\int_0^32πx\sqrt{4-\frac{4}{9}x^2}dx.\tag{10}$$

At this point, all of the hard work is done and we just need to solve the definite integral in equation (10) and then multiply our answer by \(2\) to get the volume of the oblate spheroid illustrated in Figure 3. We can solve the integral in Equation (10) by using \(u\)-substitution. If we let \(u=4-\frac{4}{9}x^2\), then

$$\frac{du}{dx}=\frac{-8}{9}x$$

$$du=\frac{-8}{9}xdx$$

$$dx=\frac{-9}{8}\frac{1}{x}du.\tag{11}$$

Substituting \(u\) and Equation (11) into (10), we have

$$\text{Volume of paraboloid}=\int_{?_1}^{?_2}(2πx)\biggl(\frac{-9}{8}\frac{1}{x}\biggr)u^{1/2}du$$

and

$$\text{Volume of paraboloid}=\frac{-9}{4}π\int_{?_1}^{?_2}u^{1/2}du.$$

When \(x=0\), \(u=4\) and when \(x=3\), \(u=4-\frac{4}{9}(3)^2=4-4=0\). Substituting the limits of integration into the integral above and solving the integral, we have

$$\text{Volume of paraboloid}=\frac{-9}{4}\biggl[\frac{2}{3}u^{3/2}\biggr]_4^0$$

$$=\frac{9}{4}π(\frac{2}{3}(4)^{3/2})=\frac{3}{2}π(4)^{3/2}$$

$$=\frac{-9}{4}π\biggl[\frac{2}{3}u^{3/2}\biggr]_4^0=\frac{-3}{2}π\biggl[(4-\frac{4}{9}x^2)^{3/2}\biggr]_0^3$$

$$=\frac{-3}{2}π\biggr[(4-4)-(4-0)^{3/2}\biggr]=12π.$$

Thus we have shown that the volume of the paraboloid is \(12π\) units squared. Multiplying this result by \(2\), we find that the volume of this oblate spheroid is given by

$$\text{Volume of oblate spheroid}=24π.\tag{12}$$

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