Overview
We have discussed in previous lessons how definite integrals can be used to compute the areas of regions underneath curves using the Fundamental Theorem of Calculus:
\(F(b)-F(a)=\int_a^bf(x)dx.\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(1)\)
But definite integrals have many other uses; for example, finding the volume of certain solids. Albeit, definite integrals cannot be used to compute the volume of every kind of solid but they can be used to compute the volume of a particular subset of solids known as solids of revolution. If you take a curve \(y=f(x)\) and rotate it about an axis, you create a two-dimensional surface which encloses a solid or region of space—this region of space is called a solid of revolution. For example, if you rotate the quarter-circle \(y=f(x)\) in Figure 2 about the \(x\)-axis, this will create a hemispherical, bowl-shaped surface; the solid of revolution enclosed by that surface is called a hemisphere. (All a hemisphere is is the two left over pieces one would have after chopping a sphere in half.) By computing the volume of this solid, we can determine the volume enclosed within a hemisphere; then, by multiplying this result by two, we can determine the volume enclosed within a sphere.
Figure 1: The "vase" shaped surface is obtained by revolving a curve about the vertical axis. The region of space that the surface encloses is called a solid of revolution.
Setting up the Riemann sum
Figure 2: A graph of the quarter circle \(x^2+y^2=r^2\) for \(x≥0\) and \(y≥0\). An \(n\) number of rectangles, \(f(x_i\)Δx\), are drawn underneath the curve.
Figure 3: By rotating the \(i^{th}\) rectangle about the \(x\)-axis, the cylinder depicted above is obtained. By doing this for each rectangle, an \(n\) number of cylinders are obtained. All of these cylinders fit inside of a hemisphere of radius \(r\) and the sum of the volumes of all of those cylinders approximate the volume of the hemisphere.
Before finding the exact volume, let’s first come up with an estimate. If we subdivide the hemisphere into an \(n\) number of cylinders, by adding up the volumes of all the cylinders we can obtain a good estimate for the volume of the hemisphere if \(n\) is large. The bigger \(n\) is, the better the estimate. But how do we compute the volume of each cylinder? Let’s subdivide the line \(AB\) in Figure 2 into an \(n\) number of subdivision of equal widths, \(Δx\). We’ll label the \(x\)-value associated with each interval \(x_i\) where \(i=1,…,n\). Let’s now construct a rectangle at each \(x\)-coordinate, \(x_i\). By rotating each rectangle about the \(x\)-axis, we can construct an \(n\) number of cylinders enclosed inside of the hemisphere. The height and width of an arbitrary rectangle under the curve \(y=f(x)\) are \(f(x_i)\) and \(Δx\), respectively. As the rectangle rotates about the \(x\)-axis, it traces out a cylinder with radius \(f(x_i)\) and width \(Δx\) as illustrated in Figure 3. Since the volume of a cylinder is given by the formula
$$\text{Volume of cylinder = π × }(radius)^2\text{ × width},$$
we know that the volume of each cylinder must be \(π(f(x_i))^2Δx\). By summing the volume of each cylinder, we get
$$S_n= π(f(x_1))^2Δx+…+ π(f(x_n))^2Δx.\tag{2}$$
Notice how Equation (2) looks very similar to the Reimann sum of the form,
$$S_n’=f(x_1)Δx+…+f(x_n)Δx,\tag{3}$$
that we used to approximate the area underneath an arbitrary function \(f(x)\). Indeed, if we think of each term \(π(f(x_1))^2\) in Equation (2) as some function \(h(x_i\) by letting \(h(x_i)≡π(f(x_1))^2\), Equation (2) becomes identical to Equation (3). This suggests that Equation (2) is related to definite integrals. This becomes especially apparent if we consider the limit of \(S_n\) as \(n→∞\). If we take this limit then \(Δx→dx\), \(x_i→x\), and the sum in Equation (2) becomes infinite. Thus, the limit of \(S_n\) as \(n→∞\) is given by
$$\lim_{n→∞}S_n=\int_0^r π(f(x))^2dx,\tag{4}$$
Figure 4: By taking the infinite sum of the volumes, \(π(f(x))^2dx\), of every cylinder from \(x=0\) to \(x=R\), we can obtain the volume of half of the sphere depicted above. By multiplying our answer by two, we can obtain the volume of the whole sphere. Image credit: https://www.miniphysics.com/uy1-calculation-of-moment-of-inertia-of-solid-sphere.html
Where the upper limit of integration, \(x=R\), is the radius of a sphere and \(f(x)\) is the height of the curve \(CB\) (see Figure 2). Equation (4) gives the exact volume enclosed within a hemisphere; but to use this equation to calculate the volume, we must first determine what \(f(x)\) is. The formula of a circle is given by
$$f(x,y)=x^2+y^2=R^2.\tag{5}$$
If we restrict the domain of \(f(x,y)\) by considering only the positive values of \(x\) and \(y\), then the graph of this function is a quarter-circle. The function \(f(x)\) in Equation (4) is associated with the \(y\)-coordinate, \(y=f(x)\), of each point along the quarter-circle in Figure 1a. We can express Equation (5) in terms of the y-coordinate associated with each point along a quarter circle by subtracting \(x^2\) from both sides to get
$$y=\sqrt{R^2-x^2}.\tag{6}$$
Since \(f(x)\) in Equation (4) is also the \(y\)-coordinate associated with each point along a quarter circle, it follows that
$$f(x)=\sqrt{R^2-x^2}.\tag{7}$$
Substituting Equation (7) into (4), we have
$$\text{Volume enclosed within hemisphere}=\int_0^r π(R^2-x^2)dx.\tag{8}$$
Now that the integrand is expressed only in terms of the single variable \(x\), we can solve the definite integral. Computing the integral in Equation (8), we have
$$\int_0^rπ(R^2-x^2)dx=\biggl[πR^2x-\frac{π}{3}x^3\biggr]_0^r$$
$$=πr^2(R)-\frac{π}{3}R^3=\frac{3πr^3}{3}-\frac{πR^3}{3}$$
$$=\frac{2}{3}πR^3.$$
Thus, the volume enclosed within a hemisphere is given by
$$\text{Volume enclosed within hemisphere}=\frac{2}{3}πR^3.\tag{9}$$
If we multiply this result by a factor of \(2\), then we’ll obtain the familiar formula for the volume enclosed within a sphere:
$$\text{Volume of sphere}=\frac{4}{3}πR^3.$$
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Sources: Wikipedia, Mini Physics