Calculating Volumes - Washer/Disk Method

Before we look into calculating volumes of shapes, we will first precisely define volume in terms of calculus as follows:

Definition: Let $S$ be a solid between the lines $x = a$ and $x = b$. Let $A(x)$ be the continuous cross-sectional area function of $S$ in the plane $P_x$, through $x$, and perpendicular to the $x$-axis. Furthermore, let $\Delta x$ be the thickness of the cross sections. Then the Volume of $S$ is defined to be $V = \lim_{n \to \infty} \sum_{i=1}^n A(x_i^*) \Delta x = \int_a^b A(x) \: dx$.

From our definition of the volume of this solid, we see that the volume formula comes from summing up the values of $A(x)$. We will now also derive a method for calculating $A(x)$.

• If $A(x)$ generates circular disks, then we must find the area of each washer by finding it's radius. Therefore, $A(x) = \pi r^2$ where $r$ is the radius of the disk. If our object is rotated on the $x$-axis or $y-axis$, then $r = f(x)$ or $r = f(y)$. If our object is rotated on the line $y = a$, then our radius is $r = a - f(x)$. If our object is rotated around the line $x = a$, then our radius is $r = a - f(y)$.

• If $A(x)$ generates a washer (or ring), then we must find the area of the outer disk and subtract the area of the inner disk to get the area of the washer. Both of these disks will be circular, so we get that $A_{\mathrm{washer}} = A_{\mathrm{outer disk}} - A_{\mathrm{inner disk}}$. We calculate the areas of the outer and inner disks in the same way mentioned above.

Example 1

Determine the volume of the solid of revolution around the $x$-axis bounded by the function $f(x) = \cos x + 2$ on the interval $[0, 2\pi ]$.

Applying our formula for volume we obtain:

(1)

\begin{align} \mathbf{Volume} = \pi \int_{0}^{2\pi} (\cos x + 2) ^2 x \: dx \\ \mathbf{Volume} = \pi \int_{0}^{2\pi} \cos ^2 x + 4\cos x +4 \: dx \\ \end{align}

Using the trig substitution that $cos^2x = \frac{1 + \cos (2x)}{2}$ we obtain:

(2)

\begin{align} \mathbf{Volume} = \pi \int_{0}^{2\pi} \frac{1 + \cos (2x)}{2} + 4\cos x +4 \: dx \\ \quad \mathbf{Volume} = \pi ( \frac{1}{2} \int_{0}^{2\pi} 1 + \cos (2x) \: dx + 4 \int_{0}^{2\pi} \cos x \: dx + 4 \int_{0}^{2\pi} 1 \: dx ) \\ \mathbf{Volume} = \pi ( \frac{x}{2} + \frac{\sin (2x)}{4} + 4 \sin x + 4x ) \bigg |_{0}^{2\pi} \\ \mathbf{Volume} = 9\pi^2 \end{align}

Example 2

Calculate the volume generated by rotating the area between the curves $f(x) = \sqrt{x}$ and $g(x) = x$ around the $x$-axis.

We first note that $f(x) = g(x)$ when $x = 0$ and $x = 1$, thus, we have found our limits of integration. Furthermore, we note that on the interval $[0, 1]$, $f(x) ≥ g(x)$. The diagram below illustrates the area we're rotating about the $x$-axis:

We note that this will result in washers, so $A_{\mathrm{washer}} = A_{\mathrm{outer disk}} - A_{\mathrm{inner disk}}$. We note that the area of the larger disk cross-sections will be $\pi (\sqrt{x})^2 = \pi x$, while the area of the smaller disk cross-sections will be $\pi (x)^2 = \pi x^2$ Therefore, $A_{\mathrm{washer}} = \pi(x - x^2)$. Now we evaluate the following integral:

(3)

\begin{align} V = \int_{0}^{1} \pi(x - x^2) \: dx \\ V = \pi \left ( \frac{x^2}{2} - \frac{x^3}{3} \right ) \biggr \rvert_{0}^{1} \\ V = \pi \left (\frac{1}{2} - \frac{1}{3} \right ) \\ V = \frac{\pi}{6} \end{align}

Example 3

Calculate the volume generated by rotating the area between the curves $f(x) = x^2$ and $g(x) = 1$ and $x = 2$ around the $x$-axis.

The diagram below illustrates the area we're rotating:

We note that our lower bound is $1$ and our upper bound is $2$. Furthermore, we note that the outer radius is $x^2$ and the inner radius is $1$, therefore we get $A_{\mathrm{outer disk}} = \pi(x^2)^2 = \pi x^4$, and $A_{\mathrm{inner disk}} = \pi (1)^2 = \pi$

(4)

\begin{align} V = \int_{1}^{2} \pi [ x^4 - 1] \: dx \\ V = \left ( \frac{x^5}{5} - x \right ) \biggr \rvert_{1}^{2} \\ V = \left ( \frac{32}{5} - 2 \right ) - \left ( \frac{1}{5} - 1 \right ) \\ V = \frac{31}{5} - 1 \\ V = \frac{26}{5} \end{align}