Calculating Volumes with Disks/Washers Examples 1

We will now look at some more examples of calculating volumes via the Cylindrical Shells. It is important to first recall the following formulae, the first for volume calculated with disks where $A(x) = \pi [f(x)]^2$ if $f$ is rotated around the $x$-axis, or $A(x) = \pi[a - f(x)]^2$ if $f$ is rotated around the line $y = a$. Similarly, if the volume is calculated with washers where $f(x) > g(x)$, then $A_{\mathrm{washer}} = A_{\mathrm{outer\:disk}} - A_{\mathrm{inner\:disk}}$. Both formulae can be summarized in the following integral:

(1)

\begin{align} \int_a^b A(x) \: dx \end{align}

Example 1

Calculate the volume of the solid obtained by taking the area trapped between the curve $f(x) = \frac{1}{x}$, the $x$-axis, and the vertical lines $x = 1$ and $x = 2$ about the $x$-axis.

From the graph, we can see that the radius of our disk will be $\frac{1}{x}$, and therefore, we set up our integral as follows:

(2)

\begin{align} V = \int_{1}^{2} \pi \left ( \frac{1}{x} \right )^2 \: dx \\ V = \int_{1}^{2} \pi x^{-2} \: dx \\ V = \pi \left [- \frac{1}{x} \right ]_{1}^{2} \\ V = \pi \left (- \frac{1}{2} \right ) - (-1) \\ V = \frac{\pi}{2} \end{align}

Example 2

Calculate the volume of the solid obtained by taking the area trapped between the curves $f(x) = x^2$ and $g(x) = \sqrt{x}$ about the line $y = 1$.

We must first calculate the points of intersection between $f(x)$ and $g(x)$. We note that $f(x) = g(x)$ when $x = 0$ and when $x = 1$, so we have found our bounds of integration. We note that we will be generating washers. The outer disk radius is $1 - f(x) = 1 - x^2$, while the inner disk radius is $1 - g(x) = 1 - \sqrt{x}$, therefore, we must evaluate the following integral:

(3)

\begin{align} V = \int_{0}^{1} \pi \left [ (1 - x^2)^2 - (1 - \sqrt{x})^2 \right ] \: dx \\ \: V = \int_{0}^{1} \pi \left [ 1 - 2x^2 + x^4 - \left (1 - 2\sqrt{x} + x \right ) \right ] \: dx \\ \: V = \pi \int_{0}^{1} \left [ - 2x^2 + x^4 + 2x^{1/2} - x \right ] \: dx \\ \: V = \pi \left [ -2\frac{x^3}{3} + \frac{x^5}{5} + 2\frac{x^{3/2}}{3/2} + \frac{x^2}{2} \right ]_{0}^{1} \\ \: V = \pi \left [ -\frac{2}{3} + \frac{1}{5} + \frac{1}{3} + \frac{1}{2} \right ] \\ \: V = \frac{11\pi}{30} \end{align}

Example 3

Calculate the volume of the solid of revolution obtained by taking the area trapped between the curves $f(x) = \sin x$ and $g(x) = \cos x$, the lines $x = 0$ and $x = \frac{\pi}{4}$, about the line $y = -1$.

We note that our limits of integration are $0$ and $\frac{\pi}{4}$. We note that the area of our cross-sections of our solid will be washers. The radius of the outer disk is $1 + g(x) = 1 + \cos x$, while the radius of the inner disk is $1 + f(x) = 1 + \sin x$. Therefore, we must evaluate the following integral:

(4)

\begin{align} V = \int_{0}^{\frac{\pi}{4}} \pi[(1 + \cos x)^2 - (1 + \sin x)^2] \: dx \\ \: V = \int_{0}^{\frac{\pi}{4}} \pi[1 + 2 \cos x + \cos ^2 x - \left (1 + 2\sin x + \sin ^2 x\right)] \: dx \\ \: V = \int_{0}^{\frac{\pi}{4}} \pi[2 \cos x + \cos ^2 x - 2\sin x - \sin ^2 x] \: dx \\ \: V = \int_{0}^{\frac{\pi}{4}} \pi[2 \cos x + \cos ^2 x - 2\sin x - (1 - \cos^2 x)] \: dx \\ \: V = \int_{0}^{\frac{\pi}{4}} \pi[2 \cos x + \cos ^2 x - 2\sin x - 1 + \cos^2 x] \: dx \\ \: V = \int_{0}^{\frac{\pi}{4}} \pi[2\cos ^2 x + 2 \cos x - 2\sin x - 1] \: dx \\ \: V = \pi \left [ \sin x \cos x - 2x + 2 \sin x + 2 \cos x \right ]_{0}^{\frac{\pi}{4}} \\ \quad \: V = \pi \left [ \left ( \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\pi}{2} + 2 \frac{\sqrt{2}}{2} + 2 \frac{\sqrt{2}}{2} \right ) - \left ( 0 \cdot 1 - 0 + 2 \cdot 0 + 2 \right )\right ] \\ \quad V = \pi \left [ \frac{1}{2} - \frac{\pi}{2} + 2 \sqrt{2} - 2\right ] \\ \quad V = \frac{\pi (4 \sqrt{2} -3 - \pi)}{2} \end{align}