Calculating Volumes with Disks/Washers Examples 2

# Calculating Volumes with Disks/Washers Examples 2

We will continue to look at more examples of calculating volumes via the Washer/Disk Method. Please review the Calculating Volumes with Disks/Washers Examples 1 if you haven't already!

## Example 1

Calculate the volume of the solid obtained by taking the area trapped between the curves $f(x) = x^3$, $y = 27$, and $x = 0$, rotated around the $y$-axis. We will first rewrite $f(x)$ in terms of $y$, that is, $x = y^{1/3}$. We note that we want to integrate from $y = 0$ to $y = 27$. Additionally, we get that our radius is $y^{1/3}$ and therefore, we must evaluate the following integral:

(1)
\begin{align} V = \int_{0}^{27} \pi [y^{1/3}]^2 \: dy \\ V = \int_{0}^{27} \pi y^{2/3} \: dy \\ V = \pi \left [ \frac{3y^{5/3}}{5} \right]_{0}^{27} \\ V = \frac{729 \pi}{5} \end{align}

## Example 2

Calculate the volume of the solid obtained by taking the area trapped between the curves $f(x) = x^3$ and $g(x) = x$ where $x ≥ 0$ around the line $x = 0$. We note that $f(x) = g(x)$ when $x = 0$ and $x = 1$ (we exclude $x = -1$ since $x ≥ 0$). Let's first rewrite $f(x)$ and $g(x)$ in terms of $y$, that is $x = y^{1/3}$ and $x = y$. Our limits of integration are $y = 0$ and $y = 1$. We note we will be generating washers as our cross-sections and that our outer radius will be $f(y)$ while our inner radius will be $g(y)$ and therefore:

(2)
\begin{align} V = \int_{0}^{1} \pi [ y^{2/3} - y^2 ] \: dy \\ V = \pi \left [ \frac{3y^{5/3}}{5} - \frac{y^3}{3} \right ]_{0}^{1} \\ V = \pi \left ( \frac{3}{5} - \frac{1}{3} \right) \\ V = \frac{4\pi}{15} \end{align}