Question
Consider a triangle in the xy-plane with vertices at A = (0, 1), B = (2, 3), and C = (3, 1). Let R denote the region that is bounded by the triangle shown in the figure below.
(A) Find the volume of the solid obtained by rotating R about the x-axis.
(B) Find the volume of the solid obtained by rotating R about the y-axis.
(C) Find the volume of the solid having R as its base while cross sections perpendicular to the x-axis are squares.
Answer/Explanation
(A) The line segment $A B$ is defined by the equation $y=x+1$ for $0 \leq x \leq 2$, the line segment $B C$ is given by $y=-2 x+7$ for $2 \leq x \leq 3$, while $A C$ is simply $y=1$.
Note that each cross section perpendicular to the $x$-axis is a washer with inner radius 1 , outer radius $x+1$ for $0 \leq x \leq 2$, and outer radius $7-2 x$ for $2 \leq x \leq 3$. Therefore, the volume of the solid is given by the integral:
$
\begin{aligned}
\pi\left(\int_0^2\left[(x+1)^2-1^2\right] d x+\int_2^3\left[(7-2 x)^2-1^2\right] d x\right) & =\pi\left(\left[\frac{(x+1)^3}{3}-x\right]_0^2+\left[-\frac{(7-2 x)^3}{6}-x\right]_2^3\right) \\
& =\pi\left(\left[(9-2)-\frac{1}{3}\right]+\left[\left(-\frac{1}{6}-3\right)+\left(\frac{27}{6}+2\right)\right]\right) \\
& =\pi\left(\left[\frac{21}{3}-\frac{1}{3}\right]+\left[\left(-\frac{1}{6}-\frac{18}{6}\right)+\left(\frac{27}{6}+\frac{12}{6}\right)\right]\right) \\
& =\pi\left(\frac{20}{3}+\frac{20}{6}\right)=\frac{30 \pi}{3}=10 \pi
\end{aligned}
$
(B) Observe that $A B$ may be written as $x=y-1$ and $B C$ as $x=\frac{7-y}{2}$ for $1 \leq y \leq 3$.
Therefore, revolving $R$ about the $y$-axis is the value of the following integral:
$
\pi \int_1^3\left(\frac{7-y}{2}\right)^2-(y-1)^2 d y=\pi\left(\frac{1}{4} \int_1^3(7-y)^2 d y-\int_1^3(y-1)^2 d y\right)
$
Next, you let $u=7-y$ in the first integral above, noting that under this substitution $-d u=d y$. Therefore, the integral becomes
$
\begin{aligned}
\pi\left(-\frac{1}{4} \int_{y=1}^y u^2 d u-\int_1^3(y-1)^2 d y\right) & =\pi\left(-\frac{1}{4}\left[\frac{1}{3} u^3\right]_{y=1}^{y=3}-\left[\frac{1}{3}(y-1)^3\right]_1^3\right) \\
& =\pi\left(\frac{1}{4}\left[-\frac{1}{3}(7-y)^3\right]_1^3-\left[\frac{1}{3}(y-1)^3\right]_1^3\right) \\
& =\pi\left(-\frac{1}{12}\left[4^3-6^3\right]-\frac{1}{3}\left[2^3\right]\right) \\
\text { (Note: } \left.4^3-6^3=4 \cdot 4^2-4 \cdot 9 \cdot 6 .\right) & =\pi\left(-\frac{1}{12}\left[4\left(4^2-9 \cdot 6\right)\right]-\frac{8}{3}\right) \\
& =\pi\left(-\frac{1}{3}(16-54)-\frac{8}{3}\right) \\
& =\pi\left(\frac{38}{3}-\frac{8}{3}\right)=10 \pi
\end{aligned}
$
(C) The area of each square cross section from $x=0$ to $x=2$ is $A_1(x)=(x+1-1)^2=$ $x^2$, while from $x=2$ to $x=3$, the area is given by $A_2(x)=(-2 x+7-1)^2=(6-2 x)^2=4\left(x^2-\right.$ $6 x+9)$. Therefore, the volume of the described solid is
$
\begin{aligned}
\int_0^2 A_1(x) d x+\int_2^3 A_2(x) d x & =\int_0^2 x^2 d x+4 \int_2\left(x^2-6 x+9\right) d x \\
& =\left[\frac{1}{3} x^3\right]_0^2+4\left[\frac{1}{3} x^3-3 x^2+9 x\right]_2^3 \\
& =\left(\frac{8}{3}\right)+4\left[(9-27+27)-\left(\frac{8}{3}-12+18\right)\right] \\
& =\left(4 \cdot \frac{2}{3}\right)+4\left[9-\frac{26}{3}\right] \\
& =4\left(9-\frac{24}{3}\right)=4[9-8]=4
\end{aligned}
$