The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney.
Problem:
Find the volume generated by revolving the following region: The triangle with vertices $(1,1)$, $(1,2)$ and $(2,2)$ about the
y-axis.
Answer:
Since we are going around the y-axis, our integral will integrate with respect to $y$ not $x$. Observe that the points$(1,1)$ and $(2,2)$ are on the line $y = x$. Let $v$ be the volume we seek.\begin{align*}
v &= \pi \int_1^2 2^2 - y^2 \,\, dy = \pi \left( 4y - \frac{y^3}{3} \right) \Big|_1^2 \\
v &= \pi \left( 8 - \frac{8}{3} - \left( 4 - \frac{1}{3} \right) \right) = \pi \left( 8 - \frac{8}{3} - 4 + \frac{1}{3} \right) \\
v &= \pi ( 4 - \frac{7}{3 } ) \\
v &= \frac{5\pi}{3} \\
\end{align*}The book's answer is $\frac{4\pi}{3}$. What did I do wrong?
2 Answers
$\begingroup$When the triangle is rotated, a "washer" of outer radius $y$ and inner radius 1 is formed. Thus, the volume should be given by $\pi \int^{2}_{1} y^2 - 1 dy$, which should evaluate to $4 \pi/3$.
$\endgroup$ $\begingroup$$$V=\pi\int\limits_1^2(y^2-1)dy=\pi\left(\frac{y^3}{3}-y\right)\big{|}_1^2=\pi\left(\frac{8}{3}-2-\left(\frac{1}{3}-1\right)\right)=\pi\left(\frac{2}{3}+\frac{2}{3}\right)=\frac{4\pi}{3}.$$
I think, the mistake is that you rotated around $x$-axis.
$\endgroup$ 2
