The question comes from the 1991 AP Calculus test.
Let $R$ be the region between the graphs of $y = 1 + sin(\pi x) $ and $y = x^2$ from $x = 0$ to $x = 1$.
c) Set up, but do not integrate an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the y-axis
The answers are
$$V = 2\pi\int_{0}^{1}x(1 + sin(\pi x) – x^2)dx $$
or
$$ V = \pi\int_{0}^{1}y dy + \pi\int_{1}^{2}(1 – \frac{1} \pi arcsin(y-1))^2 – (\frac{1} \pi arcsin(y-1))^2dy $$
I am confused on how they reached these two answers. I was able to do the question before this which involved rotating the same area but around the x-axis. In school, we learned to write the equations as $x = …$ when dealing with rotations around the y-axis but I'm not sure how that applies here.
Best Answer
1) Using the shell method, $\displaystyle V=\int_0^1 2\pi r(x)h(x)dx=\int_0^1 2\pi x(1+\sin(\pi x)-x^2)\;dx$
$\hspace{.2 in}$since $r(x)=x-0=x$ and $h(x)=y_1-y_2=1+\sin(\pi x)-x^2$.
2) Using the disc method, we can divide the region into 2 parts using the line segment from (0,1) to (1,1).
Then $V=\displaystyle\int_0^1 \pi((R_1(y))^2-(r_1(y))^2)\;dy+\int_1^2 \pi((R_2(y))^2-(r_2(y))^2)\;dy$ where
$\hspace{.4 in}y=x^2\implies R_1(y)=x=\sqrt{y}\;$ and $\;r_1(y)=0$.
Solving $y=1+\sin(\pi x)$ for $x$ gives $\sin(\pi x)=y-1$, so
a) $0\le x\le\frac{1}{2}\implies 0\le \pi x\le\frac{\pi}{2}\implies \pi x=\sin^{-1}(y-1)\implies x=\frac{1}{\pi}\sin^{-1}(y-1)$
$\displaystyle\hspace{3.4 in}\implies r_2(y)=\frac{1}{\pi}\sin^{-1}(y-1)$
b) $\frac{1}{2}\le x\le1\implies\frac{\pi}{2}\le \pi x\le \pi\implies \pi x=\pi-\sin^{-1}(y-1)$
$\displaystyle\hspace{2.05 in}\implies x=1-\frac{1}{\pi}\sin^{-1}(y-1)\implies R_2(y)=1-\frac{1}{\pi}\sin^{-1}(y-1)$