set up an integral when the following functions revolve around the $x$, $y$ and $y=\frac{1}{2}$

Question

$y=\sin x$, $y=\frac{1}{2}$, $x=0$ i got the same integral which is $$ \int_0^{ \frac{5 \pi}{6}} \pi \left(\sin^2(x)- \frac{1}{4}\right) \, \mathrm d x. $$ Anyone help

Answer 1

You have a few issues when you rotate around $x$ axis.

1. The upper limit for integration is $\pi/6$. That's the first value where $\sin x=1/2$.
2. The order of functions in the integral is reversed. The outer radius is $1/2$, the inner radius is $\sin x$.

Therefore $$V_{y=0}=\pi\int_0^{\frac \pi 6}\left(\frac 14-\sin^2 x\right)dx$$

Similarly, if you rotate around axis $y=\frac 12$, you have outer radius $\frac 12-\sin x$ and inner radius $0$. Therefore $$V_{y=\frac 12}=\pi\int_0^{\frac\pi6}\left(\frac 12-\sin x\right)^2dx$$

When you rotate around the $y$ axis ($x=0$), the inner radius is $0$, the outer radius is the value of $x$ where $\sin x=y$, so $x=\arcsin y$. Then $$V_{x=0}=\pi\int_0^{\frac 12}\arcsin^2 y dy$$