The Issue
I've been attempting to complete the below problem for a solid 45 minutes, and I am really stuck. I've looked online, used calculators, and scoured the Stack Exchange, and I haven't found anything that can help me.
The Question
Find the volume of the region bounded by the graph of $y=2\sin x$ and the x-axis on $[0, \pi]$, that is revolved about the line $y=-2$. Type an exact answer using $\pi$ as needed.
Based on the problem, I came up with $V=\int_{0}^\pi\pi\left((2\sin x)^2-(-2)^2\right)dx$ for the volume equation. However, I keep working myself into a dead end after this point. The work I currently have is: $$V=\int_{0}^\pi\pi\left((2\sin x)^2-(-2)^2\right)dx$$ $$V=\pi\int_{0}^\pi\left(4\sin^2x-4\right)dx$$ $$V=-4\pi\int_{0}^\pi\left(1-\sin^2x\right)dx$$ $$V=-4\pi\int_{0}^\pi\left(\cos^2x\right)dx$$ But, that's as far as I can get. I'm hoping someone can give me some tips or helpful pointers about whether or not my above process is correct, and what steps I can take next to solve the problem.
You're getting a negative volume, so that can't be right. The problem is that you're using a formula for rotation about the $x$-axis, but in the question we're not rotating about the $x$-axis, we're rotating about $y=-2$.
To get around this problem, translate everything up $2$ units, so that we are rotating the region bounded by $y=\sin x+2$ and $y=2$ from $x=0$ to $x=\pi$, completely about the line $y=0$ (the $x$-axis).
Translating everything, including the axis of rotation, does not change the volume. (It's like we're just shifting the coordinate axes; the size and shape of the region are unchanged.)
So the volume is:
$$V=\pi\int_{0}^\pi\left[(2\sin x+2)^2-2^2\right]\mathrm dx$$
Integrating this involves integrating $\sin^2 x$, which can be done using the identity $\cos 2A\equiv\cos^2 A-\sin^2 A\implies \sin^2 A\equiv\frac{1-\cos 2A}{2}$.
This works out to be $V=16\pi+2\pi^2\approx 70.0047$.