In my textbook, I had a question:
The region bounded by the $x^2+(y-1)^2=1$ is rotated about the y-axis. Find the volume of resulting solid using any method.
I did using shell method:
Radius: $-y$.
Circumference: $-2\pi y$.
Height: $x=\sqrt{1-(y-1)^2}$.
So, The Volume is:
$$V=-2\pi \int_{-1}^1y\sqrt{1-(y-1)^2}$$ or $$-2\pi \int_{-1}^1y\sqrt{y(2-y)}$$
So, I stuck here. Using substitution ($z=y(2-y)$) did not make any sense to calculate this integral. What can I do next or are there any alternative methods to solve this problem?
Best Answer
If you are rotating about the $y$ axis, then you need only the right half of the circle and you know it will be a sphere with volume $\frac{4\pi}{3}$. By the shell method
$$ 2\pi rh=4\pi x\sqrt{1-x_2} $$
$$ \int_0^12\pi rh\,dx=2\pi\int_0^12x(\sqrt{1-x^2})\,dx =\left[-\frac{4\pi}{3}(1-x^2)^{3/2}\right]_0^1=\frac{4\pi}{3}$$