area of part of sphere inside cylinder

Question

Question is to find area of part of sphere of radius $a$ centered at origin that lies inside cylinder $x^2+y^2=ax$.

By drawing figure I thought it is exactly half part of sphere so answer must be $2\pi a^2$ but actual answer has one additional term $-4a^2$. From where this term come?

Answer 1

Look at the figure:

The request volume is less than an half sphere. This is a classical exercize an I think that you can solve it starting from this symilar ( a bit more general) example: Calculate the volume of the solid bounded laterally.

Answer 2

It looks something like this

You can integrate it with x going from $0$ to $a$, making slices of the area. Each slice has width of $dx\frac{a}{\sqrt{a^2-x^2}}$ and length of $2x\arcsin{\sqrt{1-\frac{x^2}{a^2}}}$

$\int_0^a\frac{a}{\sqrt{a^2-x^2}}2x\arcsin{\sqrt{1-\frac{x^2}{a^2}}}dx$

putting $x=a\cos t$ it becomes

$\int_0^{\pi/2}2a^2\cos ttdt=2a^2((t\sin t+\cos t)|_0^{\pi/2}=a^2(\pi-2)$

EDIT this was one area, upper for example, so just multiply by 2. It is $P=2a^2(\pi-2)$