Find by double integration volume of sphere $x^2 + y^2 + z^2 = a^2$ cut off by the plane $z = 0$ and the cylinder $x^2 + y^2 = ax$.
I proceded like this :
$x = r\cos(\theta)$
$x = r\sin(\theta)$
$r^2 = x^2 + y^2 $
$V = \int^ {\pi /2}_0\int^{a\cos(\theta)}_a {(a^2-r^2)}^{1/2}r~drd\theta$
$V = \int^ {\pi /2}_0[ {(a^2-r^2)}^{3/2}]^{a\cos(\theta)}_a~d\theta$
following the same path and solving the integral further I got certain ans. but it is wrong. I cant find out what is wrong. Please help.
Your definition of $r$ is missing a square root. But that's probably just a typo. The lower limit of your inside integral should be $0$, not $a$.
But we can't really tell you what you did wrong if you don't post the bit of your work that you think is wrong.