I came across this question in double integral.
What I did: I converted the equation $x^2+y^2-2ax=0$ to polar form $r=2acos\theta$ and $z=\frac{x^2+y^2}{2a}$ to $z = \frac{r^2}{2a}$ thus got the double integral as:
$$\int_{-\pi/2}^{\pi/2} \int_{0}^{2acos\theta}\frac{r^2}{2a}\cdot r\cdot dr\cdot d\theta$$ which I calculated to be $\frac{3\pi a^3}{4}$.
The problem is:The answer given in the book is $3\pi a^3$. I can't figure out where I did wrong.