Integrate function $f(x,y) = x^2+y^2$ in circle $(x-a)^2+y^2 < a^2$

Question

I am asked to integrate function $f(x,y) = x^2+y^2$ in circle $(x-a)^2+y^2 < a^2$

My answer is $\frac{3 \pi a^4}{2}$ but for some reason the answer of the textbook is $\frac{32 \pi a^4}{2}$.

Does the textbook's answer contain a typo?

My integral set-up is

$$ \int_{- \pi/2}^{\pi/2} \int_{0}^{2a cos(\theta)} r^3 dr d\theta $$

Answer 1

Use change of variables with $z = x-a$

Then the function will be $f = (z+a)^2 + y^2$ and the domain will be $z^2 + y^2 < a^2$

Again $f = z^2 + y^2 + 2az + a^2$, now apply polar coordinates $z = rcos(\theta)$ and $y = rsin(\theta)$

$f(r,\theta) = r^2 + 2a\cos(\theta) + a^2$

The rest is trivial resulting $\frac{3.\pi . a^4}{2}$

Answer 2

I also get $$\int_{-\pi/2}^{\pi/2}\int_0^{2a\cos{\theta}}~r^3drd\theta=\color{red}{\frac{3\pi a^4}{2}}$$