I have the following problem to solve:
$\large\mathbf{4.18}\;$ Given that $g(x)\geq 0$ has the property that $$\int g(x)\,dx=1,$$ show that $$f(x,y)=\frac{2g\big(\sqrt{x^2+y^2}\big)}{\pi\sqrt{x^2+y^2}},\tag*{$x,y>0$}$$ is a pdf.
After I switch to polar coordinates I get the following expression
$$ \int_0^{\infty}\int_0^{2\pi} \frac{2g(r)r}{\pi r}\,d\theta\,dr $$ which obviously doesn't integrate to 1. If the limit of the inner integral is $\pi/2$ then the last expression integrates to $1$ like it's in the Solution Manual.
My question is, why do we integrate to $\pi/2$ and not $2\pi$ when we make a transformation to polar coordinates.
Thanks in advance!