# Double integral in polars

I have to calculate the following integral in polar coordinates:

$$\int_D y\sqrt{x^2+y^2}dxdy$$

where $D=\{(x,y)\in\mathbb{R}^2:a^2-x^2 \geq y \geq x^2-a^2\}$, with $a>0$.

Well, I know that $D$ is the surface bounded by two parabolas. As the vertex of those it isn't coincident with the origin, I thought that I can do a traslation like (I only write one parabola):

$$\left\{ \begin{array}{ll} Y=y-a^2 \\ X=x \end{array} \right.$$

with jacobian $J=1$. Can I do that? What integral would I obtain? Is there an easier way?

• Is the integral $0$ by inspection? Commented Apr 6, 2016 at 23:24
• I don't know. Why do you think so? Commented Apr 6, 2016 at 23:26
• The region is symmetric about the $x$-axis, and the integrand is odd w.r.t. $y$. Commented Apr 6, 2016 at 23:26
• You're right. Thank you. Commented Apr 6, 2016 at 23:34

Observe that $$\int_D f\,dA = \int_{-a}^a \int_{x^2 - a^2}^{a^2 - x^2} f\,dy\,dx$$ where $f = y\sqrt{x^2 + y^2}$. Since $f$ is odd w.r.t. $y$, the inner integral is $0$.