# Limits of integration, double integral

$$\int_Y y\sqrt z\sqrt{4x^2+4y^2+1}dS$$

$Y$ is given by $z=x^2 + y^2$ and $x\leq0$, $y\leq0$, $1\leq x^2+y^2 \leq 9$

I'm having a bit of trouble with this one, not sure if I'm getting the limits right. I was thinking I should use cylindrical coordinates and that the integral should be $0$ to $\pi/2$ over $\theta$ and $1$ to $3$ over $r$. But I keep getting the answer wrong, so I would appreciate any help to see if it's the limits or simply my calculations that's off.

• Since x and y are both negative, the angle should range over the 3rd quadrant of the plane, not the 1st as you have done – David H May 21 '14 at 9:09
• Of course! How did I not think about the quadrants! Thank you David, you just saved me a lot of self doubt. – Robin May 21 '14 at 9:18
• Assuming you did everything else correctly, integrating over the 3rd quadrant instead of the first should only change your answer by a minus sign. – David H May 21 '14 at 9:22
• I had managed to drop an r somewhere along the way from the dS, but once I found that and used the limits of the 3rd quadrant instead of the 1st everything seemed to be in order. – Robin May 21 '14 at 9:49

Without your calculations, one cannot say if they are correct. My own calculations are shown in attachment so that you could compare. 