- #1

- 139

- 0

^{2}+y

^{2}) and on the disk x

^{2}+y

^{2}< 4. Use polar coordinates.

Graphing x

^{2}+y

^{2}< 4, I get a circle centered at 0,0 with radius of 2

So theta goes from 0 to 2pi

Also, since x

^{2}+y

^{2}< 4

This means that r^2 < 4

so -2 < r < 2

[tex]\int_{0}^{2\pi}\int_{-2}^{2}r\sqrt{x^{2}+y^{2}} dr d\theta=\int_{0}^{2\pi}\int_{-2}^{2}r^{2} dr d\theta

= \frac{32\pi}{3}[/tex]

but the answer is 16pi/3

Can anyone help me