Volume using double integral and polar coordinates

[fishingspree2]

Find the volume under the cone z = sqrt ( x²+y² ) and on the disk x²+y² < 4. Use polar coordinates.

Graphing x²+y² < 4, I get a circle centered at 0,0 with radius of 2
So theta goes from 0 to 2pi

Also, since x²+y² < 4
This means that r^2 < 4
so -2 < r < 2

$$\int_{0}^{2\pi}\int_{-2}^{2}r\sqrt{x^{2}+y^{2}} dr d\theta=\int_{0}^{2\pi}\int_{-2}^{2}r^{2} dr d\theta<br /> = \frac{32\pi}{3}$$
but the answer is 16pi/3

Can anyone help me

 

[radou]

Your integration boundaries for r are not correct.

 

[cronxeh]

radius can't be negative, its length is still 2
$$\int_{0}^{2\pi}\int_{0}^{2}r^{2} dr d\theta$$

(although if you negate the radius, you increase theta by pi):
$$\int_{pi}^{2\pi}\int_{-2}^{2}r^{2} dr d\theta$$

(probably just easier to think of radius as a non-negative variable)