# Volume using double integral and polar coordinates

1. Apr 26, 2010

### 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

$$\int_{0}^{2\pi}\int_{-2}^{2}r\sqrt{x^{2}+y^{2}} dr d\theta=\int_{0}^{2\pi}\int_{-2}^{2}r^{2} dr d\theta = \frac{32\pi}{3}$$

Can anyone help me

2. Apr 26, 2010

Your integration boundaries for r are not correct.

3. Apr 26, 2010

### 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)

Last edited: Apr 26, 2010