Volume using double integral and polar coordinates

fishingspree2 · Apr 26, 2010

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

[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

radou · Apr 26, 2010

Your integration boundaries for r are not correct.

cronxeh · Apr 26, 2010

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

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

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

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Volume using double integral and polar coordinates

fishingspree2

radou

cronxeh