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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
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