Find the volume under the cone z = sqrt ( x(adsbygoogle = window.adsbygoogle || []).push({}); ^{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

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# Volume using double integral and polar coordinates

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