Volume of solid within sphere and cone using polar coordinates

[amcavoy]

I posted this in the Calculus section also, so I apologize for double-posting.

"Find the volume of the solid that lies within the sphere $$x^2+y^2+z^2=4$$, above the x-y plane, and below the cone $$z=\sqrt{x^2+y^2}$$."

I set this up in polar coordinates as follows:

$$V=\int_{0}^{2\pi}\int_{0}^{\sqrt{2}}r^2drd\theta+\int_{0}^{2\pi}\int_{\sqrt{2}}^{2}r\sqrt{4-r^2}drd\theta$$

and then solved it coming up with $$V=\frac{8\pi\sqrt{2}}{3}$$.

Is this the correct way to set this up? Also, is there a way I could do this with one double integral rather than adding two of them together?

Thanks for your help.

 

[ehild]

I posted this in the Calculus section also, so I apologize for double-posting.

"Find the volume of the solid that lies within the sphere $$x^2+y^2+z^2=4$$, above the x-y plane, and below the cone $$z=\sqrt{x^2+y^2}$$."

I set this up in polar coordinates as follows:

$$V=\int_{0}^{2\pi}\int_{0}^{\sqrt{2}}r^2drd\theta+\int_{0}^{2\pi}\int_{\sqrt{2}}^{2}r\sqrt{4-r^2}drd\theta$$

and then solved it coming up with $$V=\frac{8\pi\sqrt{2}}{3}$$.

Is this the correct way to set this up? Also, is there a way I could do this with one double integral rather than adding two of them together?

Thanks for your help.

It is correct, and I think that was the simplest way to solve.

ehild