Volume Question (Can someone check my work?)

  • Thread starter amcavoy
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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 [tex]x^2+y^2+z^2=4[/tex], above the x-y plane, and below the cone [tex]z=\sqrt{x^2+y^2}[/tex]."

I set this up in polar coordinates as follows:

[tex]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[/tex]

and then solved it coming up with [tex]V=\frac{8\pi\sqrt{2}}{3}[/tex].

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.
 

Answers and Replies

  • #2
ehild
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apmcavoy said:
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 [tex]x^2+y^2+z^2=4[/tex], above the x-y plane, and below the cone [tex]z=\sqrt{x^2+y^2}[/tex]."

I set this up in polar coordinates as follows:

[tex]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[/tex]

and then solved it coming up with [tex]V=\frac{8\pi\sqrt{2}}{3}[/tex].

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
 

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