- #1
amcavoy
- 665
- 0
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.
"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.