Volume Problem (Can someone check my work?)

[amcavoy]

"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.

 

[HallsofIvy]

I think that will work but the way I would analyze it is this: On the sphere, z= \sqrt{4- x^2-y^2}= \sqrt{4- r^2}. On the cone, z= \sqrt{x^2+ y^2}= r. So the "height" at each point is \sqrt{4- r^2}- r. You want to integrate that, times dA= rdrd\theta, over the disk inside the intersection of the two figures projected down to the xy-plane.
The sphere and cone intersect when \sqrt{4- r^2}= r or 4- r^2= r^2[/tex] so r²= 2 or r= \sqrt{2} just as you have.<br /> <br /> The volume is \int_{\theta= 0}^{2\pi}\int_{r=0}^{\sqrt{2}}\left(\sqrt{4- r^2}- r\right)rdrd\theta.<br /> <br /> <br /> Personally, I would put it in spherical coordinates and do it as a single triple integral:<br /> \theta goes from 0 to 2\pi, \phi goes from 0 (vertical) to \frac{\pi}{4} (the cone) and \rho goes from 0 to 2 (the sphere). Integrate dV= \rho sin(\phi)d\theta d\phi d\rho over that.

 

[amcavoy]

Ok that makes sense. Thanks for your help HallsofIvy

 

[TD]

Ivy, I computed your approach as well and I don't find the same as your first solution, and they're both different from apmcavoy's answer.

I get:

\int_{\theta= 0}^{2\pi}\int_{r=0}^{\sqrt{2}}\left(\sqrt{4- r^2}- r\right)rdrd\theta = \frac{-8\,\left( -2 + {\sqrt{2}} \right) \,\pi }{3}

While:

\int\limits_0^{2} {\int\limits_0^{{\raise0.5ex\hbox{$\scriptstyle \pi $}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 4$}}} {\int\limits_0^{2 \pi} {\rho \sin \left( \phi \right)} } } d\theta d\phi d\rho = -2\,\left( -2 + {\sqrt{2}} \right) \,\pi

Any idea what's wrong?

 

[amcavoy]

Ivy, I computed your approach as well and I don't find the same as your first solution, and they're both different from apmcavoy's answer.

I get:

\int_{\theta= 0}^{2\pi}\int_{r=0}^{\sqrt{2}}\left(\sqrt{4- r^2}- r\right)rdrd\theta = \frac{-8\,\left( -2 + {\sqrt{2}} \right) \,\pi }{3}

While:

\int\limits_0^{2} {\int\limits_0^{{\raise0.5ex\hbox{$\scriptstyle \pi $}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 4$}}} {\int\limits_0^{2 \pi} {\rho \sin \left( \phi \right)} } } d\theta d\phi d\rho = -2\,\left( -2 + {\sqrt{2}} \right) \,\pi

Any idea what's wrong?

I may be wrong, but doesn't \int_{\theta= 0}^{2\pi}\int_{r=0}^{\sqrt{2}}\left(\sqrt{4- r^2}- r\right)rdrd\theta represent the volume above the cone and within the sphere?

 

[shooter]

Ivy, I computed your approach as well and I don't find the same as your first solution, and they're both different from apmcavoy's answer.
I get:
\int_{\theta= 0}^{2\pi}\int_{r=0}^{\sqrt{2}}\left(\sqrt{4- r^2}- r\right)rdrd\theta = \frac{-8\,\left( -2 + {\sqrt{2}} \right) \,\pi }{3}
While:
\int\limits_0^{2} {\int\limits_0^{{\raise0.5ex\hbox{$\scriptstyle \pi $}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 4$}}} {\int\limits_0^{2 \pi} {\rho \sin \left( \phi \right)} } } d\theta d\phi d\rho = -2\,\left( -2 + {\sqrt{2}} \right) \,\pi
Any idea what's wrong?

try changing from psin to p^2sin. :)