Volume Problem (Can someone check my work?)

"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?

Last edited:

HallsofIvy
Homework Helper
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 $4- r^2= r^2[/tex] so r2= 2 or [itex]r= \sqrt{2}$ just as you have.

The volume is $$\int_{\theta= 0}^{2\pi}\int_{r=0}^{\sqrt{2}}\left(\sqrt{4- r^2}- r\right)rdrd\theta[/itex]. Personally, I would put it in spherical coordinates and do it as a single triple integral: $\theta$ goes from 0 to $2\pi$, $\phi$ goes from 0 (vertical) to $\frac{\pi}{4}$ (the cone) and $\rho$ goes from 0 to 2 (the sphere). Integrate $dV= \rho sin(\phi)d\theta d\phi d\rho$ over that. Last edited by a moderator: Ok that makes sense. Thanks for your help HallsofIvy TD Homework Helper Ivy, I computed your approach as well and I don't find the same as your first solution, and they're both different from apmcavoy's answer. I get: [tex]\int_{\theta= 0}^{2\pi}\int_{r=0}^{\sqrt{2}}\left(\sqrt{4- r^2}- r\right)rdrd\theta = \frac{-8\,\left( -2 + {\sqrt{2}} \right) \,\pi }{3}$$

While:

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

Any idea what's wrong?

TD said:
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:

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

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?

TD said:
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:
$$\int\limits_0^{2} {\int\limits_0^{{\raise0.5ex\hbox{\scriptstyle \pi } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 4}}} {\int\limits_0^{2 \pi} {\rho \sin \left( \phi \right)} } } d\theta d\phi d\rho = -2\,\left( -2 + {\sqrt{2}} \right) \,\pi$$
Any idea what's wrong?

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