Double integral to find the volume of a solid bound by a sphere and cone

**flawless** · May 17th 2008, 11:27 PM

Find the volume of the solid bounded by the upper hemisphere
x^2+y^2+z^2=8 z>0

and the cone z=squareroot(x^2 + y^2).
What is the double double integral i need to find the volume, what would the limits be? It didnt seem that hard until i actually started doing it. thanks

**flyingsquirrel** · May 18th 2008, 03:48 AM

Hi

Without using a double integral, it is possible to find the volume by splitting it into two volumes which can be easily computed :

The intersection of the sphere and of the cone is given by $\left\{ \begin{array}{l} x^2+y^2+z^2=8 \\ z=\sqrt{x^2+y^2}\end{array}\right.\implies z=2$

If $z\in[0,\,2]$ , $z=\sqrt{x^2+y^2}\in [0,\,2]$ hence $x^2+y^2+z^2\leq 8$ : the part of the cone such that $z\in[0,\,2]$ is inside the sphere.

If $z>2$ , $x^2+y^2=8-z^2<4$ hence $\sqrt{x^2+y^2}<2$ : the part of the sphere such that $z>2$ is inside the cone.

Thus the volume we're looking for is the sum of the volume bounded by the cone for $z\in[0,\,2]$ (red surface, see attached images) and of the volume bounded by the sphere for $z>2$ . (green surface) Can you take it from here ?

**galactus** · May 18th 2008, 04:45 AM

I can make things worse and show you a triple integral in spherical coordinates.

In the spherical, the equation of the sphere is ${\rho}=2\sqrt{2}$

And the cone is ${\rho}cos{\phi}=\sqrt{{\rho}^{2}sin^{2}{\phi}cos^{ 2}{\theta}+{\rho}^{2}sin^{2}{\phi}sin^{2}{\theta}}$

I know this looks menacing, but it whittles down to

${\rho}cos{\phi}={\rho}sin{\phi}$

If we divide both sides by ${\rho}cos{\phi}$ , then we get

$tan{\phi}=1, \;\ {\phi}=\frac{\pi}{4}$

So, we have the triple integral:

$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{0}^{2 \sqrt{2}}{\rho}^{2}sin{\phi}d{\rho}d{\phi}d{\theta }$

It's not that bad to do. Don't ya' wish they covered quadruple integrals in clac class

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