# Thread: another triple integral problem..

1. ## another triple integral problem..

Calculate the volume of the solid that lies within $x^2+y^2+z^2=49$ above the xy=plane and below $z=\sqrt{\frac{1}{3}x^2+\frac{1}{3}y^2}$

Having trouble visualizing what that cone would look like.

2. It should look something like this:

I would convert to polar co-ordinates.

EDIT: Scratch that.

3. Originally Posted by ANDS!
It should look something like this:

I would convert to polar co-ordinates.

EDIT: Scratch that.
Thanks though. Yeah I was considering spherical, my trouble is obtaining the angle phi for the cone.

4. It resembles a snowball.

$z^2 = \frac{1}{3}(x^2 + y^2)$

$\rho^2\cos^2{\phi} = \frac{1}{3}(\rho^2\sin^2{\phi}\cos^2{\theta} + \rho^2\sin^2{\phi}\sin^2{\theta})$

$\rho^2\cos^2{\phi} = \frac{1}{3}\rho^2\sin^2{\phi}$

$3= \tan^2{\phi}$

you can finish up from here to find phi?

5. Originally Posted by 11rdc11
It resembles a snowball.

$z^2 = \frac{1}{3}(x^2 + y^2)$

$\rho^2\cos^2{\phi} = \frac{1}{3}(\rho^2\sin^2{\phi}\cos^2{\theta} + \rho^2\sin^2{\phi}\sin^2{\theta})$

$\rho^2\cos^2{\phi} = \frac{1}{3}\rho^2\sin^2{\phi}$

$3= \tan^2{\phi}$

you can finish up from here to find phi?
Sorry I am not sure how to solve that last line.
With phi solved would the volume then be described by this:
$\int_?^{\frac{\pi}{2}}\int_0^{2\pi}\int_0^7p^2sin{ \phi} dpd{\theta}d{\phi}$

6. Originally Posted by Em Yeu Anh
Sorry I am not sure how to solve that last line.
With phi solved would the volume then be described by this:
$\int_?^{\frac{\pi}{2}}\int_0^{2\pi}\int_0^7p^2sin{ \phi} dpd{\theta}d{\phi}$
$3 = \tan^2{\phi}$

$\sqrt{3} = \tan{\phi}$

$\arctan{\sqrt{3}} = \phi$

$\phi = \frac{\pi}{3}$

$\int_{0}^{\frac{\pi}{3}}\int_0^{2\pi}\int_0^7p^2si n{\phi} dpd{\theta}d{\phi}$

7. Originally Posted by Em Yeu Anh
Calculate the volume of the solid that lies within $x^2+y^2+z^2=49$ above the xy=plane and below $z=\sqrt{\frac{1}{3}x^2+\frac{1}{3}y^2}$

Having trouble visualizing what that cone would look like.
$z= \sqrt{\frac{1}{3}x^2+ \frac{1}{3}y^2}$, restricted to the xz-plane so that y= 0, is $z= \pm\sqrt{\frac{1}{3}}x$, two lines with slopes $\pm\sqrt{\frac{1}{3}}$.

In spherical coordinates, then, $\phi= cot^{-1}\left(\sqrt{\frac{1}{3}}\right)= \frac{\pi}{3}$.

Blast! 11rdc11 beat me again!