Calculate the volume of the solid that lies within $\displaystyle x^2+y^2+z^2=49$ above the xy=plane and below $\displaystyle z=\sqrt{\frac{1}{3}x^2+\frac{1}{3}y^2}$
Having trouble visualizing what that cone would look like.
Calculate the volume of the solid that lies within $\displaystyle x^2+y^2+z^2=49$ above the xy=plane and below $\displaystyle z=\sqrt{\frac{1}{3}x^2+\frac{1}{3}y^2}$
Having trouble visualizing what that cone would look like.
It resembles a snowball.
$\displaystyle z^2 = \frac{1}{3}(x^2 + y^2)$
$\displaystyle \rho^2\cos^2{\phi} = \frac{1}{3}(\rho^2\sin^2{\phi}\cos^2{\theta} + \rho^2\sin^2{\phi}\sin^2{\theta})$
$\displaystyle \rho^2\cos^2{\phi} = \frac{1}{3}\rho^2\sin^2{\phi}$
$\displaystyle 3= \tan^2{\phi}$
you can finish up from here to find phi?
$\displaystyle z= \sqrt{\frac{1}{3}x^2+ \frac{1}{3}y^2}$, restricted to the xz-plane so that y= 0, is $\displaystyle z= \pm\sqrt{\frac{1}{3}}x$, two lines with slopes $\displaystyle \pm\sqrt{\frac{1}{3}}$.
In spherical coordinates, then, $\displaystyle \phi= cot^{-1}\left(\sqrt{\frac{1}{3}}\right)= \frac{\pi}{3}$.
Blast! 11rdc11 beat me again!