Find the volume of the given solid: above the cone
z = ((x^2) + (y^2))^0.5 and below the sphere (x^2) + (y^2) + (z^2)=1.
Thanks for any help.
As a triple integral, you can use cylindrical coordinates.
Note that $\displaystyle r\leqslant z\leqslant \sqrt{1-r^2}$ and $\displaystyle 0\leqslant\theta\leqslant2\pi$
Now to find r, find where the surfaces intersect.
This would be where $\displaystyle z=\sqrt{r^2}\implies r^2+(\sqrt{r})^2=1\implies r^2=\tfrac{1}{2}\implies r=\frac{\sqrt{2}}{2}$. Thus, $\displaystyle 0\leqslant r\leqslant \frac{\sqrt{2}}{2}$
Therefore, your triple integral would be $\displaystyle \int_0^{2\pi}\int_0^{\frac{\sqrt{2}}{2}}\int_r^{\s qrt{1-r^2}}r\,dz\,dr\,d\theta$
Does this make sense?
Once again we use cylindrical coordinates:
For the cone we have, in cylindrical coordinates: r = z,
For the sphere we have: r = $\displaystyle \sqrt[]{1-z^2}$
Note that the cone is inverted, so the volume makes an ice-cream cone shape. Where exactly does the cone run into the sphere?
Thats when z = $\displaystyle \sqrt[]{1-z^2}$, or z = $\displaystyle \pm \, \sqrt[]{\frac{1}{2}}$. This is not quite true since, as it is defined, the cone doesn't extend into the negative z range. So we stick with positive $\displaystyle \sqrt[]{\frac{1}{2}}$.
This time we break it up into two integrals. We have to change over from the cone to the sphere when we reach z = $\displaystyle \sqrt[]{\frac{1}{2}}$
Volume = $\displaystyle \int_{0 }^{ \sqrt[ ]{ \frac{ 1}{2 } } } \int_{0 }^{2\pi } \int_{ 0}^{ z} rdrd \theta dz + \int_{ \sqrt[ ]{ \frac{1 }{2 } } }^{ 1} \int_{0 }^{2 \pi } \int_{0 }^{ \sqrt[ ]{ 1-z^2} } rdrd \theta dz$
Note how in the first integral we integrate r from 0 to z, corresponding to the cone. In the second we integrate corresponding to the sphere.
I got $\displaystyle -\frac{1}{3} \left(-2+\sqrt{2}\right) \pi \approx .613$ for the volume.
I would recommend spherical coordinates because I see much more spherical symmetry than cylindrical symmetry!
The sphere $\displaystyle x^2+ y^2+ z^2= 1$ in spherical coordinates is just $\displaystyle \rho= 1$.
The cone $\displaystyle z= \sqrt{x^2+ y^2}$ is $\displaystyle \rho cos(\phi)= \rho sin(\phi)$ or $\displaystyle tan(\phi)= 1$ so $\displaystyle \phi= \pi/4$.
The volume is given by
$\displaystyle \int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/4}\int_{\rho= 0}^1 \rho^2 sin(\phi)d\rho d\phi d\theta$$\displaystyle = \left(\int_0^{2\pi} d\theta\right)\left(\int_0^{\pi/4} sin(\phi)d\phi\right)\left(\int_0^1 \rho^2 d\rho\right)$