I have a volume $V$ bounded by the following equations:
$ x^2 + y^2 + z^2 = 1 $
$ z^2 = (x^2 +y^2) {\sqrt2} $
and I have to find out the mass of the volume given that:
$\rho(x,y,z) = z$
where $\rho$ is the density
I tried it but cannot do it. Can anyone help?
On
Assuming the upper and lower mass is the same, we can find the mass in the first quadrant with x, y and z greater than 0. In this case it is easier to use spherical coordinate : $$ \begin{split} x &= r\cos\theta \cos\phi\\ y &= r\sin\theta \cos\phi\\ z &= r\sin\phi \end{split} $$ Hence the equations become $$r = 1$$ $$\tan^2\phi = \sqrt{2} \implies \tan\phi = 2^{\frac{1}{4}}$$ $$\cos2\phi = 2\cos^2\phi - 1 = \frac{1 - \sqrt{2}}{1 + \sqrt{2}}$$
This is the angle where the cone intersects the sphere.
Hence the mass is $$ \begin{split} \int\int\int\text{density}\cdot r^2\cos\phi dV&= 8\int_0^{\frac{\pi}{2}}\int_{\tan^{-1}2^{\frac{1}{4}}}^{\frac{\pi}{2}}\int_0^1r\sin\phi r^2\cos\phi drd\theta d\phi\\ & = \int_0^{\frac{\pi}{2}}\int_{\tan^{-1}2^{\frac{1}{4}}}^{\frac{\pi}{2}}\sin2\phi d\phi d\theta\\ & = \int_0^\frac{\pi}{2}\bigg(\frac{1}{2} + \frac{1}{2}\frac{1 - \sqrt{2}}{1 + \sqrt{2}}\bigg)d\theta \\ & = \frac{\pi}{2}\big(\sqrt{2} - 1\big) \end{split} $$
To find the mass you should integrate $\rho (x,y,z)$ around the volume described by your equation.
$$m=\int_V \rho (x,y,z)\ dV$$
I think that there is something wrong with your question.
Your first case is a sphere of radius $1$ centered at origin.
For each $\rho\ dV=(z)\ dV$ at any point on positive $z$-octants belonging to the sphere, there is a corresponding $\rho\ dV=(-z)\ dV$ on negative $z$-octants belonging to the sphere. Therefore they cancel out so that the net integral (i.e. mass) is zero.
This doesn't make sense. This is because in your question it has been said that $\rho (x,y,z)=z$ and thus at negative $z$-octants, the mass density is negative. So in your question, I think you should replace the equation $\rho (x,y,z)=z$ to $\rho (x,y,z) = \left|z\right|$.