trippel integer of sqrt(x^2+y^2+z^2)

[green_tea]

Hi!
So I'm stuck on a problem, I have to calculate a trippel integer of f(x,y,z)=sqrt(x^2+y^2+z^2) over a volume D that is described by x^2+y^2+z^2<=1 , z>=sqrt(x^2+y^2)

I thought that I could do int(f) dz first from z=sqrt(x^2+y^2) to z=sqrt(1-x^2-y^2) and then do a double integer of what i get for x and y. But I'm stuck on the first one! How do I integrate sqrt(x^2+y^2+z^2) for z? I tried it in an online integrator, and I got something kind of complicated, and I'm worried that the doubble integer over x and y will be even harder to do...

So maybe there is another, easier way to solve this problem? I also thought of using space polar coordinates(x=rcos(v)sin(u), y=rsin(v)sin(u), z=rcos(u)) , but I don't really know how to describe the volume D in those coordinates.

Can someone please help me with this?

 

[galactus]

Let's try this in polar and spherical.

\(\displaystyle f(x,y,z)=\sqrt{x^{2}+y^{2}+z^{2}\). We want the region bounded inside the sphere \(\displaystyle x^{2}+y^{2}+z^{2}=1\) and above the cone \(\displaystyle z=\sqrt{x^{2}+y^{2}}\)

POLAR:

Note, that \(\displaystyle z^{2}=x^{2}+y^{2}\).

Sub into f(x,y,z) and get \(\displaystyle \sqrt{2z^{2}}=\sqrt{2}\cdot z\)

Knowing that \(\displaystyle z^{2}=r^{2}=x^{2}+y^{2}\), we have

\(\displaystyle \sqrt{2}\int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}}\int_{r}^{\sqrt{1-r^{2}}}rz \;\ dz \;\ dr \;\ d{\theta}\)


SPHERICAL:

Same as above, except note that \(\displaystyle z={\rho}cos(\phi)\)

Since the sphere has radius 1, then \(\displaystyle \rho = 1\) and \(\displaystyle z=\frac{1}{\sqrt{2}}\)

\(\displaystyle {\phi}=cos^{-1}(\frac{1}{\sqrt{2}})=\frac{\pi}{4}\)

Thus, we have:

\(\displaystyle \sqrt{2}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{0}^{1}{\rho}^{2}sin(\phi)\cdot {\rho}cos(\phi) \;\ d{\rho}d{\phi}d{\theta}\)

RECTANGULAR:

Try picturing the cone intersecting the sphere and projecting that region onto the xy-plane. It will be a circle of radius \(\displaystyle \frac{1}{\sqrt{2}}\)

\(\displaystyle \sqrt{2}\int_{\frac{-1}{\sqrt{2}}}}^{\frac{1}{\sqrt{2}}}\int_{-\sqrt{\frac{1}{2}-x^{2}}}^{\sqrt{\frac{1}{2}-x^{2}}}\int_{\sqrt{x^{2}+y^{2}}}^{\sqrt{1-x^{2}-y^{2}}}z \;\ dzdydx\)

This mess is why it is sometimes better to use polar or spherical.

 

[green_tea]

Thanks a lot!