Spherical coordinates find volume

[fk378]

Homework Statement

Use spherical coordinates to find the volume of the solid that lies above the cone z= sqrt(x^2 + y^2) and below the sphere x^2 + y^2 + z^2 = z.

The Attempt at a Solution

I'm having trouble solving for rho (p). I know it starts from 0, and it reaches to the sphere and to the cone. But I don't know where to go from there.

 

[Knissp]

Assuming you meant \rho_o, the question becomes one of finding \phi.
In this case, you have to look at your cone. Essentially, it says z = r, where r is in cylindrical coordinates. Now use the conversions to spherical, z=\rho cos(\phi) and r = \rho sin (\phi). Equating those gives you cos(\phi) = sin(\phi), or \phi = \pi/4. So 0<\phi<\pi/4.

 

[fk378]

It is a sphere, though. If you move the z over to the left side of the equation, complete the square, you get a sphere with center (0,0,1/2). Also, I need to find rho, not phi.

 

[Knissp]

Sorry.
x^2 + y^2 +z^2 = \rho^2

z = \rho cos(\phi)

x^2 + y^2 +z^2 = z implies \rho^2 = \rho cos(\phi)

or \rho = cos(\phi)

so 0 < \rho < cos(\phi)

 

[fk378]

Why are we just looking at the sphere though and not the cone boundary?

 

[Knissp]

Because you said "below the sphere x^2 + y^2 + z^2 = z" which implies the sphere is the upper bound, so its value for \rho is the maximum value that \rho can take on, thus establishing the upper limit of integration. Hope that made sense :)

 

[fk378]

That makes sense...but we are also looking at the cone too. And the value of p for the cone doesn't depend on theta...it is constant. How do we know if we need to divide the triple integral expression into 2 separate ones, one for each different p?

 

[Knissp]

Okay, this took me a while to find, but here's an example of a time when you need to split the expression for different values of \rho.

There are several evident differences between your problem and this one. Most importantly, notice that your solid is "above the cone z= sqrt(x^2 + y^2)" and the example I show is enclosed by the planes. There is a uniquely different value of \rho that the function takes on for different values of \phi namely because the solid that is being integrated must satisfy either being inside a cone or inside a sphere.

Essentially, in the example I give, there are uniquely different values of \rho for different values of \phi, whereas in your case, no such thing occurs. If you try to find a different \rho, you get what I did in an earlier post:

Essentially, it says z = r, where r is in cylindrical coordinates. Now use the conversions to spherical, z=\rho cos(\phi) and r = \rho sin (\phi). Equating those gives you cos(\phi) = sin(\phi), or \phi = \pi/4. So 0<\phi<\pi/4.

Notice that the \rho cancels, thus showing that there is no other unique value to make different bounds of integration.



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