Thread: 3 Dimensional Volume Integral

Find:

triple integral ( ?-?; ?-?;?-R) x^2 + y^2 + z^2 dV

where R is the region that lies above the cone z =c sqrt (x^2 + y^2) and inside the sphere x^2 + y^2 + z^2 = a^2


no clue really except it needs to be set up in polar form

Originally Posted by hammer

Find:

triple integral ( ?-?; ?-?;?-R) x^2 + y^2 + z^2 dV

where R is the region that lies above the cone z =c sqrt (x^2 + y^2) and inside the sphere x^2 + y^2 + z^2 = a^2


no clue really except it needs to be set up in polar form

Not polar form. It needs to be spherical form.

Let us assume a>0 and c>0.

Then the we need to consider the following:
1)How much does theta move?
2)What is the radius?
3)What much does phi move?

Answers:
1)Theta is from 0 to 2pi because it is a complete revolution.
2)The radius is a, because x^2+y^2+z^2=a.
3)We have a cone, the equation of the cone is simply, phi=constant.
But we need to find what that constant is.
We know the equation of cone in rectangular coordinates is,
z=c*sqrt(x^2+y^2)
z=c*sqrt(r^2)
z=c*r
Convert to Spherical,
rho*sin(phi) = c*rho*cos(phi)
sin(phi) = c*cos(phi)
tan(phi)=c
phi=arctan(c).

Look at picture below.

I will do the integral in the next post.

Hier.