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

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- Apr 14th 2007, 06:09 PMhammer3 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 - Apr 14th 2007, 07:20 PMThePerfectHacker
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. - Apr 14th 2007, 07:34 PMThePerfectHacker
Hier.