Converting Triple Integrals to Spherical Coordinates.

Question

X goes from 0 to 2

Y goes from 0 to sqrt(4-x^2)

Z goes from 0 to sqrt(4-x^2-y^2).

 

The function is zsqrt(4-x^2-y^2)dzdydx.

 

I know the conversion from rectangular to spherical and the jacobian. I don't know how to convert this specific rectangular function into spherical. I also don't know how to calculate the limits of integration in spherical for this question. Please help and thanks!

Answer 1

Consider the change of variables: x=rcosθsinφ, y=rsinθsinφ, z=rcosφ, with 0<φ<π/2, 0<θ<π/2, and 0<r<2. The range of the angles are determined by the limits of integration, which is a triple integral over a spherical region in the first octant (x>0,y>0,z>0, x²+y²+z²<2).

The Jacobian of the change of variables is known as r²sinφ.

The original integral becomes

∫_{0}^{π/2}∫_{0}^{π/2}∫_{0}^{π/2}rcosφ√(4–r²sin^·2φ)·r²sinφdφdθdr

=∫_{0}^{π/2}∫_{0}^{π/2}∫_{0}^{π/2}r³cosφsinφ(4–r²sin^·2φ)dφdθdr

Here we used LaTeX commands to write the integrals.