(879,#31) Evaluate

,

where E lies above the paraboloid z = x² + y² and below the plane z = 2 y.Use either the Table of Integrals or a computer algebra system to evaluate the integral.

Solution: First, the intersection of the parabola and the plane lies on the surface x² + (y-1)² = 1, which is obtained by equating x² + y² = 2y and completing the square.Second, the polar form of x² + (y-1)² = 1 is r = 2 sin(theta). Thus, the limits of integration for the volume are z = r² to z = 2 r sin(theta), r = 0 to r = 2*sin(theta), and theta = 0 to Pi.

That is, the integral is given by

Its value is 5Pi/6:

lprint( int(r*z, z) );
lprint( int(r*z, z = r^2 .. 2*r*sin(t)) );
1/2*r*z^2
2*r^3*sin(t)^2-1/2*r^5

lprint( int(int(r*z, z = r^2 .. 2*r*sin(t)), r) );
lprint( int(int(r*z, z = r^2 .. 2*r*sin(t)), r = 0 ..2*sin(t)) );
1/2*r^4*sin(t)^2-1/12*r^6
8/3*sin(t)^6
lprint( int(int(int(r*z, z = r^2 .. 2*r*sin(t)), r = 0 ..2*sin(t)),t) );
lprint( int(int(int(r*z, z = r^2 .. 2*r*sin(t)), r = 0 ..2*sin(t)),t = 0..Pi) );
-4/9*sin(t)^5*cos(t)-5/9*sin(t)^3*cos(t)-5/6*cos(t)*sin(t)+5/6*t
5/6*Pi

by:sh