(856,#15) Use polar coordinates to find the volume of the solid under the paraboloid z = x² + y² above the disk
x² + y² = 9.

The Geometry:

f := (x,y) -> x^2 + y^2:
Graph_rdrdt(f,0,2*Pi,0, 3, 7,4);

The value of the double integral is (81 Pi)/2 :

Integrate_rdrdt(f,0,2*Pi,0,3);

echo:f(r*cos(t), r*sin(t))*r = (r^2*cos(t)^2+r^2*sin(t)^2)*r c(t) = 0 d(t) = 3

int(f(r*cos(t), r*sin(t))*r, r ))= 1/4*(cos(t)^2+sin(t)^2)*r^4

int(f(r*cos(t), r*sin(t))*r, r = c(t)..d(t))= 81/4*cos(t)^2+81/4*sin(t)^2

int(int(f(r*cos(t), r*sin(t))*r, r = c(t)..d(t)),t)= 81/4*t

int(int(f(r*cos(t), r*sin(t))*r, r = c(t) ..d(t)), t = a..b) = 81/2*Pi

by:sh