(856,#1) Evaluate

where R is the disk with center the origin and radius 5 by changing to polar coordinates.

The Geometry:

f := (x,y) -> x:
Graph_rdrdt(f,0,2*Pi,0, 5, 11,6);

Note: Since the region of integration is symmetric with respect to the x-axis and since the inegrand, x, takes values of opposite sign on either side of the y-axis, the value of the kintegral will be 0.

This can also be verified by direct computation:

Integrate_rdrdt(f,0,2*Pi,0,5);

echo: f(r*cos(t), r*sin(t))*r = r^2*cos(t) c(t) = 0 d(t) = 5

int(f(r*cos(t), r*sin(t))*r, r ))= 1/3*r^3*cos(t)

int(f(r*cos(t), r*sin(t))*r, r = c(t)..d(t))= 125/3*cos(t)

int(int(f(r*cos(t), r*sin(t))*r, r = c(t)..d(t)),t)= 125/3*sin(t)

int(int(f(r*cos(t), r*sin(t))*r, r = c(t) ..d(t)), t = a..b) = 0

by:sh