F = (x^2) i + (y^2) j + (z) k, S is the cone z = (x^2+y^2) ^ (1/2), with x^2 +y^2 <= 1, x >= 0, y >= 0, oriented upward.
All of the above
The Attempt at a Solution
My attempted solution is 0. But other students claim that the answers is (2/3)pi.
My work is this:
double integral ( (x^2) i + (y^2) j + (z) k ) . ( -(x / (x^2+y^2)^ (1/2) i -(y / (x^2+y^2)^ (1/2) j + (x^2 + y^2) ^ (1/2) k ) dx dy
Then finding common denominator and simplifying:
double integral ( (- x - y ) / (x^2+y^2)^(1/2) ) dx dy
Changing to polar coordinates:
(first integral 0 to 2pi)(second integral 0 to 1) (-cos(theta) - sin(theta) / (r^2) ) * rdrd(theta)
Finally, doing those integrals I have the result of 0.