Surface integral from vector calculus

[m453438]

Homework Statement

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.

Homework Equations

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.

 

[HallsofIvy]

I do not get 0 as the integral.

 

[tiny-tim]

Welcome to PF!

Hi m453438! Welcome to PF!

(have a theta: θ and a pi: π and a √ and a ∫ and a ≤ and ≥, and try using the X² tag just above the Reply box )

Useful tip: when you do it again, try putting r = √(x² + y²) at the start …

it makes it a lot easier to write, which means you're less likely to make a mistake.