# Surface Integrals: Flux of F across S

1. Nov 30, 2009

### ryoonc

1. The problem statement, all variables and given/known data
Evaluate the surface integral ∫∫S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.

F(x, y, z) = xy i + yz j + zx k

S is the part of the paraboloid z = 3 - x$$^{2}$$ - y$$^{2}$$ that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation.

2. Relevant equations
$$\int \int F\cdot dS = \int \int \left( -P \frac{\partial g}{\partial x} -Q \frac{\partial g}{\partial y} +R \right) dA$$

3. The attempt at a solution
I've gone over the examples in my calculus book two or three times now and I get confused about a couple things:

dA is r*dr*d$$\theta$$, and I replace the x's and y's with rcos$$\theta$$ and rsin$$\theta$$, respectively, and substituting z with the given equation, but I still get a wrong answer. Here's my attempt:

$$\int \int ( -y(-2x) -x(-2y) +3-x^{2}-y^{2})dA$$

...and after substituting x and y and z for their polar coordinates and then simplifying, I get:

$$\int ^{\pi / 2}_{0} \int ^{1}_{0} (3+4r^{2}cos(\theta)sin(\theta)-r^{2})rdrd\theta$$

I was kind of hoping this would get me the right answer, but it's not (I end up with $$\frac{5 \pi}{8} + \frac{1}{2}$$), and I think it has to do with the r domain that I used, or perhaps my entire equation. In any case I'm not sure how to implement the square of length 1 in xy plane domain into my equation. Where have I gone wrong?

2. Nov 30, 2009

### LCKurtz

This is your P(x,y)i + Q(x,y)j + R(x,y)k

This is your z = g(x,y)

Are your P(x,y), Q(x,y) and R(x,y) terms here correct?

Why in the world would you try polar coordinates at this point? The integrand is simple polynomial terms in x and y and you are integrating over a square.​

Last edited: Nov 30, 2009