- #1

ryoonc

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## Homework Statement

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[tex]^{2}[/tex] - y[tex]^{2}[/tex] that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation.

## Homework Equations

[tex]\int \int F\cdot dS = \int \int \left( -P \frac{\partial g}{\partial x} -Q \frac{\partial g}{\partial y} +R \right) dA[/tex]

## 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[tex]\theta[/tex], and I replace the x's and y's with rcos[tex]\theta[/tex] and rsin[tex]\theta[/tex], respectively, and substituting z with the given equation, but I still get a wrong answer. Here's my attempt:

[tex]\int \int ( -y(-2x) -x(-2y) +3-x^{2}-y^{2})dA[/tex]

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

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

I was kind of hoping this would get me the right answer, but it's not (I end up with [tex] \frac{5 \pi}{8} + \frac{1}{2}[/tex]), 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?

Thanks in advance