Part of the surface, S, is: $z=x^2+y^2$ above the disk $ \ x^2+y^2 = 1 \ $ oriented in the $\vec k$ direction. I need to set up an integrated integral to calculate the flux of $\vec F = yz\vec i+xz\vec j-y^2\vec k$ through S.
I am wanting to make sure I am setting up the flux integral properly before I begin to calculate it.
$$\int_S \vec F \cdot dA = \int_S \vec F(x,y,f(x,y)) \cdot dA $$
First I found dA:
$$dA = (-f_x\vec i-f_y\vec j+\vec k)d xd y=(-2x\vec i-2y\vec j+\vec k)d xd y$$
Then found $\vec F(x,y,f(x,y))$:
$$\vec F(x,y,f(x,y)) = yz\vec i+xz\vec j-y^2\vec k=y(x^2+y^2)\vec i +x(x^2+y^2)\vec j-y^2\vec k$$
Then I changed to polar coordinates:
$$dA=(-2r\cos\theta\vec i-2r\sin\theta\vec j+\vec k)rd rd \theta$$
$$\vec F(r,\theta)=r^3\sin\theta\vec i +r^3\cos\theta\vec j-r^2\sin^2\theta\vec k$$
Did the dot product of the two vectors obtaining:
$$(-4r^4\cos\theta \sin\theta-r^2\sin^2\theta)$$
Thus,
$$\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1} (-4r^4\cos\theta \sin\theta-r^2\sin^2\theta)r dr d\theta$$
Does this seem right? I'm working off the example in the book, and as usually not very helpful with intermediate steps.
Best Answer
I think you have your thoughts in the right direction. However, I would be careful about a couple of things:
1) Generally we abuse notation by writing $d \vec{S} = \vec{n} \cdot dS$ denoting the oriented infinitesimal surface element, with orientation given by the unit outward normal $\vec{n}$. Therefore, your $dA$ should been written different. Also, do not write $\delta x \, \delta y$ for $dx \, dy$.
2) I would switch to polar coordinates only after I've completely set up the double integral in the plane. Then you exploit the circular symmetry by switching into polar coordinates.
The rest looks okay. Like James, I haven't really checked your substitutions but I considered these points relevant enough to write an answer.