# Calculate flux through a surface

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.

• Your $dA$ does not look very infinitesimal... I think that's actually the normal vector field... but in the end it looks right. I haven't checked the arithmetic. – James S. Cook Nov 25 '13 at 21:12
• I made a small edit to the statement of the problem, since it did not indicate the radius of the disk, which only finally appeared in the integration limit near the end of the post. Note, incidentally, that $\ \nabla \cdot \vec{F} \ = \ 0 \$ . – colormegone Apr 27 '14 at 5:14

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 $dx \, dy$.