Surface Integral of Flux on Plane with Normal Vector and Cylinder

[kasse]

Homework Statement

Evaluate the surface integral of F*n (the flux) where n is the upward-pointing unit normal vector to the plane z = 3x+2 that lies within the cylinder x^2 + y^2 = 4. F = <0, 2y, 3z>

2. The attempt at a solution

The normal vector is <3,0,1> so that F*n = 3z. dS equals sqrt(10). I transform to polar coordinates and integrate 3*sqrt(10)*(3r^2*cos(theta) + 2r) over the cylinder. The answer I get is 24sqrt(10)*pi, while the correct answer is 24*pi. Does this mean that dS=1? Why?

 

[Dick]

You said n is a unit vector. Then you said n=<3,0,1>. That's not a unit vector. <3,0,1>/sqrt(10) is.

 

[HallsofIvy]

Time for me to rush to battle again! I fight constantly against the very notation "\vec{f}\cdot\vec{n}d\sigma" because if you follow that literally, computing \vec{n} and d\sigma separately you wind up calculating the length of a vector twice and then watch them cancel out! (Unless you forget one of them!)

In this problem, we can write the position vector of a point on the plane as \vec{r}(x,y)= x\vec{i}+ y\vec{j}+ (3x+ 2)\vec{k}.
Differentiating: \vec{r}_x= \vec{i}+ 3\vec{k} and \vec{r}_y= \vec{j}. The cross product of those two vectors, -3\vec{i}+ \vec{k}, is the "fundamental vector product" for the plane and the vector differential of area, d\vec{\sigma} is (-3\vec{i}+ \vec{k})dxdy.

Now \int\int \vec{F}\cdot\vec{d\sigma} = \int\int (2y\vec{j}+ 3z\vec{k})\cdot(-3\vec{i}+ \vec{k})dxdy = \int\int 3z dxdy= 3\int\int (3x+ 2)dxdy.

Integrate that over the circle of radius 2.

Obviously the way to do that is to switch to polar coordinates. We could do that right at the beginning:
write \vec{r}(r,\theta)= rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}+ (3rcos(\theta)+ 2)\vec{k}, differentiate with respect to r and \theta and the "r" you need for rdrd\theta pops out automatically!

[Here endeth the sermon]

 

[Dick]

Amen. It is clearer to keep the volume and the direction information in the same package.