A surface integral

The problem:

Let S be the portion of the cylinder in the first octant bounded by z=0, z=3, x=0 and y=0, and oriented by the unit normal which points away from the z-axis. Let F= x i + 2z j + y k. Compute the flux of F through the surface S.

First, here is a picture of the surface and the vector field:

> with(plots): A:=fieldplot3d([x,2*z,y],x=-0.2..2.2,y=-0.2..2.2,z=-0.2..3.2,grid=[8,8,8],arrows=THICK,style=patch):

> B:=plot3d([2*cos(v),2*sin(v),u],u=0..3,v=0..Pi/2):

> display3d({A,B});

You can see that most or all of the arrows point out of the surface, so the surface integral is likely to be positive.

To set up the integral, we paramatrize the surface as follows: as x goes from 0 to 2 and z independently goes from 0 to 3. So we are considering y as a function of x and z (even though there are no z's in the formula for y).

Next, we have to calculate the unit normal vector to the surface. To do this, we write the equation of the surface as and take the gradient of this to get that a normal vector is 2x i + 2y j . Since on the surface, the length of this vector is 4. So the unit normal vector n is

n = x/2 i + y/2 j = x/2 i + / 2 j .

So we have F dot n = x^2/2 + z .

The element of surface area is dx dz = dx dz = dx dz.

So we need to integrate:

.

According to Maple, this gives:

> Int(Int((x^2/2+z*sqrt(4-x^2))*sqrt(4/(4-x^2)),z = 0 .. 3),x = 0 .. 2)= int(int((x^2/2+z*sqrt(4-x^2))*sqrt(4/(4-x^2)),z = 0 .. 3),x = 0 .. 2);

>

That wasn't much help!

First of all, there are two terms in the integral. The first is apparently equal to: . The second is which is easily seen to be equal to 18. The first integral is done by substituting x=2 sin(t) as t goes from 0 to . This turns the integral into:

> 3*Int(4*sin(t)^2/(2*cos(t))*2*cos(t),t=0..Pi/2);

which is integrated via the substitution to get

So the answer to the problem, being the sum of the two terms, is 18 + .