# Surface integrals and flux

1. Dec 4, 2008

### CalculusSandwich

1. The problem statement, all variables and given/known data
calculate the upward flux of f(x,y,z) = <yz,2x+y,y^2+z>

Let S be the portion of the cylinder z=4-y^2 lying in the first octant to the right of the plane y=4.
a parametrization into the u v plane is:r(u,v)=(u,v,4-v^2)
region is a rectangle in the uv plane with bounds, (0,0) , (0,2) and (4,0)

2. Relevant equations
$$\int$$$$\int$$ F $$\bullet$$(ru $$\times$$rv) dA

3. The attempt at a solution

ru x rv = 0i + 2vj + 1k

so then i have $$\int$$ <4v-v^3, 2u+v, v^2 + 4 - v^2 > x <0 + 2v + 1 >

$$\int$$ (2v^2 + 4uv + 4)dudv
4$$\int$$ (v^2/2 + uv + 1) dudv

i know i can't go to polar so i integrate

is this the correct integral?

0 to 2, outside integral, 4* 0 to 4 inside integral ( v^2/2 + uv + 1 ) dudv ?

2. Dec 4, 2008

anyone?

3. Dec 5, 2008

### Defennder

You're not writing it out properly. It's supposed to be (0 2v 1)^T There isn't any "+".

How did you get v^2 ? And I still don't understand your notation. Why is there 2 separate integrals instead of a single double integral?

4. Dec 5, 2008

### CalculusSandwich

I'm sorry there's not. That's just the next step to solving the integral.

then i take the dot product of <4v-v^3, 2u+v, v^2 + 4 - v^2 > dot <0, 2v,1 >

I get < 0, 4uv + 2v^2, v^2 -v^2 + 4 > which simplifies into 2v^2 + 4uv + 4

so i integrate between...

$$\int$$ 0 to 2 and $$\int$$ 0 to 4

integrand of : 2v^2 + 4uv + 4 (dudv)

5. Dec 5, 2008

### Defennder

Well ok, I can get that. So what's the problem now?

6. Dec 5, 2008

### HallsofIvy

Staff Emeritus
So it is
$$\int_{v= 0}^2\int_{u= 0}^4 (2v^2+ 4uv+ 4)dudv$$?

Looks pretty straightforward to me. What is the problem? That can, of course, be separated into
$$2\int_0^4 du\int_0^2 v^2dv+ 4\int_0^4 udu\int_0^2 vdv+ 4\int_0^4du\int_0^2dv$$

7. Dec 5, 2008

### Defennder

For some reason the "u=0" and "v=0" isn't showing up properly in Latex.