# Surface Integral

1. Nov 30, 2008

1. The problem statement, all variables and given/known data
Evaluate the surface integral $$\vec{F}\cdot\vec{n}\, dS$$

where $\vec{F}=<-y,x,0>$ and S is the part of the plane $z=8x-4y-5$ that lies below the triangle with vertices at (0,0,0,), (0,1,0,) and (1,0,0). The orientation of S is given by the upward normal vector. answer: 2

I am not sure if I am just making a careless mistake or a conceptual one.

3. The attempt at a solution

i.) Parametrizing S gives $\vec{r}(x,y)=<x, y, 8x-4y-5>$

ii.) Finding $$\frac{\partial r}{\partial x}\times \frac{\partial r}{\partial y}=<1,0,8>\times<0,1,-4>=-8,4,1$$

iii.) Thus, $\vec{F}(\vec{r}(x,y))\cdot (\frac{\partial r}{\partial x}\times \frac{\partial r}{\partial y})=<-y,x,0>\cdot<-8,4,1>=<4x+8y>$

iv) Therefore $$I=\int\int_D (4x+8y)\, dA$$

$$=\int_{x=0}^1 \int_{y=0}^x (4x+8y)\,dy\, dx$$

I belive that if I made an error, it was made somewhere in here and not in my integration.

Any major blunders here?

Thanks,
Casey

2. Nov 30, 2008

### Dick

Look at your integral limits. You are integrating over the triangle formed by (0,0,0), (0,1,0) and (1,1,0).

3. Nov 30, 2008

Not sure I follow. My x bounds are correct right?

4. Nov 30, 2008

Is my upper y bound supposed to be (1-x)?

5. Nov 30, 2008

### Dick

It sure is!!

6. Nov 30, 2008