# Finding a surface form the intersection of two surfaces- Stokes' Thrm.

1. May 31, 2013

### coljnr9

1. The problem statement, all variables and given/known data

Let $\vec{F}=<xy,5z,4y>$

Use Stokes' Theorem to evaluate $$\int_c\vec{F}\cdot d\vec{r}$$

where $C$ is the curve of intersection of the parabolic cylinder $z=y^2-x$ and the circular cylinder $x^2+y^2=36$

2. Relevant equations

Stokes' Theorem, which says that $$\int_c\vec{F}\cdot d\vec{r}=\int\int_s ∇×\vec{F}\cdot d\vec{S}$$

3. The attempt at a solution

Because we are in the chapter of Stokes' Theroem, I am supposed to integrate over a surface that is defined by the intersection of the two surfaces given in the problem. However, I am having a bear of a time coming up with what that surface would look like, and what an equation for it would be.

I tried plugging the equations into a LiveMath module to see the intersection, but it would break as soon as I zoomed out far enough to see what I needed.

When I have to do this with other problems (intersection of a parabloid and a cylinder), I can look at the graph, or just imagine it, and fairly quickly see the general shape, and then do some algebra to get the coefficients.

If I could have some help coming up with the equation of that surface, I would be able to do some dot-and-cross products and be on my merry way.

Thanks!

2. May 31, 2013

### LCKurtz

It is the bounding curve that is defined by the intersection of the surfaces. But the surface that is bounded by that curve is just your $z = y^2 - x$ parabolic surface. So you have that slanted parabolic surface whose domain in the xy plane is the the disk $x^2+y^2\le 36$.

You could set up the surface integral in terms of x and y and change the resulting integral to polar coordinates or parameterize the parabolic surface in terms of $r,\theta$ in the first place. Does that get you going?

Last edited: May 31, 2013
3. May 31, 2013

### coljnr9

Yes, that was perfect. I used the surface $z=g(x,y)=y^2-x$, and got a normal vector from $<-\frac{\partial g} {\partial x}, -\frac{\partial g}{\partial y}, 1>$. Did my curl and dot product calculations to get
$$\int\int_d <1,-2y,1>\cdot<-1,0,-x>dA=\int\int_d -x-1dA$$
I converted to polar using
$$x=rcos(\theta)$$
$$y=rsin(\theta)$$

and finally had an integral of
$$\int_0^{2\pi}\int_0^6 (-1-rcos(\theta))rdrd\theta$$

Thanks so much for your help!

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