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

[coljnr9]

Homework Statement

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 $x^2+y^2=36$

Homework Equations

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

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!

 

[LCKurtz]

Homework Statement

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 $x^2+y^2=36$

Homework Equations

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

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.

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?

 

[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!