# Path Integral problem

Compute the path integral where f(x,y,z) = x^2 and the path C is the intersection of the sphere x^2+y^2+z^2=1 and the plane x+y+z=0.

I found the intersection to be x+y-(1/sqrt(2))=0 (not sure if that's right) but I am not sure how to parametrize it in terms of t.

Any help would be appreciated.

Anyone?

I don't think that's too easy. You sure? You can see what it looks like below.

First find the intersection.

You can do that. Just solve for z in both of them up there, equate and you should get:

$$1=2x^2+2xy+2y^2$$

See what I mean. That requires a rotation of axes cus' of that xy in there. It's not hard. Just follow the book. So you should get:

$$x=u\cos(\alpha)-v\sin(\alpha)$$
$$y=u\cos(\alpha)+v\sin(\alpha)$$

That's already tough. But turn the crank and it turns out to be an ellipse that looks like:

$$\frac{u^2}{a^2}+\frac{v^2}{b^2}=1$$

Now you know for that ellipse, the parametric representation is:

$$u=a\cos(t)$$
$$v=b\sin(t)$$

Now the line integral in 3D is:

$$\int_C f(x,y,z)\sqrt{(g')^2+(h')^2+(k')^2}dt$$

where:

$$x(t)=g(t)$$
$$y(t)=h(t)$$
$$z(t)=k(t)$$

but that's in x, y and z. I believe you would have to convert that integral in terms of u, v, and z.

#### Attachments

• 15.4 KB Views: 284
How would I convert it in terms of u,v,z?

I should have said convert it to an integral in t. You have:

$$\int_C x^2\sqrt{g'^2+h'^2+k'^2}dt$$

and I stated what x and y were in terms of u and v and from the intersection equation, z=-(x+y), and I showed what u and v were in terms of t. So make those substitutions, integrate from 0 to 2pi. Would be nice if you already knew what the anwere was so that we could check it numerically first, then worry about actually evaluating the integral analytically once we know we have the right form.

The answer is suppose to be 2PI/3

The answer is suppose to be 2PI/3
I get 1.9 but I tell you what, if this was a final exam question and the answer we're suppose to get is 2pi/3, I would turn in the answer for 1.9. That's just me though.

What are a and b though?