Path Integral problem

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:

[tex]1=2x^2+2xy+2y^2[/tex]

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:

[tex]x=u\cos(\alpha)-v\sin(\alpha)[/tex]
[tex]y=u\cos(\alpha)+v\sin(\alpha)[/tex]

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

[tex]\frac{u^2}{a^2}+\frac{v^2}{b^2}=1[/tex]

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

[tex]u=a\cos(t)[/tex]
[tex]v=b\sin(t)[/tex]

Now the line integral in 3D is:

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

where:

[tex]x(t)=g(t)[/tex]
[tex]y(t)=h(t)[/tex]
[tex]z(t)=k(t)[/tex]

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

Not entirely sure about this. Nice plot though huh?.

 

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

[tex]\int_C x^2\sqrt{g'^2+h'^2+k'^2}dt[/tex]

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.

 

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.