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so the curl is -3i-7j-4k

If you want to go from line integral to surface integral, shouldn't you need the anticurl?

What is the anticurl?

Nevermind, they gave the anticurl

I see

I mean A such that \[
B = \nabla \times A
\]

But apparently they gave it to you already.

So to use stokes theorem, you need to parametrize the surface.

They don't want you to set up the line integral, but I can help you do that if you want.

The surface is the triangle?

I want to be able to do it using stoke's theorem/how they want it

You can easily find two parallel vectors for the triangle right?

the yz-plane?

No, you do it by finding a vector between two points.

Between two points. So would I have to use points not contained in the triangle?

oh wow, you're right.

yeah so we want the bounds on the yz plane.

oh so it is a projection onto the yz plane

of the triangle, right?

|dw:1362708042969:dw|

You could parametrize it as \[
\Phi(u,v) = (6,u,v),\quad 0\le u \le 5,\quad u\le v \le 5
\]

How many times do I have to say you don't take the curl?

Sorry I keep looking up at the equation. Okay. What would be the next step in my process?

\[
\mathbf{F}\circ \Phi(u,v) = \mathbf{F}(x(u,v),y(u,v), z(u,v))
\]

\(D\) corresponds to the bounds of \(u,v\).

okay so it is the vector field f in terms of u and v dotted with the normal of the parametrization?

Yes.

Okay I'm going to work it out really quick.

You have to consider the orientation.

|dw:1362709722289:dw|

Yes it should be negative.

the normal?

yeah.

Do you know what the answer is?