Let F=(2x,2y,2x+2z). Use Stokes' theorem to evaluate the integral of F around the curve consisting of the straight lines joining (1,0,1),(0,1,0), and (0,0,1).
I have to find the unit normal and the curl of F (okay, finding the curl is easy, I can do that) and evaluate the integral.
Okay, so I've got -sqrt2 as my integrand, but what should my bound of integration be?
If the bounds are -1 to 1, that would make the integral 0, right?
Anonymous
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By Stroke's Theorem:
∫C F • dS = ∫∫S F • dS.
In order to evaluate the RHS, we need a surface that has the curve in question as its boundary (it doesn't matter what surface we choose as long as it has the boundary in question). Such a curve is the plane that passes through (1, 0, 1), (0, 1, 0), and (0, 0, 1). It shouldn't be too difficult to show that the equation of such a plane is:
y + z = 1 ==> z = 1 - y.
So, we can parametrize the surface as follows:
s(x, y) = <x, y, 1 - y>, for 0 <= x, y <= 1.
With s_x(x, y) = <1, 0, 0> and s_y = <0, 1, -1>, the normal vector is:
<1, 0, 0> x <0, 1, -1>,
which should be an easy cross product. Find the curl of F, dot it with this, and evaluate the resulting integral.
I hope this helps!
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Your bounds will be the ranges of x and y, which are -1 <= x, y <= 1.