Okay so this is another Stokes' theorem problem. You're given a line integral with the vector
field and the curve and we're asked to solve using Stokes' theorem. Again like last time since we're
given the curve, we're going to want to evaluate it as a surface integral using this equation. The
curve integral equals the curl of the vector field dot n ds and we're going to have to calculate
these three parts. Okay so let me sketch the curve here on the side. The curve is a triangle
going counterclockwise. It's 1 on Y and 2 on the Z. The surface is going to be a plane if
we imagine the curve as a surface and the normal vector goes out kind of like that. All right so
probably one of the first things that we can do is solve for the curl of the vector field. That
equals del crossed with the vector field. That's going to be a matrix ijk. Del is partial partial
X partial, partial partial Y, partial partial Z. And just to simplify the notation we can call these the
components P Q R. Just to make it easier to write in the matrix, so PQR. That's going to equal partial
R partial Y minus partial Q partial Z minus the j component, which is partial R partial X minus
partial P partial Z plus the k component, which is partial Q partial X minus partial P partial
Y. All right, so let me just solve that over here. Partial R partial Q is 0 minus partial Q partial
Z, that's going to be 6. Minus the j component. J is partial R partial X. It's 3 minus 0 plus
the k component. Partial Q partial X, 0 again, and partial P partial Y is 1. The curl of the
vector field is <-6, -3, -1>. Next we're going to want to find the unit normal
vector to do the dot product. We're going to want to take another cross product. The points
that we're given were (2, 0, 0), (0, 1, 0), and (0, 0, 2). From these points, we want to make two lines. Again
let's just call this PQR. Two lines would be for example PQ and PR. And again the idea
is we're going to take two lines from the plane, do the cross-product. That'll give us a vector
and then we want to find the unit vector of that. Let's just subtract from the coordinates to get
the lines. So 0 minus 2 is going to be -2, 1 minus 0 is 1, and 0. PR is 0 minus 2,
-2, 0, and 2. Okay so now let's take the cross product of PQ and line PR. We can imagine it like that
and again we know that that will give us a normal vector. So ijk again. That's going to be
-2, 1, 0; -2, 0, 2. All right so that will give us 2 minus 0 minus the j component, which is
-4 minus 0 plus k component 0 minus -2. That's going to give us <2, 4, 2>.
To find the unit normal vector, we're going to take V over the magnitude of V. V is just this vector
that we found. That equals the normal vector. The magnitude will be the square root of 2 squared
plus 4 squared plus 2 squared. That's going to give us 2 over to the square root
of 24, 4 over the square root of 24, and 2 over the square root of 24. Okay so we have the curl of the
vector field. We have the normal vector. The next part of the equation that we want to worry about
is the dS. In this case, we're going to use what we learned from previous sections, which is that dS
can equal the square root of 1 plus Z partial X plus Z partial Y squared dA. What we have
to do next is find a plane equation of the form Z equals so that we can take the partial derivatives.
We already have that the normal vector is <2, 4, 2>. You can put, if you consider the normal
vector ABC, it's A X plus B Y plus C Z equals D and that will give you the equation of a plane as long
as you have a point, which we do. We have three points. We can go ahead and plug everything
in. It's going to be 2 X plus 4 Y plus 2 Z equals D. Let's use the point, for example, (2, 0, 0). It doesn't
matter which one we use. That's going to give us 4 plus 0 plus 0 equals D. We have the equation 2X
plus 4Y plus 2Z equals 4. And from that equation we can just solve for Z pretty simply. We're going to
subtract 4 minus 2X minus 4Y and then divide by 2. Z equals 2 minus X minus 2Y and now we can
plug that into this formula. We have 1 plus, the partial of Z with respect to X is 0 minus
1 plus 0 squared plus partial of Z with respect to Y will be -2 squared dA. And that will give
us 1 plus 1 plus 4, which will give us the square root of 6 dA. Now we are again ready to plug
everything into our formulas. We have <-6, -3, -1> dot with <2 over square root
24, 4 over square root 24, and 2 over square root 24>. Then we have square root of 6 d--oops, let me scoot
over so you have some more room. It's going to be square root of 6 dA. And that's all a double
integral over a plane. To make this easier, I would just distribute this square root of 6 over the
square root 24 and then if you do that, you just get 1/2. This all simplifies if you just put a 2
under the denominator and this will again simplify to <1, 2, 1>. I would just do that before doing the
cross product because it makes it easier in the end. And this surface integral is pretty simple. If
we just consider when Z equals 0, we have from the plane equation we already have, we have 0 equals
2 minus X minus 2 Y. We can solve for either one. I went ahead and solved for Y. So 2 Y equals 2 minus
X. Y equals 1 minus 0.5 X. If we plug that in here, we have Y goes from 0 to 1 minus 0.5 X and X
goes from 0 to 2. Again, we know that from the points. Okay, so if we do the dot product, that's
going to give us -6 minus 6 minus 1 and that gives us -13 from 0 to 1 minus 0.5
X and from 0 to 2. That's going be dy/dx. Then again that's a very simple double integral. If we
plug that in, we just get -13, 1 minus 0.5 X, and that's from 0 to 2, dX. Let's distribute. Plus
13 over 2X, 0 to 2, dX. Take the integral. That's gonna be -13X plus 13X squared over 4
and that's from 0 to 2. Plugging in 2, we're going to get -26 plus 13 times 4 over 4.
That's going to be -13 minus--plugging in 0 we get 0--so our final answer is -13.