If your solution to stated problem does not match the given solution, you should self-critique per instructions at

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Your solution, attempt at solution.  If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises. 

Question:  Verify Stokes' theorem given F = 2xyi + z^2k, and the surface S being the portion of the paraboloid y = x^2 + z^2 with y <= 4.

Your solution: 

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Given Solution: 

The paraboloid intersects the plane y = 4 on the curve 4 = x^2 + z^2, which is the circle of radius 2, centered at (0, 4, 0), within the plane y = 4.

Stokes' Theorem is just Green's Theorem generalized from regions of the x-y plane, where it relates an integral over a region with respect to area to a line integral around the bounding curve. 

Specifically Stokes' Theorem says that the integral of the curl of a vector function F over a surface is equal to the line integral of the tangential component of F taken over the bounding curve.

In the present example the bounding curve is the circle x^2 + z^2 = 4, y = 4.  The surface is the portion of the given paraboloid between its vertex at the origin and the plane y = 4.

The curl of our function is -2x k . 

Our function z = f(x, y) = sqrt( y - x^2), so f_x = -x / sqrt( 4 - x^2) and f_y = 1 / (2 sqrt(y - x^2).  A normal vector is therefore f_x i + f_y j + k = -x / sqrt( y - x^2) i +1 / (2 sqrt( y - x^2) j + k , and a unit normal is equal to this vector divided by its magnitude sqrt( 1 + (4 x^2 + 1) / (4 ( y - x^2) ) ).

Our expression curl F X n is - 2 x

We need to integrate this over our surface.  Our surface area increment dS is equal to sqrt(1 + f_x^2 + f_y^2) dA = sqrt(1 + x^2 / (4 - x^2) + 1 / ( 4 ( y - x^2) ), where dA is our area increment in the xy plane. 

So we will be integrating -2 x sqrt(1 + f_x^2 + f_y^2) dA = -2 x sqrt(1 + x^2 / (4 - x^2) + 1 / ( 4 ( y - x^2) )

We first integrate over the upper half of the paraboloid, which we describe by 0 <= x <= 2, 0 <= y <= x^2, 0 <= z <= sqrt(4 - x^2 + y).  Our integral is

int ( int ( - 2 x sqrt(1 + x^2 / (4 - x^2) + 1 / ( 4 ( y - x^2) ) dy, x^2, 4) dx, 0, 2)

This integral is somewhat complicated, but doable.  The integral with respect to y is of the form int( a ( b + c / (y - d)) dy), where a, b, c and d are expressions involving x.  An antiderivative is a * c ln ( y - d ) + a * b * y. 

However we don't need to do that.  Remember that we need to integrate over the entire paraboloid, and this integral is just for the top half.  A symmetry argument shows that the integral must be zero.

Now we integrate the tangential component of F around the bounding curve:

We can parameterize the curve as x = cos(t), z = sin(t), y = 4, 0 <= t <= 2 pi.  We get dx = - sin(t) dt, z = cos(t) dt and dy = 0.  On our curve our function is F = 2xyi + z^2k, = 2 cos(t) * 4 i + sin^2(t) k.  and F dot ds = 8 cos(t) * (- sin(t) dt) + sin^2(t) cos(t) dt.

Thus our integral is

int( ( 8 sin(t) cos(t) + sin^2(t) cos(t) ) dt, 0, 2 pi ) = 0.

This agrees with the result of our symmetry argument.

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Question:  Use Stokes' theorem to evaluate the line integral Int[x^3y^2 dx + dy + z^2 dz, C] where C is the circle x^2 + y^2 = 1 in the plane z = 1 traversed counterclockwise when viewed from the top.

Your solution: 

Confidence rating:
 

Given Solution: 

The integrand is F dot ds for F  = x^3 y^2 i + j + z^2 k and ds = dx i + dy j + dz k.

The circle bounds the surface x^2 + y^2 = 1, z = 1, which can be described by -1 <= x <= 1, -sqrt(1 - x^2) <= y <= sqrt(1 - x^2), z = 1.

So we can find the desired integral by integrating curl F over this surface.

The k vector is normal to the surface.

curl F = -2 x^3 y k, so curl F dot N = -2 x^3 y.

Integrating this expression over the surface we get

-2 int ( int( x^3 y dy, -sqrt(1-x^2), sqrt(1-x^2)) dx, -1, 1)

= 0.

To confirm, using the parameterization x = cos(t), y = sin(t), z = 1 our integral becomes

int (- cos^3(t) sin^3(t) + cos(t), t, 0, 2 pi), which is also zero.

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Question:  Use Stokes' theorem to evaluate the line integral Int[y dx + z dy + y dz, C] where C is the intersection of the sphere x^2 + y^2 + z^2 = 4 and the plane x + y + z = 0, traversed counterclockwise when viewed from above the origin.

Your solution: 

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Given Solution: 

Substituting this expression for z in the equation x^2 + y^2 + z^2 = 4 we get

x^2 + y^2 + (-(x+y))^2 = 4 so that

2 x^2 + 2 y^2 - 2 x y = 4.

This describes an ellipse in the x y plane, rotated 45 degrees with respect to the x axis.  The curve lies above this ellipse, subject still to the constraint z = - (x + y).  We could use rotation of coordinates to parameterize this curve, but it's easier to evaluate a surface integral.

Our surface

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