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

   http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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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:  Evaluate the surface integral Int[Int[2xy dS, S]] where S is the surface described by z = 10, and x^2/4 + y^2 <= 1.

Your solution: 

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Given Solution: 

z = f(x, y) = 10, so f_x and f_y are both zero.  The area 'multiplier' sqrt( 1 + f_x^2 + f_y^2) is therefore equal to 1 at all points.

The region x^2 / 4 + y^2 <= 1 is an ellipse with semiaxes parallel to the x and y axes, centered at the origin, intercepting the axes at (2, 0), (-2, 0), (0, 1) and (0, -1).  It can be described by -2 <= x <= 2, -sqrt(1 - x^2 / 4) <= y <= sqrt( 1 - x^2 / 4).

The desired integral is therefore

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

= int( 2 sqrt(1 - x^2 / 4) dx, -2, 2).

This integral is easily evaluated (let x = 4 cos(theta), dx = - 4 sin(theta) dTheta, etc.).  The result is 2 pi.

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Question:  Write out but do not evaluate the surface integral Int[Int[(x^2 + y^2) dS, S]] where S is the surface described by z = xy, x^2 + y^2 <= 4, x > 0, y >= 0.

Your solution: 

Confidence rating:
 

Given Solution: 

The surface is z = f(x, y) = x y, so f_x = y and f_y = x.  The area factor is therefore sqrt(1 + f_x^2 + f_y^2), so that dS = sqrt( 1 + y^2 + x^2) dA.

The region is 0 <= x <= 2, 0 <= y <= sqrt( 4 - x^2).

Our integral is thus

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

However, this integral is fairly messy, involving trig substitutions and a lot of bookkeeping.  Not particularly difficult, but it can get confusing.

At some point in the process we would be likely to note that x^2 + y^2 occurs in both factors of the integral.  This might inspire us to use polar coordinates, replacing x^2 + y^2 with r^2.  Our region becomes 0 <= r <= 4, 0 <= theta <= pi / 2.

Our integrand would become r^2 sqrt( 1 + r^2).  That doesn't look too bad, but it gets even better when we use our polar coordinate area increment r dr dTheta.

We end up with the integral

int( int( r^2 * sqrt(1 + r^2) * r dr, 0, 2) dTheta, 0, pi/2).

Our inner integral has integrand r^3 ( 1 + r^2).  Substituting u = r^2 we get du = 2 r dr, so that r dr = du / 2.  Writing r^3 ( 1 + r^2) dr as r^2 ( 1 + r^2) ( r dr ), our change of variables becomes obvious.  Our integrand is

u sqrt( 1 + u) du / 2.  One more simple change of variable helps:  let w = 1 + u so that u = w - 1.  du = dw so our integrand is

(w - 1) sqrt(w) dw / 2 = w^(3/2) - w^(1/2) = 2/5 w^(5/2) - 2/3 w^(3/2). 

Changing the variable back to u then to r, we get our antiderivative

1/5 (1 + r^2) ^ (5/2) - 1/3 (1 +  r^2) ^ (3/2)

At r = 0 this is 0, and at r = 2 it is 1/5 ( 5 )^(5/2) - 1/3 ( 5 ) ^ (3/2)

Our integral for theta is then

int(1/5 ( 17 )^(5/2) - 1/3 ( 17 ) ^ (3/2) dTheta, 0, pi/2) = pi/2 ( 1/5 ( 5 )^(5/2) - 1/3 ( 5 ) ^ (3/2))

We get approximately 11.7

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Question:  Evaluate Int[Int[ F dot N dS, S]], where F = xi + 2yj + zk, S is the surface of the cube bounded by the planes x = 0, x = 1, y = 0, y = 1, z = 0, z = 1. N is the outward directed normal field.

Your solution: 

Confidence rating:
 

Given Solution: 

int(int(1 dy, 0, 1) dz, 0, 1) = 1.

The face intersecting the vertical plane x = 0 can be described similarly, except that the outward unit normal is - i and F dot N = (0 i + 2 y j + z k) dot (- j ) = 0, so we have

int(int(0 dy, 0, 1) dz, 0, 1) = 0.

Similar analysis yields analogous results for the vertical faces intersecting the planes y = 0 and y = 1, yielding integrals 0 and 2, respectively, and for the horizontal faces intersecting the planes z = 0 and z = 1, on which we get integrals - and 1.

Thus the total surface integral is 4.

This can be interpreted as the flux of the field F through the cube.

Note that we have div F = 4.  Integrating this through the volume of the cube, which has volume 1, our result is just 4 * 1 = 4.  Thus the volume integral of the divergence of F is equal to the flux of F through the surface.

Note furthermore that the curl of F is zero, so that F is the gradient of a scalar potential function, and therefore a conservative function.  We can easily find the scalar potential function 1/2 x^2 + y^2 + 1/2 z^2

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