Query135Mth

#$&*

course Mth 277

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.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Int[Int[2xy dS, S]]

z = 10, and x^2/4 + y^2 <= 1

Int[Int[g(x,y,z)dS, S]]

Int(Int( g(x,y,f(x,y))sqrt(f_x(x,y)^2 + f_y(x,y)^2 + 1) dA

x^2/4 + y^2 <= 1 -2<=x<=2

-sqrt(1-(x^2/4))<=y<= sqrt(1-(x^2/4))

Int[Int[g(x,y,z)dS, S]]

Int(Int( g(x,y,f(x,y))*`sqrt(f_x(x,y)^2 + f_y(x,y)^2 + 1) dA

Int[Int[2xy dS, S]]

Int(Int(2xy *`sqrt(0^2 + 0^2 + 1) dA

Int(Int(2xy dy,-`sqrt(1-x^2/4), `sqrt(1-x^2/4)) dx,-2,2)

Int(2xy dy) = xy^2d

x(`sqrt(1-x^2/4)^2) - (x(-`sqrt(1-x^2/4)^2))

Int(x(2 - x^2/2) dx) = x^2 - x^4/8

(2^2 - 2^4/8) - ((-2^2) - (-2^4/8))

(4 - 2) - (4 - 2) = 0

confidence rating #$&*:

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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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Self-critique (if necessary):

I must have made a mistake somewhere

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Self-critique rating:OK

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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.

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Your solution:

Int[Int[(x^2 + y^2) dS, S]]

z = xy, x^2 + y^2 <= 4, x > 0, y >= 0

Int[Int[(x^2 + y^2) dS, S] = Int(Int((x^2+y^2)*`sqrt(y^2 + x^2 + 1) dA)

Int(Int((x^2+y^2)*`sqrt(y^2 + x^2 + 1) dA)

Int(Int((x^2+y^2)*`sqrt(y^2 + x^2 + 1) dy,0,`sqrt(4-x^2) dx,0,2)

confidence rating #$&*:

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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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Self-critique (if necessary):OK

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Self-critique rating:OK

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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.

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Your solution:

Int[Int[ F dot N dS, S]]

F = xi + 2yj + zk

x = 0, x = 1, y = 0, y = 1, z = 0, z = 1

Int(Int(F dot N dS))

Int(Int(F (x,y,f(x,y)) dot <-0,-0,1> dA)

Int(Int(F (x,y,f(x,y)) dot <-0,-0,1> dA)

Int(Int([0+0+(f(x,y)*1)] dA)

Int(Int( 1 dy,0,1) dx,0,1)

confidence rating #$&*:

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Given Solution:

Consider the face which intersects the vertical plane x = 1. On this face we have x = 1, 0 <= y <= 1, 0 <= z <= 1. The outward unit normal vector is i . F dot N = (1 i + 2 y j + z k) dot i = 1, so the surface integral is

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

Self-critique (if necessary)

I think I am on the right track but need a little more to get to where I need to be.

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Self-critique (if necessary):

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Self-critique rating:

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