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course Mth277

5/9 3

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[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

So because our surface is described z = 10 we also have f(x,y) = 10 as given by our definition of surface integral.

x^2/4 + y^2 <= 1, is of the form of an ellipse with -2<=x<=2 and -sqrt(1-(x^2/4))<=y<= sqrt(1-(x^2/4))

So that

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

Int(2xy dy) = xy^2, evaluated

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

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

(2^2 - 2^4/8) - ((-2^2) - (-2^4/8)) = (4 - 2) - (4 - 2) = 0

????How is that possible????

I must have messed up somewhere, checking solution.

@& 2 x y is positive in two quadrants and negative in two. The ellipse breaks into quarters with one quarter in each quadrant. The integral over the first quadrant 'cancels' that over the second, and the integrals over the third and fourth quadrants are also equal and opposite.*@

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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 understand everything except why we didn’t have to account for our fn which defined our surce area or the g(x,y,f(x,y)) fn. I thought that when given the form Int[Int[2xy dS, S]] we had to mult. our fn(which in this case was 2xy) by the “area multiplier”. It seems your integral was based simply on the “area mult” and the boundary curves. So do we not have to incorporate that fn if we know the boundary curve, that doesn’t make sense to me but not much does these days ????????

@& My solution failed to integrate the function 2 x y. My integral only evaluated the area.

The correct integral would have been

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

The details are straightforward, and the result is zero, as you found.*@

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

Because we have the restrictions x^2+y^2<=4, with x>0,y>0, we know this is a part of a circle found in Quad I. Therefore

= 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), which I’m pretty sure we could rewrite as

= Int(Int((r^2)*`sqrt(r^2 + 1) *r dr,0,2 dtheta,0,`pi)

You could approach to solve by then making trig sub with r = tan(theta), so that sqrt(x^2+1) becomes sqrt(tan^2 + 1) = sqrt(sec(theta)^2), with r =tan, dr = sec^2……

Anyway its set up I believe , checking solution.

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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):

Very same approach I had except I was thinking trig sub, but you approach was much simpler. Not that you need me to tell you that. Again I’m confused about why we took into account the “x^2+y^2” or the g(x,y,f(x,y)) fn this time but not previously?????

@& My oversight on the previous.*@

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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:

F = xi + 2yj + zk

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

????having trouble finding f(x,y), but here is what I’m thinking. Since we are given this cube we would’nt have a f(x,y) or z = so and so(in other words be able to parameterize our surface with z component being a fn of x an y). But that would make sense because we are not looking at ONE surface we are looking at 6(duh, makes sense now). So this is what we will do

We have surface one

For this surface(lets say top surface) we have z = 1, so f(x,y) = 1, f_x = 0, f_y = 0,so that

Int(Int(F dot N dS)) = Int(Int(F (x,y,f(x,y)) dot <-0,-0,1> dA), with both x and y between 0 and 1

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 dA)

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

Int( 1 dy,0,1) = y, evaluated = 1

Int(1 dx,0,1) = x, evaluated = 1, so flux across surface is 1

We would now do this for the other 5 surfaces, lets do bottom surface where z = 0

Int(Int(F (x,y,f(x,y)) dot <-0,-0,1> dA) = Int(Int([0 + 0 + 0] dA), which before wasting any time integral of zero will Always(I’m pretty sure) be 0.

Now looking surface across x = 0, 0<=y<=1 and 0<=z<=1, so we have

Int(Int(F dot N dS)) = Int(Int(F (x,y,f(x,y)) dot <-f_x,-f_y,1> dA)

???Sort of confused on how to exactly set this up, it’s really late and I’ve been working at this for probably for 48 out of the last 96hrs. I know normal vector must be in direction i , never mind I believe I just figured it out. I got it correct in previous solutions but I wasn’t using formula correctly. That’s what happens when you try and follow a formula????????

Int(Int(F (x,y,f(x,y)) dot <-f_x,-f_y,1> dA) = Int(Int(F (0,2y,z) dot <1,0,0> dA) = 0.

Integral therefore = 0.

Previous Integrals should have looked like

For z = 0

Int(Int(F (x,y,f(x,y)) dot <-f_x,-f_y,1> dA) = Int(Int(F (x,2y,0) dot <0,0,-1> dA) = 0

For z = 1

Int(Int(F (x,y,f(x,y)) dot <-f_x,-f_y,1> dA) = Int(Int(F (x,2y,1) dot <0,0,1> dA) = 1

This is because for Normal to be by definition normal(perpendicular to plane/surface) we would have to have a vector with these properties. Reason for -1 at z=0 and 1 at z =1 because at z=0 normal vector is moving downwards with respect to cube and upwards when z=1 wrt cube.

We do this again at x = 1, and to save time and space we get integral =1

Then when done for y we get 0 at y = 0 and I’ll work it out for y=1

Int(Int(F (x,y,f(x,y)) dot <-f_x,-f_y,1> dA) = Int(Int(F (x,2,z) dot <0,1,0> dA)

= Int(Int(2 dA), with 0<=x<=1 and 0<=z<=1 we have

Int(Int(2 dx,0,1) dz,0,1)

Int(2 dx,0,1) = 2x = 2 evaluated

Int(2 dz,0,1) = 2z = 2

So all together we have a total force of 4 units which would be more than likely referred to as the flux across the surface of the cube.

I hope I did this correctly, I’ve worked and thought through this problem almost 2hrs but a lot of stuff became clear when working through it!!!!

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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

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Self-critique (if necessary):

I swear I didn’t look at your solution and it seems I was correct when I finally worked through it!

@& I can tell from your solution that you didn't look (and of course I'd take your word for that anyway). You reasoned it out. Well done.*@

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@& Very good. Check my notes.*@