Question RE: stoke's theorem. difficulty finding limits of integration

[wown]

Homework Statement

Let S be the part of the plane z=f(x,y)=4x - 8y +5 above the region (x-1)^2 + (y-3)^2 <= 9 oriented with an upward pointing normal. Use Stoke's theorem to evaluate the surface integral for the vector field <2z, x, 1>.

Homework Equations

Stoke's Theorem is surface integral of Fdot product ds = surface integral curl(F) dot product dS

The Attempt at a Solution

I figured out the curl: <0,2,1>. I am having trouble with finding the limits of integration for the normal to the surface. I see that they are asking for the surface that lies on the plane z=4x-8y+5 above the xy plane. Great.

the normal to the surface is easy as well : <-4, 8, 1>. here is where i hit a stop.
reason 1: there are no variables in either of the two vectors and this flags an alarm because I feel I did something wrong.
reason 2: I cannot figure out the limits of integration to the surface. I considered polar/spherical coordinates but since the boundary on the plane is an ellipse, these do not work (unless there is a way to make them work, in which case I do not know what that is).

One thing i considered doing was reasoning that since stoke's theorem depends solely on the boundary, I said "well we basically have a boundary of half an ellipse in the xy plane". So I solve 4x-8y+5=0 for x or y, plug that into the equation for the original circle with radius 3, and solve for the extreme values of x and y. This way I can come up with the equation for an ellipse and set one of the limits equal to the extreme values and the other limit to the upper half of the ellipse... Except that sounds way too complicated for the average problem in my class and I feel that I must have missed something.

Any guidance would be much appreciated. Thanks!

 

[Dick]

The dot of the curl and the normal is a constant. That's a good thing. So to integrate over the surface you just need to multiply it by the area of the surface. The surface is a section of a plane over a circle of radius 3 in the x-y plane. Can you think of a way to do that?

 

[wown]

right. but figuring out the area of the surface is where i am having most difficulty :(

the plane cuts through the circle when 0 = 4x - 8y +5. So I can solve for y and draw the line where the circle is cut. I also know that z= 9 when x = 1 and y = 0 and 9 is the highest value of z.

where do i go from there? How can i calculate the area of the part of the ellipse on the plane?

 

[Dick]

f(x,y)=4x - 8y +5. What's the expression in the area formula you need to integrate in the x-y plane to get the area of that surface?

 

[wown]

I really have no clue :(
have you drawn the surface? it is a very weird looking thing. the projection of the surface on the xy plane is of course the part of the circle with radius 3 and center (1,23) that is cut by the line 4x-8y+5 and lies to the left of that line. The surface itself has one edge that has the line 4x-8y+5 and has max height of z=9 when x=1, y=0. so it is an ellipse with a ray that doesn't run through the middle and is slanted.

 

[Dick]

I haven't really 'drawn' it. As I'm visualizing it, it's just a slanted ellipse. Could you answer my question about what function you should integrate over the circle in the x-y plane to get the area of the ellipse? The details of the ellipse really aren't that important.

 

[wown]

hm pi*a*b? or in this case 1/2*pi*(x or y)*z(which is 4x-8y+5)?

 

[Dick]

hm pi*a*b? or in this case 1/2*pi*(x or y)*z(which is 4x-8y+5)?

Look up a formula for the area of a surface z=f(x,y). If you haven't done that yet, you may be worrying about details of the surface that you don't have to.

 

[wown]

OOOOOOOOOOOOOOOO lol it just clicked... i really should not be doing these problems after working 10 hour days :)

but to make sure then, i would have to set 4x-8y+5 equal to the equation of the circle to figure out the limits on x and y, right?

 

[Dick]

OOOOOOOOOOOOOOOO lol it just clicked... i really should not be doing these problems after working 10 hour days :)

but to make sure then, i would have to set 4x-8y+5 equal to the equation of the circle to figure out the limits on x and y, right?

You haven't answered my question yet. You don't NEED any explicit limits. What's the integrand in the area formula??

 

[wown]

it would be the integral of f(x,y,g(x,y))*sqrt(1+4^2 (this the derivative of g in terms of x) + 8^2))dxdy

 

[Dick]

it would be the integral of f(x,y,g(x,y))*sqrt(1+4^2 (this the derivative of g in terms of x) + 8^2))dxdy

I think it's just sqrt(1+4^2+8^2). It's a constant. What's the f(x,y,g(x,y)) thing doing in there? http://mathworld.wolfram.com/SurfaceArea.html formula 3.

 

[wown]

first, thanks so much for bearing with me and being so patient :)

so, basically i have curl*normal = <0,2,1>dot prod<-4,8,1> = 17 = f(x,y,g(x,y)) which is a constant. multiply this by the sqrt of (1+16+64) = 9. So the final is a constant = 153 which i multiply times the area of the circle = pi*9 and that is my final answer?

 

[wown]

I think it's just sqrt(1+4^2+8^2). It's a constant. What's the f(x,y,g(x,y)) thing doing in there? http://mathworld.wolfram.com/SurfaceArea.html formula 3.

hm interesting. my calc book includes the f(x,y,g(x,y))... which in this case would make sense, no? because ultimatwly i need to multiply by the constant i got from stoke's formula

 

[Dick]

it would be the integral of f(x,y,g(x,y))*sqrt(1+4^2 (this the derivative of g in terms of x) + 8^2))dxdy

first, thanks so much for bearing with me and being so patient :)

so, basically i have curl*normal = <0,2,1>dot prod<-4,8,1> = 17 = f(x,y,g(x,y)) which is a constant. multiply this by the sqrt of (1+16+64) = 9. So the final is a constant = 153 which i multiply times the area of the circle = pi*9 and that is my final answer?

Well, I'm with you as far as you multipy sqrt(1+16+64) times the area of the circle to find the area of the slanted ellipse. And sure, then you multiply by curl.normal to get the final answer. I'll let you actually check the number. But the point is so many things are constant you don't need to worry about messy details like limits.

 

[Dick]

hm interesting. my calc book includes the f(x,y,g(x,y))... which in this case would make sense, no? because ultimatwly i need to multiply by the constant i got from stoke's formula

Mmm. I really don't know what the book is talking about. What's g(x,y) supposed to be?