1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Line & Surface Integrals

  1. May 5, 2008 #1
    1) (a past exam question)
    [​IMG]

    I am stuck with the parametrizations. For part b, to evaluate the resulting surface integral, I think I need to parametrize the surface. How can I parametrize the surface enclosed by the curve of intersection in part a?

    Thank you!
     
  2. jcsd
  3. May 5, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since in (b) they mention "Stokes Theorem", what is "Stokes Theorem"? That should help a lot in (a).

    If z= f(x,y) then an obvious parameterization is [itex]\vec{r}= x\vec{i}+ y\vec{j}+ f(x,y)\vec{k}[/itex].
     
  4. May 5, 2008 #3
    I am OK with part a, but I just don't know how to parametrize the surface in part b, which is the surface enclosed by the curve of intersection of z=3-2x and z=x^2 + y^2
     
    Last edited: May 5, 2008
  5. May 5, 2008 #4
    For b, the surface is part of the plane z=3-2x, so can be parametrized as g(u,v)=(u,v,3-2u) but what are the restrictions on u and v?
     
  6. May 5, 2008 #5

    Defennder

    User Avatar
    Homework Helper

    I thought if you're using Stokes theorem you would have to evaluate it as a line integral? So shouldn't you parametrise closed path which bounds the surface instead of the surface itself?
     
  7. May 5, 2008 #6
    Yes, but I am talking about part b. Using Stoke's theorem TWO times convert it back to a surface integral over a different surface, and I am stuck with parametrizing that surface. Could someone help me, please?
     
  8. May 5, 2008 #7

    Defennder

    User Avatar
    Homework Helper

    Oh, you're right. This time the surface is a flat plane. I misread, sorry.
     
  9. May 6, 2008 #8

    Defennder

    User Avatar
    Homework Helper

    You can represent the equation x^2 + y^2 = 3 - 2x in the form of an ellipse. Then once you do that, you have the value of the semimajor axis with which you can find the limits of x, which corresponds to u in your parametrisation. v, which is y, can be expressed in terms of u, for its limits.
     
  10. May 6, 2008 #9
    Setting x^2 + y^2 = 3 - 2x, I get (x+1)^2 + y^2 = 4 a circle with centre (-1,0) and radius 2, not really an ellipse as you said...

    The surface is part of the plane z=3-2x, so can be parametrized as g(x,y)=(x,y,3-2y), but what are the restrictions on x and y?
     
  11. May 6, 2008 #10

    Defennder

    User Avatar
    Homework Helper

    Ok, well a circle is a special kind of ellipse. But anyway, you can express y in terms of x, and that gives you the limits of y, for the negative and positive values of y. For the limits of x, aren't they simply the max and minimum values of x constrained by the equation of the circle?
     
  12. May 6, 2008 #11
    But for parametrizations, should the limits be constants?
    i.e. a<u<b, c<v<d
    a<x<b, c<y<d
    where a,b,c,d are constants

    e.g. parametrization of unit circle: x=cost,y=sint, 0<t<2pi
     
  13. May 6, 2008 #12

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, the limits do NOT have to be constants. For example, in the xy-plane, z= 0 so we can use x and y themselves as parameters but if our area were the triangle with vertices (0,0), (1,0) and (0, 1), we would have to say [itex]0\le x\le 1[/itex], [itex]0\le y\le 1-x[/itex], or, equivalently, [itex]0\le y\le 1[/itex], [itex]0\le x\le 1-y[/itex] just as we would for an integral. It is only rectangles, for rectangular coordinates and circles for polar coordinates that have constant limits on the parameters.

    Here, the only constraint on your paraboloid is [itex]z\le 3- 2x[/itex]. Since z= x2+ y2, the boundary, in terms of x and y, is x2+ y2= 3- 2x. x2- 2x+ y2= 3, x2- 2x+ 1+ y2= 4, (x- 1)2+ y2= 4, a circle of radius 2 with center at (1, 0). You could do it as: [itex]-1\le x\le 3[/itex], [itex]-\sqrt{4- (x-1)^2}\le y\le \sqrt{4-(x-1)^2}[/itex].

    The change of variable, u= x-1, v= y, will convert that into a circle with center at (0,0) in the uv-plane: [itex]-2\le u\le 2[/itex], [itex]-\sqrt{4- u^2}\le v\le \sqrt{4-u^2}[/itex].

    And you could convert that to polar coordinates: taking [itex]u= r cos(\theta)[/itex], [itex]v= r sin(\theta)[/itex], [itex]0\le r\le 2[/itex], [itex]0\le \theta\le 2\pi[/itex] although then your integrand might be messy.
     
  14. May 6, 2008 #13

    Defennder

    User Avatar
    Homework Helper

    You are performing a double integration with respect to x and y right? Not r,theta. The limits of the 'inner' integrand needn't be constants, they can be expressed in terms of the variable with which you are integrating the 'outer' integrand with respect to. But the limits of the 'outer' integrands should be constants because your final answer is a numerical value.

    See for example:

    http://www.math.oregonstate.edu/hom...usQuestStudyGuides/vcalc/surfint/surfint.html

    EDIT: HallsofIvy beat me to it.
     
  15. May 6, 2008 #14
    Thanks for clarifying! Now I understand it!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Line & Surface Integrals
Loading...