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Surface Integral finding the limits

  1. May 28, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to evaulate ∫ ∫S dS where S is the surface z = x² + y², 0 ≤ z ≤ 4.

    2. Relevant equations

    dS = √( 1 + ƒ²x + ƒ²y)dxdy


    3. The attempt at a solution

    dS = √( 1 + 4x² + 4y²)dxdy

    here's the problem

    what are the limits to the surface integral? no clue.. dx means i should find a limit for x but i see no obvious way to, also i could change this to cylindrical coordinates but still dont know what the limits would be.

    ∫ ∫ √( 1 + 4x² + 4y²)dxdy
     
  2. jcsd
  3. May 28, 2012 #2

    sharks

    User Avatar
    Gold Member

    Hi Spout

    The first thing you should do is draw the graph, which is a paraboloid.

    Put z = 4 in the equation of the surface, S. You will get a circle with radius 2.

    Since you are projecting onto the x-y plane, find ##z_x## and ##z_y## hence find the surface area:
    $$\int \int \sqrt{(z_x)^2+(z_y)^2+1}\,.dxdy $$
    Then, as the projection is a circle, it is easier to evaluate this double integral using polar coordinates.
     
    Last edited: May 28, 2012
  4. May 29, 2012 #3
    This is quite embrassing but what does "Since you are projecting onto the x-y plane, find zx and zy hence find the surface area" mean

    up to now i only know the process of solving these problems i dont know what partial derivative represents or means, so I'm unable to comprehend this sentence could you give a tiny bit of a hint :)?
     
  5. May 29, 2012 #4
    When doing these, one crucial step that is often unavoidable is visualizing the region. I've thought about some ways to manipulate limits of integration without visualizing, but I think it would make things much harder. So you should learn to visualize the regions the describe.

    You have a bowl shape that is parabolic (paraboloid is the term in this dimension), that is, consider the y=0 plane, then z=x^2 is parabolic (the term for curves). The bowl occupies space between z=0...4. At z=0, x and y must be zero. At the z=4 plane, you have the circle x^2+y^2=4, that is, radius 2. You had the right integrand, √(1+4x^2+4y^2) dxdy.

    So to find the region that paramet(e)rizes the bowl via x and y, we need to know which values of x and y will garauntee we hit all points of the bowl exactly once. So that is the projection of the bowl down (and/or up) onto the xy plane. The projection is of course the disk, x^2+y^2≤ 4. The limits would be hard to integrate with, so our look would be much better if we use a change of coordinates, from rectangular to polar coordinate.

    So we want to map from (x,y) to (r,θ). Then the integrand would be √(1+r^2), while dxdy becomes rdrdθ. Why don't you try to find out what the limits of integration would be.
     
  6. May 29, 2012 #5
    I see

    well im guessing 0 ≤ r ≤ 2 because this covers the min and max radius

    and 0 ≤ θ ≤ 2pi because this covers all points angular-ly

    then the integral would be

    ∫(0 to 2pi) ∫(0 to 2) [ √( 1 + 4r²cos²θ + 4r²sin²θ) ] rdrdθ

    ∫(0 to 2pi) ∫(0 to 2) [ √( 1 + 4r²) ] rdrdθ

    :)? amirite
     
  7. May 29, 2012 #6
    it turned out correct, you guys are legends
     
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