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Surface area using double integrals

  1. Dec 6, 2011 #1
    Find the surface area of the triangle with vertices (0,0) (L, L) (L,-L)

    I know I have to take the double integrals of f(x,y) but I have no idea what f(x,y) is supposed to be!
  2. jcsd
  3. Dec 7, 2011 #2


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    Homework Helper

    Why do you need an integral to find the area of a triangle?
    Why not good old A = bh/2 ?
  4. Dec 7, 2011 #3
    A surface area in the form of a double integral will be in the form:

  5. Dec 7, 2011 #4
    But Delphi's answer is just as good too if you simply would like to know a double integral and not use any calculus. :]
  6. Dec 7, 2011 #5
    Oops, I mean area, not double integral.
  7. Dec 7, 2011 #6
    what's dA? I wasn't there during the lecture so I'm lost. I looked through my textbooks, online, and I can't find anything. All I think I know is that I need a f(x, y).

    And no I can't use bh/2. But it's good because I know my answer from the double integral has to be something like bh/2 so I can verify my answer.
  8. Dec 7, 2011 #7
    dA would be dxdy. You will first have to integrate with respect to x or y, then integrate with respect to the second variable. If your double integrals limits of integration are just numbers, will give you the area of a rectangle. Your limits of integration can also be functions as well though.
  9. Dec 7, 2011 #8
    but what do i integrate?

  10. Dec 7, 2011 #9
    dA is basically a very small element of the area and depends on the type of coordinate system you use as well. If you just take a very small rectangle to be dA, then your dA will be dx dy. If you're using polar coordinates and your dA is a strip defined by the radius and the angle, then dA = r dr d(theta).

    When you integrate, you hold every other variable that you're not integrating over as a constant. For example, lets say you have the double integral of y+x dydx.

    ∫x = 0..2 ∫y = 0..3 y+x dy dx

    Integrating with respect to y, holding all other variables constant:

    ∫x=0..2 y^2/2 + xy | y = 0..3 dx =

    ∫x = 0..2 4.5 + 3x dx =

    4.5x + 3x^2/2 x = 0..2 =

    9 + 6 = 15.
  11. Dec 7, 2011 #10
    I know how to do double integrals but I don't know what should come here ∫∫(??)dydx

    The image is a triangle

    It looks like this

  12. Dec 7, 2011 #11
    In your case, you will have:

    x=bx=ay = g(x)y = h(x)dydx

    Or you can have it in the form of:
    y=dy=cx= g(y)x = h(y)dxdy

    But in your case I think it would be easier to do the first. The integrand will just be 1 dydx.

    So you will have to find an upper limit function and a lower limit function. Once you have those functions and you integrate, then you integrate with respect to your other variable and use your limits a to b.

    You may want to graph the triangle. Your upper limit will be the equation of a line of form y = mx + b and your lower limit will be when y = 0. I'm not sure if I'm allowed to just give out the answer. XD Give it a try and look up some example problems on the book too. :]
  13. Dec 7, 2011 #12
    Ooh in that case, you will have your limits as y = mx +b and y = -wx + h, where m and w are separate slopes of each line and b and h are different y intercepts. You will still be integrating simply over 1 dydx.
  14. Dec 7, 2011 #13
    In summary, a double integral of 1 gives you an area, a triple integral of 1 gives you a volume.
  15. Dec 7, 2011 #14
    positive slope = 1x + 0 = x
    negative slope = -1x + 0 = -x

    not sure if i'm sticking them in the right place

    Code (Text):
     x   x
     ∫   ∫ dydx
    -x   0
    I got x2

    I guess that's the same thing as L2

    I plugged in some numbers on the bh/2 formula and they seem to agree

    wow so simple.

    am I correct now?
  16. Dec 7, 2011 #15
    Actually I think it should be the other way around

    Code (Text):
     x   x
    ∫   ∫ dydx
    0   -x
    Still comes out to x2
  17. Dec 7, 2011 #16
    Very good. :] Now imagine if you were to cut the triangle right in half through the x axis, and fit the bottom half on the top left of the top half. You have a rectangle with area of L^2.
  18. Dec 7, 2011 #17
    Yep, thanks!
  19. Dec 7, 2011 #18
    Mathematically, you would probably want to change your x limits from 0 to x to limits of 0 to x prime, or another letter.
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