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Double integral limits from an equation?

  1. Feb 13, 2014 #1
    Hi everyone,

    I've the equation x+y=6 (it's a surface equation which I'll integrate over) and the following integral limits is what I suppose to get it from the equation: [itex]\int\limits_0^6 \int\limits_0^{6-x}[/itex]

    K1V95tV.png

    What's the trick here?
     
  2. jcsd
  3. Feb 13, 2014 #2

    Stephen Tashi

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    Let's try to phrase your question coherently. My guess is that you mean:

    I'm to integrate a function [itex] f(x,y) [/itex] over the triangluar region bounded by the lines [itex] y = 0, x = 0, x + y = 6 [/itex]. The integration is [itex] \int_0^6 \int_0^{6-x} f(x,y) \ dy \ dx [/itex] How does one arrive at the limits for the integrals?

    Roughly speaking, you pretend that you cover the triangular region with vertical line segments. The segment through the point [itex] (x,0) [/itex] goes from [itex] (x,0) [/itex] to [itex] (0,6-x) [/itex] where it meets the diagonal. The integral of the function over that line segment is [itex] \int_0^{6-x} f(x,y) dy [/itex]. You integrate that integral over all possible vertical line segments, which implies you integrate withr respect to [itex] x [/itex] from [itex] 0 [/itex] to [itex] 6 [/itex].
     
  4. Feb 14, 2014 #3
    Stephen Tashi:

    The only two thing that I know is the function and the surface equations, that drawing from the solution of the problem.

    So, can I assume - in these situations - that x and y are equal to zero then solve?
     
  5. Feb 14, 2014 #4

    Stephen Tashi

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    You are either badly confused or can't express what you mean. I suggest that you ask about a specific example. State the entire problem. If you have a textbook, quote a problem from the book.
     
  6. Feb 14, 2014 #5
    Both of them actually :Z

    "Determine the flux across the surface [itex]z=6-x-y[/itex] in the first octant, where [itex]\vec{f}=(0,-1,-2)[/itex]"

    The double integral will be used to determine the flux, my question is: How to get the integration limits?
     
  7. Feb 14, 2014 #6

    Stephen Tashi

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    The integration limits will depend on whether you compute [itex] \int \int ... \ dy \ dx [/itex] or [itex] \int \int ... \ dx \ dy [/itex]. Which way do you want to work the problem?

    I think what you meant is "Can I set z equal to zero and consider how x and y vary?"

    The integrations you set up must consider all points on the surface. In this particular problem, you can do that by considering the projection of the surface on the (x,y) plane. Yes, in this projection z = 0. To consider all points on the surface you may consider how (x,y) varies over this projection and set the limits of integration accordingly.

    However, the integrand you use will express z as a function z(x,y) = 6 - x - y. So you aren't actually assuming z = 0 when we evaluate the integrand. Within the integrand, z varies as a function of x and y. So the integrand involves points on the surface, not points on the projection of the surface.
     
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