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Double Integral in an Enclosed Region

  1. Nov 7, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int\int_B 1 dx dy[/tex] where B is the region enclosed by x^2 + y^2 = 9?

    What if B is the region eclosed by y = x +3, y = 5 - x, and y=8

    2. Relevant equations



    3. The attempt at a solution

    So for x^2 + y^2 = 9, does it mean that both x and y go from -3 to 3?

    bot sure about y = x +3, y = 5 - x, and y=8.

    Thank You
     
  2. jcsd
  3. Nov 7, 2008 #2

    Dick

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    Draw a picture of the regions. Start with first one. It's a circle. Yes, you can let y go from -3 to 3. But now the x limits at a specific value of y will depend on y, won't they?
     
  4. Nov 7, 2008 #3
    so y can go from -3 to 3, which means x can go from 0 to 3 or -3 or 0?

    [tex]\int^3_{-3} ( \int^3_0 1 dx + \int^0_{-3} 1 dx) dy[/tex]
     
    Last edited: Nov 7, 2008
  5. Nov 7, 2008 #4

    Dick

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    Draw a circle and pick a y value. The range of x then depends on y. Solve x^2+y^2=9 for x to figure out what they are. The integral you've written is going to give you the area of a 3x3 square, which is not what you want.
     
  6. Nov 7, 2008 #5

    HallsofIvy

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    You asked a question. You got a response saying that the answer is "no" and that the limits of integration on the first integral must be functions of y. Then you just asked the same question again!

    Let me repeat Dick's response: No, no, no! Draw a picture. The circle has center at (0,0) and radius 3. In order that we completely cover that circle we must go from the the lowest point in the circle, y= -3, to the highest point y= 3. Now, for each y imagine a horizontal line at that value of y. In order to cover the circle, we must go from the left point at which the line crosses the circle, [itex]x= \sqrt{9- y^2}[/itex], to the right point [itex]x= \sqrt{9- y^2}[/itex]. Those will be the limits of integration.
     
  7. Nov 7, 2008 #6
    Sorry, I act pretty stupid at times; I didn't realize that x and y could be functions/equations, always thought they were a single value, but I get what you're saying.

    at the right side, should [itex]x= - \sqrt{9- y^2}[/itex] since it's on the negative side?
     
  8. Nov 7, 2008 #7

    gabbagabbahey

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    The left side would be the negative side wouldn't it?
     
  9. Nov 7, 2008 #8
    Is finding the region of y = x +3, y = 5 - x, and y=8 similiar?

    I know y = 8, so x = y - 3 = 8 - 3 = 5 and x = 5 - y = 5 - 8 = -3?
     
  10. Nov 7, 2008 #9

    gabbagabbahey

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    y isn't 8 everywhere in that region!...sketch a graph of the region; what are the points of intersection of those boundary lines?
     
  11. Nov 7, 2008 #10
    points of intersections are: (5,8),(-3,8) and (1,4)
     
  12. Nov 7, 2008 #11

    Dick

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    Good! So y ranges from 4 to 8. Given a y, what's the lowest and highest value of x as a function of y? Yes, it's EXACTLY similar the other question. Except the region is triangle instead of a disk.
     
  13. Nov 7, 2008 #12
    so x would go from -3 (lowest value) to 5 (highest value)

    *praying I didn't say anything stupid again*
     
  14. Nov 7, 2008 #13

    Dick

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    Sorry. You did. And it's the same mistake again! Again, if you integrate y=4 to 8 and x=-3 to 5 you are integrating over a rectangle. The domain is a triangle bounded by three points. That would be SO wrong. And again, the bounds on x are determined by the value of y. It's the difference between the value of x on one line and the value on the other. Just as in the circle problem it was the difference between the values on the two arcs. Try again.
     
  15. Nov 7, 2008 #14
    If God exists, he certainly likes to ignore me.

    Lets try this again.

    I need to integrate this from x = 5 - y to x = y - 3
     
  16. Nov 7, 2008 #15

    Dick

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    Yes, yes, yes! God does exist! You got it. :) You'll probably get other problems like this and it's always the same deal. Draw a picture of the region and draw horizontal and vertical lines through it. Unless it's a dead stupid rectangle, the limit in one coordinate will depend on the value of the other coordinate.
     
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