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Another problem involving computing area with integrals

  1. Jan 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

    x+y^2=6

    x+y=0

    2. Relevant equations

    none, just using integrals...

    3. The attempt at a solution

    I sketch the two curves and find they intersect at x=-3 and x=2. Looking at it I want to integrate with respect to x.

    y=[tex]\sqrt{-x+6}[/tex] appears to be the top function, while y=-x is the lower function.

    I set up an integral from -3 to 2 that consists of: [tex]\sqrt{-x+6}[/tex] - (-x) dx.

    I evaluate using substitution and keep getting 10.16, which the homework website I'm using says is wrong. Not sure what I'm doing incorrectly here.
     
  2. jcsd
  3. Jan 30, 2009 #2

    Mark44

    Staff: Mentor

    It's a lot easier if you integrate w.r.t y (i.e., use horizontal area elements rather than vertical elements). If you do it this way there are no radicals to deal with. The area I get is 49/3.

    You can find the area using vertical area elements, but the y values at the bottoms of your area elements change. For one interval, the function at the bottom is y = -x; for the other interval, the function at the bottom is y = -sqrt(-x + 6). This means that your formula for the area of the typical element has to change, and that means you need to integrals.
     
  4. Jan 30, 2009 #3

    Mark44

    Staff: Mentor

    BTW, this problem is very similar to the one posted yesterday by vipertongn.
     
  5. Jan 30, 2009 #4
    Hey thanks a lot. I get it now. I got the correct answer.
     
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