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Math Help - Double integral region bound by curves

  1. #1
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    Double integral region bound by curves

    I need to find the volume of a function x dA where S is bound by the curves y = x^2/4, y=x, y=1,and y=2 with x greater or equal to 0

    I think x goes from 0 to 2 would y be between x and x^2/4???? Frostking
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  2. #2
    Super Member redsoxfan325's Avatar
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    I've attached a picture of the graphs of these. I'd like to help you, but I'm not sure what you're asking. Is this a rotation problem? The area bounded by the curves is \int_{1}^2 x-1\,dx + \int_{2}^{2\sqrt{2}} 2-\frac{x^2}{4}\,dx = \frac{1}{2}+\frac{2}{3}(4\sqrt{2}+5) = \frac{1}{6}(16\sqrt{2}-17).

    I hope this helps you in some way.
    Attached Thumbnails Attached Thumbnails Double integral region bound by curves-graph.jpeg  
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  3. #3
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    Clarification of double integral problem

    I am sorry I was not clear. The exact question is : FInd the double integral over the region S of x dA where S is the region bounded by the curves y = x^2/4, y = x, y =1 and y = 2 with x greater than or equal to 0.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Well, as y goes from 1 to 2, x goes from y to 2\sqrt{y}. (If y=\frac{x^2}{4}, then x=\sqrt{4y}=2\sqrt{y}).

    How about this: \int_{1}^{2} \int_{y}^{2\sqrt{y}} x\,dx\,dy?

    Integrating with respect to x gives you \frac{x^2}{2}. Plugging in the bounds gives you \frac{(2\sqrt{y})^2}{2}-\frac{y^2}{2} = 2y-\frac{y^2}{2}.

    Now the integral is \int_{1}^{2} 2y-\frac{y^2}{2} \,dy.

    You can integrate this by normal means to get y^2-\frac{y^3}{6}. Plugging in the bounds gives you 2^2-\frac{2^3}{6} - (1^2-\frac{1^3}{6}) = 4-\frac{8}{6} - 1+\frac{1}{6} = \frac{11}{6}.

    So \frac{11}{6} is the answer you're looking for.
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