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Math Help - Area enclosed by several functions

  1. #1
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    Area enclosed by several functions

    Hey

    I don't understand how to find the area enclosed by functions. I tried my text book, but it is pretty confusing. Can someone show me the steps to this problem, please?

    *R is closed by y= square root of x , x=9, and y=0

    a) Integrate with respect to the x -axis

    b) Integrate with respect to the y-axis.


    Thanks!
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  2. #2
    o_O
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    If you draw it out, you'll find that you're basically finding the area under the function from x = 0 to x = 9. So it's just a simple definite integral.
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    Uh, yeah, I still don't get what that means ;p
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  4. #4
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    Quote Originally Posted by NoiseandAttack View Post
    Uh, yeah, I still don't get what that means ;p
    If this isn't review material, then talk to your instructor.

    -Dan
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  5. #5
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    Quote Originally Posted by NoiseandAttack View Post
    Hey

    I don't understand how to find the area enclosed by functions. I tried my text book, but it is pretty confusing. Can someone show me the steps to this problem, please?

    *R is closed by y= square root of x , x=9, and y=0

    a) Integrate with respect to the x -axis

    b) Integrate with respect to the y-axis.


    Thanks!
    A sketch graph of the area showing all important features is essential.

    a) By definition: A = \int_{0}^{9} \sqrt{x} \, dx = \int_{0}^{9} x^{1/2} \, dx = .......

    b) The area is the rectangular area (3)(9) - area between curve and y-axis:

    A = (3)(9) - \int_{0}^{3} y^2 \, dy = 27 - \int_{0}^{3} y^2 \, dy = .......

    Note: Relative to the y-axis, the curve is x = y^2.
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  6. #6
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    Thanks for the help!
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  7. #7
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    Why do you do (3)(9) minus the integral from 0 to 3 of y^2?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NoiseandAttack View Post
    Why do you do (3)(9) minus the integral from 0 to 3 of y^2?
    The integral is the area between the curve and the y axis in this case, which is the "opposite" of the area you are interested in. You want the area of the 3 x 9 box less the area represented by the integral.

    -Dan
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  9. #9
    o_O
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    An illustration:



    Notice the rectangle formed by the y-axis, the x-axis, x = 9, and y = 3 as mentioned by mr fantastic. So R is the area of the rectangle minus the white area.
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  10. #10
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    Red face

    Quote Originally Posted by o_O View Post
    So R is the area of the rectangle minus the white area.
    Isn't R just equal to \int_{0}^{9} \sqrt{x} dx ? why do you need to subtract. But I can understand it if the R denotes the white area

    tsal15
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