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Math Help - Area between Curves

  1. #1
    aka
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    Area between Curves

    Not sure where to even start

    Find the area of the region in the first quadrant bounded by the curves y^2=7x, y^2=8x, x^2=4y, x^2=5y
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  2. #2
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    Quote Originally Posted by aka View Post
    Not sure where to even start

    Find the area of the region in the first quadrant bounded by the curves y^2=7x, y^2=8x, x^2=4y, x^2=5y
    start with a sketch of the graphs and identify the region ... looks like that small "parallelogram-looking" region in quad 1.

    find the x-values of intersection, then set up integral expressions to find the area.
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  3. #3
    Master Of Puppets
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    There is four points of intersection between these curves. First thing to do is find them. This will give you the terminals for your integration. Do you know how to do this?

    After this apply them to

    A= \int \sqrt{8x}-\frac{x^2}{5}~dx-\int \sqrt{8x}-\frac{x^2}{4}~dx-\int \sqrt{7x}-\frac{x^2}{5}~dx+\int \sqrt{7x}-\frac{x^2}{4}~dx
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  4. #4
    Senior Member AllanCuz's Avatar
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    Other guys beat me to it. The best thing to do in these programs is to draw out the total region, identify region as defined by our bounds, and then compute the double integral. In any case, the following is how I would solve it

    Quote Originally Posted by aka View Post
    Not sure where to even start

    Find the area of the region in the first quadrant bounded by the curves y^2=7x, y^2=8x, x^2=4y, x^2=5y
    You need to draw these out. We're going to end up with 4 equations in terms of y:

     y = \sqrt{7} \sqrt{x}

    y = \sqrt{8} \sqrt{x}

    y = \frac{x^2}{4}

    y = \frac{x^2}{5}

    This will be a parallelogram in the first quadrent. Notice how

     y = \sqrt{8} \sqrt{x} and  y = \frac{x^2}{5} intersect at the top

     y = \sqrt{7} \sqrt{x} and  y = \frac{x^2}{4} intersect at the bottem

    So let us bound our function in terms of y and in terms of width (actual values here).

    The width (x) can be found by equating the functions that intersect at the top and at the bottem and finding the difference.

    So,

     \sqrt{8x} = \frac{x^2}{5}

     5 \sqrt{8} = x^{ \frac{3}{2} } \to x_2 = 5.84

    Then,

     \sqrt{7} \sqrt{x} = \frac{x^2}{4}

     4 \sqrt{7} = x^{ \frac{3}{2} } \to x_1 = 4.82

    We can now construct our double integral.

     A = \int_{4.82}^{5.84} dx \int_{ \sqrt{7x} }^{ \sqrt{8x} } dy

    But wait! In this calculation we have included the sides to the left and to the right of our region. So we must find an expression for these areas so that we can subtract them.

    On the left, we have the value  x = 4.82 and are extending to the intersection of  \sqrt{8x} = \frac{x^2}{4} which leads to  x = 5.03

    Thus the area THAT WE DONT WANT on the left side is

     A_L = \int_{4.82}^{5.03} dx \int_{ \frac{x^2}{4} }^{ \sqrt{8x} } dy

    Similarly, on the right side we extend from the intersection of  \frac{x^2}{5} = \sqrt{7x}  which leads to  x = 5.59 , to  x = 5.84

    Thus, total area is

     A = \int_{4.82}^{5.84} dx \int_{ \sqrt{7x} }^{ \sqrt{8x} } dy  - \int_{4.82}^{5.03} dx \int_{ \frac{x^2}{4} }^{ \sqrt{8x} } dy  - \int_{5.59}^{5.84} dx \int_{ \sqrt{7x} }^{ \frac{x^2}{5} } dy
    Last edited by AllanCuz; July 5th 2010 at 03:35 PM.
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  5. #5
    aka
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    Thanks for the help. I ended up getting 1/3
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