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Math Help - Finding the Area Between 2 Curves

  1. #1
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    Finding the Area Between 2 Curves

    I have two problems I need help on:

    1. Find the area of the region lying to the right of x=(y^2)-5 and the left of x=3-(y^2).

    I determined that the limits of integration would be [-5, 3] and then I tried to integrate ((y^2)-5)-(3-(y^2)), but I keep getting a REALLY large unrealistic number,

    2. Calculate the area enclosed by x=9-y^2 and x=5 in two ways: as an integral along the y-axis and as an integral along the x-axis.

    I'm not sure how to start this one.

    Any help will be so greatly appreciated.
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  2. #2
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    Quote Originally Posted by KDonicki View Post
    I have two problems I need help on:

    1. Find the area of the region lying to the right of x=(y^2)-5 and the left of x=3-(y^2).

    I determined that the limits of integration would be [-5, 3] and then I tried to integrate ((y^2)-5)-(3-(y^2)), but I keep getting a REALLY large unrealistic number,
    First have you sketched the curves?

    Code:
    EuMathT code to sketch curves
    
    >y=-6:0.1:6;
    >
    >x1=y^2-5;
    >x2=3-y^2;
    >
    >
    >xplot(x1,y);
    >hold on;plot(x2,y);
    >hold off
    >

    Sketch attached.

    So you see that you need to integrate:

    \int_{y=-2}^2  (y^2-5)-(3-y^2) dy

    RonL
    Attached Thumbnails Attached Thumbnails Finding the Area Between 2 Curves-gash.png  
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by KDonicki View Post

    2. Calculate the area enclosed by x=9-y^2 and x=5 in two ways: as an integral along the y-axis and as an integral along the x-axis.

    I'm not sure how to start this one.
    Again, sketch the curves, and then see what you can do.

    RonL
    Attached Thumbnails Attached Thumbnails Finding the Area Between 2 Curves-gash.png  
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  4. #4
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    for the second problem I used the [-2, 2] as my limits of integration and then I integrated this:

    <br />
\int_{y=-2}^2 (9-y^2) dy<br />

    since (9-y^2) is the "upper" function (since its more on the right). However I keep getting, 92/3, but the back of my book says 32/3. What am I doing wrong?
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  5. #5
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    Why are you doing 9 - y^2?

    <br />
\int_{-2}^2 (9-y^2) - 5 dy = \int_{-2}^2 (4-y^2) dy <br />
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