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Math Help - Area Volume

  1. #1
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    Area Volume

    1. 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.


    y= sqrt(30-x) and y=-x i inserted in graphing calculator but no graph???

    2. 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.


    I graphed and is from 0 to 3.

    y= 4x^3/3 and y=1/3x(x^2+9)

    i got 54. did i do it right?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    I would integrate with respect to y, becaus if you did it with respect to x

    x=+/- sqrot(30-y^2) is not a function but a relation.


    Am I right?
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  3. #3
    No one in Particular VonNemo19's Avatar
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    For number 2

    4x^2=X^2+3

    3x^2=3

    x^2=1 so x=1,-1

    so the interval to look at is [-1,1] Right?
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  4. #4
    Senior Member Pinkk's Avatar
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    You don't need a graphing calculator for this. It's best to have your equations for the first problem as x=f(y). So we have x=30-y^{2} and x=-y Set the two equal to each other to find the points of intersection:

    30-y^{2}=-y
    0=y^{2}-y-30
    0=(y-6)(y+5)
    y=6,y=-5

    You should be able to sketch of graph of this and know that 30-y^{2} \ge -y on the interval of [-5,6], so your integral should be:

    \int_{-5}^{6}30-y^{2}+y\,\,dy



    For the second problem, it's obvious we should have our functions as y=f(x). Again, set the two equations equal to each other:

    4x^{2}=x^{2}+3
    3x^{2}=3
    x=-1,x=1

    Again, sketching the graphs of these are easy, and you can tell x^{2}+3 \ge 4x^{2} on [-1,1], so your integral would be:

    \int_{-1}^{1}3-3x^{2}\,\,dx
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  5. #5
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    Thanks for helping guys. I really appreciated.
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