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Math Help - volumes of solids using integrals

  1. #1
    Junior Member winterwyrm's Avatar
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    volumes of solids using integrals

    if the base of a solid is a square perpendicular to the x-axis enclosed by the ellipse
    (x^2)/9 + (y^2)/4 = 1


    Find it's volume

    Thanks!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by winterwyrm View Post
    if the base of a solid is a square perpendicular to the x-axis enclosed by the ellipse
    (x^2)/9 + (y^2)/4 = 1


    Find it's volume

    Thanks!
    We are finding the volume using the technique of slices.

    Recall that V(x)=\int_a^b A(x)\,dx, where A(x) is the area of each slice.

    To find out the area of each slice, we need to know the length of one side.

    Since the region is an ellipse:



    we can figure out that the distance between each side of the ellipse could be defined as S(x)=4\sqrt{1-\frac{x^2}{9}}=\tfrac{4}{3}\sqrt{9-x^2}

    Thus, the area of each square slice will be A(x)=(S(x))^2=\tfrac{16}{9}(9-x^2)

    Thus, the volume of the solid would be V(x)=\tfrac{16}{9}\int_{-3}^3 (9-x^2)\,dx=\tfrac{32}{9}\int_0^3 (9-x^2)\,dx.

    Does this make sense? If you have a question, feel free to ask.

    --Chris
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  3. #3
    Junior Member winterwyrm's Avatar
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    Thanks, took me a while to catch up to everything you did, but I definitely get it now, I just messed up in the multiplication, instead of simplifying like you did, I mixed up numbers, (I do that alot) and I have the answer, but only the number, so you can understand my frustration, I suppose.
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