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Thread: 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 $\displaystyle V(x)=\int_a^b A(x)\,dx$, where $\displaystyle A(x)$ 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 $\displaystyle S(x)=4\sqrt{1-\frac{x^2}{9}}=\tfrac{4}{3}\sqrt{9-x^2}$

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

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

    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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