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Thread: Volume by Slicing - # 2

  1. #1
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    Volume by Slicing - # 2

    The solid lies between planes perpendicular to the x axis at $\displaystyle x = -1$ and $\displaystyle x = 1$. The cross-sections perpendicular to the x axis between these planes are squares whose bases run from the semi-circle $\displaystyle y = -\sqrt{1 - x^{2}}$ to the semi-circle $\displaystyle y = \sqrt{1 - x^{2}}$
    $\displaystyle V = \int_{a}^{b} A(x) dx$

    $\displaystyle x^{2}$ is the area of a square

    $\displaystyle A = x^{2}$

    $\displaystyle V = \int_{-1}^{1} [2 \sqrt{1 - x^{2}}]^{2}$ On the right track here?

    $\displaystyle V = \int_{-1}^{1} [4 (1 - x^{2})]$

    $\displaystyle V = \int_{-1}^{1} 4 - 4x^{2}$

    The book says the answer will come out to $\displaystyle \dfrac{16}{3}$
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    Re: Volume by Slicing - # 2

    Are you supposed to rotate this region about an axis? As it is, you have described a 2 dimensional region, so there is no volume...
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    Re: Volume by Slicing - # 2

    Same idea as your other thread Volume by Slicing.

    - Hollywood
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    Re: Volume by Slicing - # 2

    The solid lies between planes perpendicular to the x axis at $\displaystyle x = -1$ and $\displaystyle x = 1$. The cross-sections perpendicular to the x axis between these planes are squares whose bases run from the semi-circle $\displaystyle y = -\sqrt{1 - x^{2}}$ to the semi-circle $\displaystyle y = \sqrt{1 - x^{2}}$
    $\displaystyle V = \int_{a}^{b} A(x) dx$

    $\displaystyle x^{2}$ is the area of a square

    $\displaystyle A = x^{2}$

    $\displaystyle V = \int_{-1}^{1} [x][2 \sqrt{1 - x^{2}}]^{2} dx$

    $\displaystyle V = \int_{-1}^{1} [x][4 (1 - x^{2})] dx$

    $\displaystyle V = \int_{-1}^{1} [x][4 - 4x^{2})] dx$

    $\displaystyle V = \int_{-1}^{1} 4x - 4x^{2})] dx$

    $\displaystyle V = \dfrac{4x^{2}}{2}- \dfrac{4x^{3}}{3})] $

    $\displaystyle V = 2x^{2} - \dfrac{4x^{3}}{3})] $
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