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Thread: Volume by Slicing

  1. #1
    MHF Contributor Jason76's Avatar
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    Volume by Slicing

    The solid lies between planes perpendicular to the x axis at $\displaystyle x = 0$ and $\displaystyle x = 4$. The cross-sections perpendicular to the x axis on the interval $\displaystyle 0 \leq x \leq 4$ are squares whose diagonals run from the parabola $\displaystyle y = -\sqrt{x}$ to the parabola $\displaystyle y = \sqrt{x}$. Find the volume
    First step?

    $\displaystyle V = \int_{a}^{b} A(x) dx$

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

    $\displaystyle A = x^{2}$

    $\displaystyle V = \int_{0}^{4} [2 \sqrt{x}]^{2} dx$

    $\displaystyle V = \int_{0}^{4} 4x dx$

    $\displaystyle V = \dfrac{4x^{2}}{2}$

    $\displaystyle V = 2x^{2}$

    $\displaystyle V = [2(4)^{2}] - [2(0)^{2}] $

    $\displaystyle V = [32] - [0] = 32 $ The book says the answer is 16. Something went wrong here.
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    Re: Volume by Slicing

    The length of the diagonal of the square is $\displaystyle 2\sqrt{x}$, so the length of a side is $\displaystyle \frac{1}{\sqrt{2}}$ times $\displaystyle 2\sqrt{x}$, which is $\displaystyle \sqrt{2x}$. So the area is 2x, not 4x.

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    MHF Contributor Jason76's Avatar
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    Re: Volume by Slicing

    How could we visualize this?
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    Re: Volume by Slicing

    The square is rotated so that the diagonal is in the x-y plane, so the sides of the square are 45 degrees from horizontal. I imagine it starting as a point (x=0) and growing as it comes toward me (x increasing) until it reaches x=4, where the vertices are (4,2,0), (4,0,2), (4,-2,0), and (4,0,-2).

    Is that what you meant by "How could we visualize this?"

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    MHF Contributor Jason76's Avatar
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    Re: Volume by Slicing

    The solid lies between planes perpendicular to the x axis at $\displaystyle x = 0$ and $\displaystyle x = 4$. The cross-sections perpendicular to the x axis on the interval $\displaystyle 0 \leq x \leq 4$ are squares whose diagonals run from the parabola $\displaystyle y = -\sqrt{x}$ to the parabola $\displaystyle y = \sqrt{x}$. Find the volume
    $\displaystyle V = \int_{a}^{b} A(x) dx$

    $\displaystyle A = [\dfrac{2 \sqrt{x}}{2}]^{2}$

    $\displaystyle A = \dfrac{4x}{4}$

    $\displaystyle A = x$

    $\displaystyle V = \int_{0}^{4} x dx$

    $\displaystyle V = \dfrac{x^{2}}{2} $

    $\displaystyle V = \dfrac{(4)^{2}}{2} - 0$

    $\displaystyle V = \dfrac{16}{2} = 8$ Again, answer in book is 16. Where am I going wrong here?

    How about this way?

    $\displaystyle V = \int_{a}^{b} A(x) dx$

    $\displaystyle A = [\dfrac{2 \sqrt{x}}{2}]$

    $\displaystyle A = \dfrac{2\sqrt{x}}{2}$

    $\displaystyle A = \sqrt{x}$

    $\displaystyle V = \int_{0}^{4} \sqrt{x} dx$

    $\displaystyle V = \int_{0}^{4} x^{1/2} dx$

    $\displaystyle V = \dfrac{x^{3/2}}{\dfrac{3}{2}}$

    $\displaystyle V = \dfrac{2}{3}x^{3/2}$

    $\displaystyle V = \dfrac{2}{3}(4)^{3/2} - 0 = \dfrac{16}{3}$ Also wrong

    Finally, how about this way?

    The solid lies between planes perpendicular to the x axis at $\displaystyle x = 0$ and $\displaystyle x = 4$. The cross-sections perpendicular to the x axis on the interval $\displaystyle 0 \leq x \leq 4$ are squares whose diagonals run from the parabola $\displaystyle y = -\sqrt{x}$ to the parabola $\displaystyle y = \sqrt{x}$. Find the volume
    $\displaystyle V = \int_{a}^{b} A(x) dx$

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

    $\displaystyle A = x^{2}$

    $\displaystyle V = \int_{0}^{4}[x] [2 \sqrt{x}]^{2} dx$

    $\displaystyle V = \int_{0}^{4} [x][4x] dx$

    $\displaystyle V = \dfrac{4x^{2}}{2} $

    $\displaystyle V = 2x^{2} $

    $\displaystyle V = 2(4)^{2} - 0 = 32$

    OK, this way looks better. But still not 16. Maybe the diagonal thing needs to be put in.
    Last edited by Jason76; Jan 26th 2014 at 09:47 AM.
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