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Math Help - Volume by Slicing

  1. #1
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    Volume by Slicing

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

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

    x^{2} is the area of a square

    A = x^{2}

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

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

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

    V = 2x^{2}

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

    V = [32] - [0] = 32 The book says the answer is 16. Something went wrong here.
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  2. #2
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    Re: Volume by Slicing

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

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  3. #3
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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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  5. #5
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    Re: Volume by Slicing

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

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

    A = \dfrac{4x}{4}

    A = x

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

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

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

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

    How about this way?

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

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

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

    A = \sqrt{x}

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

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

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

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

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

    x^{2} is the area of a square

    A = x^{2}

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

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

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

    V = 2x^{2}

    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; January 26th 2014 at 10:47 AM.
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