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Math Help - Rotating two curves

  1. #1
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    Rotating two curves

    1) Rotate the graph 4x^5 about the line x = 1 and calculate the volume of the generated figure.



    I assumed that the radius r = 1 - 4x^5 and then integrated from 0 to 1 and was going to double it

    A = \pi (1-4x^5)^2
    A = \pi (1-8x^5+16x^{10}
    \int^1_0 A(x)dx = x-\frac{4}{3}x^6 + \frac{16x^{11}}{11}|^1_0

    Which gives you 1-\frac{4}{3}+\frac{1}{11}

    2) y = 5x, y = 5\sqrt{x} Rotated about y = 5. Find the volume

    Washer method:

    A(x) = \pi (5-5x)^2-\pi (5-\sqrt{x})^2
    \pi \int^1_0 (25-50x+25x^2)-(25-50\sqrt{x}-25x)
    \pi \int^1_0 -75x+25x^2+50\sqrt{x}

    But that evaluates to 0, which doesn't make sense (and is also wrong).

    Where did I go wrong on these?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    If you are rotating R1 about x= 1 use the shell method

    Height is 4*x^5 and the distance from x to 1 is 1-x


    you have (2pi)integral[(1-x)4x^5dx]
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  3. #3
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    Quote Originally Posted by Open that Hampster! View Post
    1) Rotate the graph 4x^5 about the line x = 1 and calculate the volume of the generated figure.

    2) y = 5x, y = 5\sqrt{x} Rotated about y = 5. Find the volume
    (1) I assume the x-axis is the lower boundary for the region ...

    using cylindrical shells ...

    V = 2\pi \int_0^1 (1-x)(4x^5) \, dx

    using washers ...

    y = 4x^5 ... x = \left(\frac{y}{4}\right)^{\frac{1}{5}}

    V = \pi \int_0^4 \left[1 - \left(\frac{y}{4}\right)^{\frac{1}{5}}\right]^2 \, dy


    (2) using washers ...

    R(x) = 5 - 5x

    r(x) = 5 - 5\sqrt{x}

    V = \pi \int_0^1 (5-5x)^2 - (5 - 5\sqrt{x})^2 \, dx
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  4. #4
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    We never learned the shell method, so I'll give that a go.

    As for number 2, that's what I did skeeter, but I didn't get the right answer. Was it an Algebraic mistake?
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