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Math Help - Volume of a solid; please check

  1. #1
    Member Marconis's Avatar
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    Volume of a solid; please check

    Okay, so, I think I am getting the hang of these problems. Would anyone care to check? All we are required to do for the homework is set up the integral.

    x=\sqrt{y}, x=0, y=1, ; about the x-axis

    I set up my graph. It is the graph of a parabola with line y=1 going horizontally through it, and x=0. Seeing this, and the fact that it was solved in terms of y, I set up the integral in terms of a R and L volume. Integral is 0 to 1.

    I got:

    2\pi \int |x-1| (\sqrt{y})dy

    Am I correct?
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  2. #2
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    skeeter's Avatar
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    Quote Originally Posted by Marconis View Post
    Okay, so, I think I am getting the hang of these problems. Would anyone care to check? All we are required to do for the homework is set up the integral.

    x=\sqrt{y}, x=0, y=1, ; about the x-axis

    I set up my graph. It is the graph of a parabola with line y=1 going horizontally through it, and x=0. Seeing this, and the fact that it was solved in terms of y, I set up the integral in terms of a R and L volume. Integral is 0 to 1.

    I got:

    2\pi \int |x-1| (\sqrt{y})dy

    Am I correct?
    I'm afraid not ... when integrating w/r to a single variable, the integrand needs to be solely a function of that variable.

    here are two set ups rotating the described region about the x-axis ...

    using washers w/r to x ...

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

    using cylindrical shells w/r to y ...

    \displaystyle V = 2\pi \int_0^1 y \cdot \sqrt{y} \, dy
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  3. #3
    Member Marconis's Avatar
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    Shoot, three things I accidentally typed in:

    For one, in my absolute value, I meant to put y. Secondly, I also accidentally put 1 for k because of the y=1. Really, though, k=0. Thirdly, it should be |0-y|, not |y-0| If I didn't make three (critical) mistakes, then I would have gotten your second answer.

    Thank you!
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