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Math Help - Reverse Integrals. (Check my work?)

  1. #1
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    Reverse Integrals. (Check my work?)

    I'm doing some work - yes, on Thanksgiving - and I've been doing good, but the paper reversed things on me! lol... I have to BUILD an integral, not solve it! Anywho, please check my work!

    The problem(s) state(s):
    Set up, but do not evaluate, for the volume of the solid obtained by rotating the region bounded by the given curves around the specified axis.

    1) y=sinx,  y=0, x={2}\pi, x={3}\pi; about the y-axis

    2) x=\sqrt{sin y}, 0 \leq y \leq \pi, x=0; about y=5

    Okay, so for the first problem:

    Looking at the graph and rotating it, I get a nice little donut. Since I need my approximating squares to touch BOTH functions, I have to use the Shell Method......right?

    \int_{{2}_\pi}^{{3}^\pi}2\pi{r}{h})dx

    So, I would take my h = sin x and my r = x, giving me:

    \int_{{2}_\pi}^{{3}^\pi}2\pi({sin x})({x})dx

    Right?

    For the second problem, it's the same formula, but with r=5-y and h=\sqrt(siny), giving me:

    \int_{0}^{\pi}2\pi({5-y})({\sqrt(siny)})dx

    Thoughts?
    Last edited by Tall Jessica; November 26th 2010 at 03:41 AM. Reason: missing info
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  2. #2
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    Quote Originally Posted by Tall Jessica View Post
    I'm doing some work - yes, on Thanksgiving - and I've been doing good, but the paper reversed things on me! lol... I have to BUILD an integral, not solve it! Anywho, please check my work!

    The problem(s) state(s):
    Set up, but do not evaluate, for the volume of the solid obtained by rotating the region bounded by the given curves around the specified axis.

    1) y=sinx,  y=0, x={2}\pi, x={3}\pi; about the y-axis

    2) x=\sqrt{sin y}, 0 \leq y \leq \pi, x=0; about y=5

    Okay, so for the first problem:

    Looking at the graph and rotating it, I get a nice little donut. Since I need my approximating squares to touch BOTH functions, I have to use the Shell Method......right?

    \int_{{2}_\pi}^{{3}^\pi}2\pi{r}{h})dx

    So, I would take my h = sin x and my r = x, giving me:

    \int_{{2}_\pi}^{{3}^\pi}2\pi({sin x})({x})dx

    Right?
    yes, that is correct- well, except that your upper limit on the integral should be 3\pi, not 3^\pi!

    For the second problem, it's the same formula, but with r=5-y and h=\sqrt(siny), giving me:

    \int_{0}^{\pi}2\pi({5-y})({\sqrt(siny)})dx
    Yes, that is correct.

    Thoughts?
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  3. #3
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    Thank you! I am aware that the upper limit is 3\pi, but I am new to LaTeX! Thank you so much!
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