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Thread: 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) $\displaystyle y=sinx$, $\displaystyle y=0$, $\displaystyle x={2}\pi$, $\displaystyle x={3}\pi$; about the y-axis

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

    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?

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

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

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

    Right?

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

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

    Thoughts?
    Last edited by Tall Jessica; Nov 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) $\displaystyle y=sinx$, $\displaystyle y=0$, $\displaystyle x={2}\pi$, $\displaystyle x={3}\pi$; about the y-axis

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

    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?

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

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

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

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

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

    $\displaystyle \int_{0}^{\pi}2\pi({5-y})({\sqrt(siny)})dx$
    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 $\displaystyle 3\pi$, but I am new to LaTeX! Thank you so much!
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