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Thread: Stuck on a solid of rev problem

  1. #1
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    Stuck on a solid of rev problem

    The problem is: Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. $\displaystyle y=1/x^2 $ ,
    $\displaystyle y = 0 $, $\displaystyle x=1$,$\displaystyle x=5$ about $\displaystyle y=-3$

    This is the image I have (although mine's on paper and it shows the washer's depth decreasing) Stuck on a solid of rev problem-washer-pic-problem.jpg

    I did $\displaystyle V = \pi \int_{1}^{5} (\frac{1}{x^2})^2-(3)^2 dx $

    I've fiddled around with that a few times, but I feel like I'm now starting to go in circles. I'm hoping I could get an eyeball so I can see what I'm not visualizing correctly. I know that I need the area between y=0 and y, and I assumed the bounds were x=1 and x=5, so I figured I had enough to take the integral with that information.
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  2. #2
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    Re: Stuck on a solid of rev problem

    Is the radius of the inner circle for this problem $\displaystyle r = y -3 $ ?
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  3. #3
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    Re: Stuck on a solid of rev problem

    Quote Originally Posted by AZach View Post
    Is the radius of the inner circle for this problem $\displaystyle r = y -3 $ ?
    The inner circle is a constant. $\displaystyle \pi(3)^2*4$ where 3 is the radius, and 4=5-1 is the height.

    I think you can shift the curve 1/x^2 upwards 3 units. Find the volume of this new curve and then subtract the volume of the cylinder defining the inner circle.
    Thanks from AZach
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  4. #4
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    Re: Stuck on a solid of rev problem

    You can approach it more directly, but you have to make sure you put in the *radius*, not the x value:

    outer radius = $\displaystyle y+3 = \frac{1}{x^2}+3$
    inner radius = $\displaystyle 3$

    Area of "washer" = $\displaystyle \pi \left(\frac{1}{x^2}+3\right)^2 - \pi (3)^2$

    $\displaystyle V = \pi \int_{1}^{5} \left(\frac{1}{x^2}+3\right)^2-(3)^2\,dx $

    - Hollywood
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