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Math Help - 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.  y=1/x^2 ,
     y = 0 , x=1, x=5 about  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  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  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  r = y -3 ?
    The inner circle is a constant. \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 = y+3 = \frac{1}{x^2}+3
    inner radius = 3

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

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

    - Hollywood
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