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Homework Help: Find the volume of the solid obtained by rotating the region

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
    y = 5x, y = 5[itex]\sqrt{x}[/itex] about y = 5

    2. Relevant equations
    A(x)=∏(R2-r2)

    3. The attempt at a solution
    A(x)=∏(5x)2-(5[itex]\sqrt{x}[/itex])2)
    A(x)=∏(25 x2 - [itex]\frac{10}{3}[/itex]x[itex]\frac{3}{2}[/itex])

    V=∏[itex]\int[/itex][itex]^{1}_{0}[/itex](25x2- [itex]\frac{10}{3}[/itex]x[itex]\frac{3}{2}[/itex])dx

    V=∏([itex]\frac{25}{3}[/itex]x3-[itex]\frac{4}{3}[/itex]x[itex]\frac{5}{2}[/itex]) {0,1}

    V=∏([itex]\frac{25}{3}[/itex]-[itex]\frac{4}{3}[/itex])
    V= [itex]\frac{21}{3}[/itex]∏

    Where did I go wrong? I can't figure out the about y=5. If I am correct, you would usually subtract 5 from the two radii, but since they intersect at y=5, the just flip on that intersection point and thus we don't need to find the lost area. Bad explanation, I know, maybe someone can explain it to me.

    Thanks in advance!
     
  2. jcsd
  3. Jan 21, 2012 #2

    Mark44

    Staff: Mentor

    You've made an error right off the bat. I hope you sketched the region being rotated.

    You aren't taking into account the fact that the axis of rotation is the line y = 5. The larger radius, R, is 5 - 5x. The smaller radius, r, is 5 - 5√x.

    Also, your relevant equation is missing a factor - Δx. If you substitute the values above for R and r, you should get the right result.
     
  4. Jan 21, 2012 #3
    Ok, that's what my question at the end was about. I wasn't sure if I was supposed to subtract the radii from 5. I'll try it out and see how it goes. Thanks!
     
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