Help with yet another solid of revolution question

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Homework Statement



See the attached problem.


Homework Equations



See the attached problem.


The Attempt at a Solution



I used washer method and got an inner radius of x=y^2 and an outer radius of x=y+2, I calculated my upper limit as being 4 and my lower limit as being 0. The answer is 72π/5, but I can't seem to get that answer. Are my limits of integration wrong and/or did I use the wrong method?
 

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  • #2
haruspex
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That method works for me. Please post your detailed steps.
 
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V=π∫(y+2)^2 - (y^2)^2 dy, from 0 to 4
V= -2032π/15, which is obviously not the correct answer since volume can't be negative etc.
 
  • #4
haruspex
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V=π∫(y+2)^2 - (y^2)^2 dy, from 0 to 4
V= -2032π/15, which is obviously not the correct answer since volume can't be negative etc.
Right integrand, wrong range. x goes from 0 to 4. what's the range for y?
 
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Right integrand, wrong range. x goes from 0 to 4. what's the range for y?

I set √x=x-2, and solved the quadratic and got x=4,1 (I used the 4 as my upper limit and used 0 as my lower limit since it seemed that 0 was the lower limit from how the graph looked). Also how exactly do I go about calculating the limits of integration in regards to y?
 
  • #6
haruspex
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I set √x=x-2, and solved the quadratic and got x=4,1 (I used the 4 as my upper limit and used 0 as my lower limit since it seemed that 0 was the lower limit from how the graph looked).
Yes, but you are integrating with respect to y, so your bounds must be bounds on y, not x.
Also how exactly do I go about calculating the limits of integration in regards to y?
Sketch the curves and see where they cross.
 
  • #7
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Yes, but you are integrating with respect to y, so your bounds must be bounds on y, not x.

Sketch the curves and see where they cross.

I set the equations equal and I get my upper limit as 4 with respect to y and my lower limit as 1 with respect to y, however I still don't get the correct answer which is 72π/5.
 
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So I finally realized I was supposed to solve y^2=y+2 and get an upper and lower limit of 2 and -1 respectively. After plugging in those two values into the integrand, I finally ended up getting the correct answer. Thanks for the help.
 

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