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Integral using disk/washer method

  1. Jun 16, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid generated by y = [itex]\sqrt{x}[/itex] bounded by y = 0, and x = 3
    and revolved about the line x = 3

    2. Relevant equations

    volume generated by a disk/washer about the y-axis.


    3. The attempt at a solution

    I have pi(integral from 0 -> [itex]\sqrt{3}[/itex] of (y2)2dy)

    I'm thinking maybe my bounds change? I'm also pretty sure I don't use the Washer Method because I don't have an inner radius, I have a 3 dimensional half circle lying on top of the x-axis.
     
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  3. Jun 16, 2011 #2

    Dick

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    Did you pay attention to the "revolved about the line x = 3" part? You aren't revolving around x=0. There is an inner radius.
     
  4. Jun 16, 2011 #3
    I got the answer while playing around with the function, I did pay attention to my axis of revolution, but I don't see how it cuts out anything to be taken away; I don't see why my integrand is ( 3 - y^2)^2 my radius is 3 but if I draw a slice perpendicular to my axis of revolution, x = 3, I don't see it cutting anything out after you revolve it around the line.
     
  5. Jun 16, 2011 #4

    Dick

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    The outer radius is 3, right? It's the distance between x=0 and x=3. The inner radius is the distance between the curve and x=3. It looks like a washer to me, not a disk.
     
  6. Jun 17, 2011 #5
    Okay, I have an additional qualm with another problem I feel like asking. I have to integrate y = 2(x^2) bounded by y = 0 and x =2 and I get my volume = 0, how is this at all possible? That my volume is zero, doesn't that mean that one of my representitive elements are zero? By representitive elements, i'm attempting to say height, width, or length.

    nevermind, I can see that the volume carved out by my function and the left over inner radius volume are equal. Or am I wrong?
     
    Last edited: Jun 17, 2011
  7. Jun 17, 2011 #6

    HallsofIvy

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    No, it's a disk. You seem to be thinking about rotating the area between x= 0 and the curve around x= 3. The problem talks about the region bounded by x= y2, y= 0, and x= 3, rotated around x= 3. For a given y, that is a disk having as radius, the line from x=y2 to x= 3, a length of 3- = 3 y2. The area of such a disk would be [itex]\pi (3- y^2)^2[/itex]. The volume of such a disk, of thickness dy, would be [itex]\pi (3- y^2)^2 dy[/itex]. y goes from 0 up to [itex]\sqrt{3}[/itex].
     
  8. Jun 18, 2011 #7
    Okay, well i'm floundering with this and I want to move on to try my hand at shell method and other applications of the integral, work, hydrostatics, centers of mass/moments of inertia. And while they all utilize the same style of integration, I can't help but feel like they might help me see whatever it is i'm not seeing with the washer method. Should I just stay where I am and continue with Washers, i'm not really sure but i'm chewing up all my time on this. I've reviewed the section, reading through it and i've looked at a lot of examples on the internet and youtube which isn't helping all too much. Any additional information out there might be helpful, if anyone knows.
     
  9. Jun 18, 2011 #8

    Dick

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    Halls is right. I was wrong. I was reading 'y=0' as 'y axis'. That's wrong. It is a disk. Pay attention to Halls post, not mine.
     
  10. Jun 18, 2011 #9
    I'm clear on this example, really I do understand it. I moved on from that problem and i'm just getting maybe 15% of the problems i've answered actually correct, it's just something i'm not seeing or phrasing correctly. And instead of just having someone solve all my problems, I want to utilize external sources to learn more because my book isn't helping and everything else i've tried isn't working. I just need something additional, do you know where I can read about the washer method, all I want to practice is setting up my inner and outter radii. My book and all these internet sources, the ones i've tried so far, do all the same examples. They aren't helping me. Do you know where I can try more of these or get step-by-step walk throughs on more difficult setups?

    I appreciate your efforts with me here, but i'm too stubborn to give up on this.
     
  11. Jun 18, 2011 #10

    Dick

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    I would say post an example of a problem you are having trouble with here. Probably not everyone will give as bad an advice as I did.
     
  12. Jun 18, 2011 #11
    alright sure, y= 3/(1+x) bounded by y = 0 x = 0 and x = 3, this really shapes my region quite nicely. Integrate about the line y = 4, I have my outer radius as R(x) = 4, the total distance from my axis of revolution to the x-axis, the lower bound of my function y = 3/(1+x). I see my inner radius as the area in between y =4 and y = 3/(1+x) should I integrate to find the area between y = 4 and y = 3/(1+x) to find my inner radius and then subtract that away from the Outer Radius?
     
  13. Jun 18, 2011 #12

    Dick

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    Mmm. No, that doesn't sound right. The outer radius is 4, sure, The inner radius isn't the area between anything. The inner radius is 4-3/(1+x), isn't it? It's the distance between y=4 and y=3/(1+x). I would just integrate pi*(outer^2-inner^2) dx.
     
  14. Jun 19, 2011 #13
    Wouldn't that just mean the area under y = 4 minus the area under y = 3/(1+x). Don't I want the area under 3/(1+x)? Which would be the area under y = 4 minus the area in between y = 4 and y = 3/(1+x)

    Disregard my previous statement prior to my edit, if you got to it. I was mistaken.
     
  15. Jun 19, 2011 #14
    Phew, okay i've gotten 4pi ( 3 - ln(4) ) I don't think that's right... what are you getting, using your outer radius as 4 and your inner radius as 4 - 3/(1+x) ?
     
  16. Jun 19, 2011 #15

    Dick

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    The answer IS pi(48ln(2) - (27/4). But (4(1+x) - 3)/(1+x) is NOT 1. The integrand isn't 15 either. From what you've shown I'm still not sure where you are going wrong. Can you show the details? If I integrate pi*(4^2-(4-3/(1+x))^2) I definitely get the correct answer. What are you doing?
     
  17. Jun 19, 2011 #16
    I forgot to square my radii, i'm in the process of doing that now. What were your limits of integration?

    For my integrand, after squaring and simplifying and before I do a u-substitution, I have 24(x) + 15 / ( x^2 + 2x +1 )
     
    Last edited: Jun 19, 2011
  18. Jun 19, 2011 #17

    Dick

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    x goes from 0 to 3, I would hope you would be using those limits.
     
  19. Jun 19, 2011 #18
    well if I do a u-substitution, I normally change my limits and don't change them back.
    So after I simplify I get for my integrand, 24(x) + 15 / ( x^2 + 2x + 1 ) I let u = x^2 + 2x + 1 so that du = ( 2x + 2 ) dx and now dx = du / ( 2x + 2 ) as x: 0 -> 3 u: 1 -> 16 and I am left with something I can't work with.
     
    Last edited: Jun 19, 2011
  20. Jun 19, 2011 #19

    Dick

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    It would have been easier to work with if you had done less simplification. Try working from 24/(x+1)-9/(x+1)^2. Split the integral into two parts.
     
  21. Jun 20, 2011 #20
    alright, why can't I get the same answer using what I had? I don't understand the difference between simplifying and splitting up my integrals because I tried what you had here and it feels even more complicated than what I had before.
     
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