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Homework Help: Volume Bounded by Cylinder and Plane

  1. Dec 10, 2008 #1
    We need to find the volume of the solid bounded by the cylinder with the equation
    z^2 + y^2 = 4 and the plane x + y = 2, in the first octant (x,y,z all positive).

    Firstly, I am trying to visualize the graphs. From what I can tell, the cylinder is centered around the x-axis and has a radius of 2. It seems move along the x-axis infinitely. The plane intersects the cylinder at y=2, z=0. Visualizing this is tough and I haven't been able to find a graphing program that is sufficient.

    Now, we have performed double integrals of solids over general two dimensional regions. I just cannot even think of where to start with this problem. Any help setting me in the right direction would be extremely helpful!
     
  2. jcsd
  3. Dec 10, 2008 #2

    tiny-tim

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    Hi MichaelT! :smile:

    In a case like this, use either vertical or horizontal slices …

    I suggest vertical, perpendicular to the x-axis …

    what is the boundary of each slice? :smile:
     
  4. Dec 11, 2008 #3
    Ok this is how I went about this. I used a double integral of the cylinder over the plane. I solved the equation of the cylinder for z, and that was the integrand. This was integrated over [0,2] X [0, 2-y] , dx dy.

    After some lengthy calculations (and a very helpful table of integrals :approve:) I found the volume of the solid bounded by the cylinder and the plane to be 2pi - 8/3

    I still need to check this over. I do have one question though, something I am not sure I did correctly. The integral of (4-y2)1/2dx = x(4-y2)1/2 Is this correct?
     
  5. Dec 12, 2008 #4

    tiny-tim

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    Yes, that's right. :smile:

    (but if you want us to check the rest, you'll need to type it out in full, so we can see it! :wink:)
     
  6. Dec 12, 2008 #5

    HallsofIvy

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    Have you studied polar and cylindrical coordinates? Since this is a cylinder, cylindrical coordinates should simplify the calculations.
     
  7. Dec 12, 2008 #6
    I'm pretty sure I got it right, since a classmate of mine solved for the volume using a triple integral and we both got the same answer. So I don't think I need it checked, but if anyone wants to see the whole thing I would be happy to post it.

    As to polar and cylindrical coordinates....We have studied them but not to the extent that I would feel confident using them for this problem. Thanks for the tip though, I am always interested in other approaches to problems :biggrin:
     
  8. Dec 17, 2008 #7
    hey i would like to see how to do this problem ..can someone post a worked out solution or give me step by step directions how to complete it so i can try myself. I'm having trouble setting the problem up as a triple integral. After this point i should be good.

    Thanks!!!!!!
     
    Last edited: Dec 17, 2008
  9. Dec 17, 2008 #8
    Hey negat1ve!

    I haven't done this problem with a triple integral, but I did do it with a double integral. If you would like to use that approach (a bit easier, as long as you have an integration table handy :wink:)....Think of the plane x + y = 2 in two dimensions. What shape does this form in the first quadrant? (since we are dealing with all x, y, and z positive)
     
  10. Dec 17, 2008 #9
    yeah im graphing the equation it looks like a diamond or a parallelogram
     
  11. Dec 17, 2008 #10
    Hmmm well in 3 dimensions the plane x + y = 2 is a plane. But if you want to look at it in 2 dimensions, just plot some points in the first quadrant of the xy coordinate system. i.e when y=0, x=2.

    If you take the points and connect them, a common geometric shape will be bounded between the x axis, y axis, and the line. This will help you find the bounds of integration.
     
  12. Dec 17, 2008 #11
    hey ok so it makes a right triangle in the first quad right
     
  13. Dec 17, 2008 #12
    Sure does! Now think about the bounds of this triangle. Have you studied x-simple and y-simple regions? You can think of the triangle as both y-simple and x-simple because it ends in either a straight line or a point in both the x and y directions. So it is up to you to choose. In one direction the bounds will be integers, and in the other it will be a function.

    So what will the integrand be? It will help you determine which way to think about the triangle.
     
  14. Dec 17, 2008 #13
    So i want to use

    ok so would this be my integrand after solving the cylinder for z

    [tex]\int\int(4-y[/tex]2)1/2dxdy


    and i would integrate over the reigion R where

    R= [0,2-y] x [0,2]????

    I however see that you posted your region R rectangle opposite of what i stated here.

    Can you explain this?
     
  15. Dec 17, 2008 #14
    So basically i come down to doing this nasty integral....

    [tex]\int[/tex](4-y2)1/2 2-y dy

    Am i on the right track....

    What is this integration by parts?
     
  16. Dec 18, 2008 #15

    tiny-tim

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  17. Dec 18, 2008 #16
    Well am I at least proceeding correctly? Should I have come up to this integral to even have to calculate. If so how can i solve this integral?

    If not what did i do wrong can you please explain....

    I never said it should be done using integration by parts.... i was just saying it looked like it might be able to be solved that way... It cant be done using a u substitution. Is this where a table of integrals would come in handy:confused:

    Please im serious about wanting to know how to solve this problem....
     
  18. Dec 18, 2008 #17

    tiny-tim

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    I'm not following your formula …

    take either horizotal or vertical slices …

    I suggest vertical slices of thickness dx, each of which should then be a triangle and an arc-sector of a circle. :smile:
    I never said you said it should! … :biggrin:
     
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