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Homework Help: Usnig clylindrical coordinates to find the volume

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data
    The solid that is inside the surface r^2+z^2=20 and below the surface z=r^2


    2. Relevant equations



    3. The attempt at a solution

    I m confused how to solve this problem of get a projection and their intersection. Their intersection gives me a equation of
    r^4+r^2-20=0 then i m stuck.

    Please help :)
     
  2. jcsd
  3. Jul 19, 2010 #2

    vela

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    When you say it's inside the first surface, do you mean it's between z=0 and the surface?

    Try factoring r4+r2-20 to solve for r2.
     
  4. Jul 19, 2010 #3
    This is the exact question i have copied from the book.
    The solution of the equation gives me r=+-5^1/2 could you please tell me how to proceed with these types of problems.
    thanks
     
  5. Jul 19, 2010 #4

    vela

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    Oh, I just noticed the first surface is a sphere, so talking about its inside isn't vague. Have you sketched the surfaces?

    You solved for r incorrectly. You probably got the signs switched when you factored the polynomial.
     
  6. Jul 19, 2010 #5
    oh sorry yes i was wrong it should be r=+-2. I m confused about the sketch r^2+z^2=20 is a ellipsoid and z=r^2 is a elliptic paraboloid and i m asked to get the solid that is below the elliptic paraboloid ? If it is how can i get the limits of my integration ?
     
  7. Jul 21, 2010 #6

    vela

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    The easiest approach would be to calculate the volume of the sphere and then subtract out the volume of the protrusion of the paraboloid into the sphere. You found that the intersection of the two surfaces is a circle of radius 2, so the projection of the protrusion onto the xy-plane is a circle of radius 2 centered on the origin. From this, you can find the limits for r and θ. The limits for z are then given by the two surfaces to find the volume of the protrusion.
     
  8. Jul 22, 2010 #7
    Thanks i have got the solution by your method. But i m confused how the book solved the problem. They have formed an integral like this
    [tex]
    \int_0^{2\Pi} \int_0^2 \int_{-{\sqrt{20-r^2}}}^{r^2} rdz dr d\theta + \int_0^{2\Pi} \int_0^{\sqrt{20}} \int_{-{\sqrt{20-r^2}}}^{\sqrt{20-r^2}} rdz dr d\theta [/tex]

    could you please elaborate how this integral is formed.


    regards
     
    Last edited: Jul 22, 2010
  9. Jul 22, 2010 #8

    vela

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    That's not correct. The lower limit for r in the second term should be 2.

    [tex]\int_0^{2\pi} \int_0^2 \int_{-{\sqrt{20-r^2}}}^{r^2} r \,dz\,dr\,d\theta + \int_0^{2\pi} \int_2^{\sqrt{20}} \int_{-\sqrt{20-r^2}}^{\sqrt{20-r^2}} r \,dz\,dr\,d\theta[/tex]

    It's not clear to me what's confusing you. It seems if you could solve the problem the other way, you should be able to come up with this solution as well. Do you have a specific question about this expression?
     
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