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Triple Integral Problem

  1. Sep 20, 2016 #1
    Hey guys i've been working some triple integration problems and i've stumbled across a question that i'm having problems with
    So from the picture below my solution is incorrect and I can't seem to figure out where I went wrong. Is my setup for the integrals correct or is that where i've made my mistake.


    1. The problem statement, all variables and given/known data

    Q15 in the posted image
    (Further Problems 15 pg 658 in the solutions posted)

    2. Relevant equations


    3. The attempt at a solution
    So I found the intersection of the two surfaces z = sqrt(x^2+y^2) and z^2 = 4(x^2+y^2) and found this to be z=2a.
    On the y-z plane the solid basically leaves a region bounded by z=2|y| and the sphere. Since z=2a then y = a. I used this to calculate phi as arctan(a/2a) and rho as 2asec(phi).
    In the x-y plane is basically a circle of radius a...I used this to get my range for theta which was 0 to 2π

    Then I setup the triple integral as follows(With 1 as the integrand because it's a Volume calculation for the solid formed)

    My Right most integral for rho goes between 0 and 2asec(phi)
    Middle goes from 0 to arctan(0.5)
    Left most goes from 0 to 2 pi

    My answer was 2a^3/π
     

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    Last edited by a moderator: Sep 21, 2016
  2. jcsd
  3. Sep 20, 2016 #2

    BvU

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    Can't read it. Please type what you want us to help you with.
    And read the guidelines.
     
  4. Sep 20, 2016 #3
    ^Sorry About that...I've fixed it
     
  5. Sep 20, 2016 #4

    BvU

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    How hard is it to type the problem statement ? Anyway:

    In the x-y plane ##z=0## so x and y are 0 too. A circle of radius 0, therefore...:rolleyes:
     
  6. Sep 20, 2016 #5
    lol I thought it would have been easier to understand straight from the textbook.
    But it said z>=0 so at 0 wouldn't the solid be lying on the x-y plane?
    Because if I substitute z= 0 into the equation of the sphere then you get a circle
     
  7. Sep 20, 2016 #6

    SammyS

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    It is possible to post an image directly in a post like ...

    screenshot-2016-09-20-17-02-11-png.106284.png

    I suppose the following is the textbook solution.
    screenshot-2016-09-20-17-01-48-png.106285.png

    You need to determine the intersection of the cone and sphere, not the intersection of the sphere and x-y plane.
     
  8. Sep 20, 2016 #7
    Hmm okay so I solved for z already which I got as 2a ---> Resulting in y = a. Substituting those values back into the equation for the cone I get x = 0.
    So the curves intersect in the yz plane?

    But if you substitute z=2a in that same eqn for the cone you get a circle of radius a. So the region of intersection is a circle that's projected on the xy plane?
     
  9. Sep 20, 2016 #8

    SammyS

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    How does that give y = a?

    The intersection of those two surfaces is at z = 2a. The intersection is indeed a circle.
     
  10. Sep 20, 2016 #9
    Well I decided to use spherical coordinates from the start and normally to get my bounds for rho and phi I set x=0 to get the projection on the yz plane. So from the cone's equation I ended up with z=2y and I substituted z=2a into that which gave y=a.
     
  11. Sep 20, 2016 #10

    SammyS

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    y = a if you set x = 0 and z = 2a. More generally, y = ±a for this case

    Also, x = ±a, if you set y = 0 and z = 2a.
     
  12. Sep 20, 2016 #11
    Ohh I see.
    From there I graphed the region onto the yz plane.
    So its the graph z= 2|y| intersected by the sphere. I then used a bit of trigs to get that cos(phi) = 2a/rho giving rho as 2asec(phi) and tan(phi) as 1/2 -> phi = arctan(1/2)


    So my bounds were:

    0<rho<2asec(phi)
    0<phi<arctan(1/2)
     
  13. Sep 20, 2016 #12

    LCKurtz

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    The equation of the sphere in spherical coordinates is not ##\rho = 2a\sec \phi##. So your "bit of trigs" is wrong. Show us how you get ##\rho## for the sphere by expressing its equation in spherical coordinates.

    As an aside, since the equation of the sphere isn't particularly simple in spherical coordinates (as you will see when you work it out), this problem is actually easier to set up in cylindrical coordinates.
     
  14. Sep 21, 2016 #13
    This was the diagram I used to get phi from the yz plane.

    I just did cos(phi) which is 2a/rho and found rho from that to get 2asec(phi). But I understand where you're coming from...if I change the eqns I get z = 2r and z=sqrt(2a^2-r)+a.

    But for spherical...I can't use the diagram like that to get rho since its the distance from the origin to a point on the sphere?

    EDIT: I think i saw where I went wrong. I only set x of the cone to 0.
    Setting x=0 in the cylinder gives y(sq)+(z-a)(sq) = 2a(sq)
    expanding and simplifying gives y^2+z^2-2az = a^2.

    From the line z=2y, I can make a substitution into that and solve for either y or z but that would get messy...I guess this is why I should have used cylindrical coordinates then???
     

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    Last edited: Sep 21, 2016
  15. Sep 21, 2016 #14

    BvU

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    i hope you switched the z and the y in the picture by accident ?
     
  16. Sep 21, 2016 #15
    Woa haha yeah I did sorry about that
     
  17. Sep 21, 2016 #16

    BvU

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    Now: is there any difference with the corresponding picture of the x-z plane ? Or, for that matter, with the picture of the r-z plane (##\ r^2=x^2+y^2\ ##) ?
     
  18. Sep 21, 2016 #17

    LCKurtz

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    You haven't responded to this. You have ##x^2+y^2+(z-a)^2 = 2a^2## given as the equation of the sphere. You don't need any geometry to express that in spherical coordinates. Expand it out and use the appropriate equations to change ##x^2+y^2+z^2## and ##z## to spherical coordinates. Then solve the resulting equation for ##\rho##. Do it and show your equations here. Getting the proper equation for ##\rho## is all you have left to do to set it up in spherical coordinates.
     
  19. Sep 22, 2016 #18
    Sorry about that...didn't get to respond because of assignments etc for other courses.
    But I worked that out already but I didnt post it so here it is:

    x(sq)+y(sq) + z(sq)-2az+a(sq) = 2a(sq)
    rho(sq)-2az = a(sq)
    rho(sq)-2a(rho)(cos(phi)) = a(sq)
    rho(sq)-2a(rho)(cos(phi)) -a(sq) = 0

    Using the quadratic formula I got rho = acos(phi)+/-asin(phi)
     
  20. Sep 22, 2016 #19

    SammyS

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    Those look like arcsine and arccosine functions.

    Probably better as rho = (a)cos(phi) ± (a)sin(phi) .

    However, I'm pretty sure you made a mistake in applying the quadratic formula.
     
  21. Sep 22, 2016 #20
    oops made a sign error

    Im getting cos(phi) +/- sqrt(1+cos^2(x) )

    And btw is it wrong to use geometry to get rho and phi in certain cases???

    For example this qn(Not switching the topic of the thread btw...just found a question where I think geometry may be applicable)
     

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