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Triple integrals again

  1. May 6, 2009 #1
    1. The problem statement, all variables and given/known data

    evaluate the integral y2z2dV over W, which is the region bounded by x = 1 - y2 - z2 adn the plane x = 0

    2. Relevant equations



    3. The attempt at a solution

    since x = 0, that makes y2 + z2 = 1, unit circle in the yz plane right?

    so would the answer be the area of the circle times y2z2
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 6, 2009 #2

    tiny-tim

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    uhh? the answer is a number … how can ∫∫∫ y2z2dV have any y or z in it? :confused:
     
  4. May 6, 2009 #3

    Cyosis

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    Draw the region. You will see that it is a paraboloid with a maximum at y=z=0 and x=1. The paraboloid is cut off at x=0 so the region looks like the top of paraboloid. Which is not just a circle.
     
  5. May 11, 2009 #4

    Cyosis

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    Re: Which method is best for triple integration

    Those integration limits are not correct. For A) if your integration order is dzdydx there will be a z left in your answer. Every other order of integration has a similar problem so it can't be right.

    For B) you don't integrate over x,y and z, but over r, theta and x. This means you've to find limits for r, theta and x. The polar coordinates in this case would be [itex]x=x, y=r \cos \theta, z=r \sin \theta,y^2+z^2=r^2[/itex]. Try to find the correct boundaries for both A and B. Realize the region is a paraboloid with an extremum at 1 on the x axis and cut off by the yz-plane.

    I will leave it for you to decide which one is easier.
     
  6. May 11, 2009 #5

    HallsofIvy

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    Re: Which method is best for triple integration

    You might try first swapping x and z so that the integrand is [itex]x^2y^2dV[/itex] and the bounding surfaces are [math]z= 1- x^2- y^2[/math] and z= 0.

    In cylindrical coordinates that would be [itex](r^2cos^(\theta)r^2sin^2(\theta) rdrd\theta dz= r^5 sin^2(\theta)cos^2(\theta)[/itex] and the boundaries [math]z= 10- r^2[/math] and z= 0.
     
  7. May 11, 2009 #6
    Re: Which method is best for triple integration

    Well I was originally thinking polar was easier but i think ive changed my mind.

    BTW im not using x2y2 and z = 1 - x2 - y2, z = 0

    I need help setting up my limits

    z.. 0 to 1 - x2 - y2

    x... -(1-y2)1/2 to 1-y2)1/2

    y... -1 to 1

    whats right and whats wrong
     
  8. May 13, 2009 #7

    HallsofIvy

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    Re: Setting up Integration limits

    So you are integrating over the region bounded by [itex]z= 1- x^2- y^2[/itex] and z= 0?

    I would recommend converting to cylindrical coordinates: [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex] so that [itex]z= 1- r^2[/itex]. Don't forget that the "differential of volume" in cylindrical coordinates is [itex]dV= r dr d\theta[/itex].

    If you are required to do this in rectangular coordinates (or just want to), then, yes, your limits of integration are correct. Projecting the parboloid onto the z= 0 plane, you get the circle [itex]x^2+ y^2= 1[/itex]. y going from -1 to 1 will cover that and, for any given y, x will go from [itex]-\sqrt{1- y^2}[/itex] to [itex]\sqrt{1-x^2}[/itex]. Finally, for any given x and y, z going from 0 to [itex]1- x^2- y^2[/itex] will cover the solid.
     
  9. May 13, 2009 #8
    Ok, i tried it in cylindrical but i got my limts wrong

    z is from 0 to 1-r2

    r is from 0 to [tex]\sqrt{1-z}[/tex]

    theta is from 0 to pi
     
  10. May 13, 2009 #9

    Cyosis

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    You switched z and x around right? You can't have a z in your limits, it will mean you end up with a variable. The radii of the circular slices vary from the base r=1 to the top r=0. Also to integrate over the full circular slice you need to adjust your [itex]\theta[/itex] limits.
     
  11. May 13, 2009 #10
    im integrating over dz,dr,dQ Q = theta

    dz = 0 to 1-r2

    dr = 0 to 1

    dQ = 0 to 2pi

    [tex]\int\int\int[/tex]r4cos2Q sin2Q r dz dr dQ
     
  12. May 13, 2009 #11

    Cyosis

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    Yes that should give you the same answer as the Cartesian method.
     
  13. May 13, 2009 #12
    ok i got stuck integrating

    in integrated z and pluged in the limits and now im integrating r

    [tex]\int[/tex](r5 - r7)cos2Q sin2Q dr = 1/24 .........

    1/24 [tex]\int[/tex] cos2Q sin2Q dQ =

    1/24 [tex]\int[/tex] ( - /sin2Q) sin2Q dQ =

    1/24 [tex]\int[/tex] sin2Q - sin4Q dQ =

    Now im stuck... my calculater gave me pi/4, but how do i do it by hand
     
  14. May 13, 2009 #13

    HallsofIvy

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    [itex]sin^2(Q)- sin^4(Q)= sin^2Q(1- sin^2(Q))= sin^2(Q)cos^2(Q)[/itex]
    Use trig identities [itex]sin^2(Q)= (1/2)(1- cos(Q))[/itex], [itex]cos^2(Q)= (1/2)(1+ cos(Q))[/itex]. You may need to use those twice.
     
  15. May 13, 2009 #14
    i havent those identities in a while

    1/24 [tex]\int[/tex]cos2Q sin2Q dQ =

    1/24 [tex]\int[/tex](1/2 + 1/2 cosQ)(1/2 - 1/2 cosQ) dQ =

    1/96 [tex]\int[/tex](1 + cos2Q) dQ =

    1/96 [tex]\int[/tex](1 + 1/2 + 1/2 cosQ) dQ =

    1/96 [tex]\int[/tex](3/2 + 1/2 cosQ) dQ =

    1/96 (3Q/2 + 1/2 sinQ)] from 0 to 2pi = pi/32

    my calc said it was pi/4 do u see any problems
     
  16. May 13, 2009 #15

    Cyosis

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    I would personally start out with [itex]\cos^2 \theta \sin^2 \theta=(\cos \theta \sin \theta)^2=\frac{1}{4}\sin^2 (2\theta)[/itex]. This way you can skip a few steps.

    You used Halls identities which are incorrect the correct ones are [itex]\cos^2x=\frac{1}{2}(1+\cos 2x),\sin^2x=\frac{1}{2}(1-\cos 2x)[/itex]. Note the double angle.
     
  17. May 13, 2009 #16
    ok when i integrated with the last identity i got pi/96 did u get this too
     
  18. May 13, 2009 #17

    Cyosis

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    I got pi/96.
     
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