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Triple integral and cartesian coordinates

  1. Dec 28, 2011 #1
    we all know that triple integral can be solved by either cartesian coordinates , spherical ,or cylindrical coordinates
    i just need like some advice in knowing when the variable used is constant and when it is not
    for example : r in cylindrical coordinates can it be constant or not?? because i always see it in terms of y and z like that
    and any advice would be very helpful and thank you
     
  2. jcsd
  3. Dec 28, 2011 #2

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    Hi queenstudy!

    Can you give a couple of examples of the integrals you mean?

    Taken literally, if you keep r in cylindrical coordinates constant, you have a double integral, meaning you integrate over a cylindrical surface instead of a cylindrical volume.
     
  4. Jan 1, 2012 #3
    by fixing the plane Z=z change the domain of definitoin of the triple integral f(x,y,z) where D is
    D=(x,y,z) belongs to R cube and it is the intersection of the sphere and cone (icecream cone) sphere xsquare +y square + z square =2 and z=xsquare + y square
    z is positive
    in cartesian coordinates
    my question here is when i do my projection i will have a circle , and if i want it in cartesian i need like 8 integrals because the plane Z=z should be cut twice and for each domain the intersection is a circle and in each circle we have y and x positive x and y negative and one positive and one negative so can i replace those with r and theta and save time and work and still in cartesian or not??
    do you have any extra double and triple integral because i solved on google and want some more and thank you thank you twice serena for helping me out
     
  5. Jan 1, 2012 #4

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    Hmm, as yet I'm having trouble understanding what you write...?


    You appear to refer to the integral (polar):
    $$
    \int_{r=0}^{r=R} \int_{\theta=0}^{\theta=2\pi} \int_{\phi=0}^{\phi={\pi \over 4}} ... d\phi d\theta dr
    $$

    Or cartesian:
    $$
    \int_{x=-\frac R 2 \sqrt 2}^{x=+\frac R 2 \sqrt 2} \int_{y=-\sqrt{R^2-x^2}}^{y=+\sqrt{R^2-x^2}} \int_{z=\sqrt{x^2+y^2}}^{z=\sqrt{R^2-x^2-y^2}} ... dz dy dx
    $$


    Does this look like what you mean?
     
    Last edited: Jan 1, 2012
  6. Jan 1, 2012 #5
    what you wrote is true but i am supposed to write z as two constants, and when you study x and y , you must write in terms of constant K , thus in both answers you wrote they arent correct ,
    let me ask a simpler question:
    i should write the domain in cartesian may i write x and y in polar coordinates and still call them cartesian because if i keep them in cartesian , x and y will have 4 domains when:
    1) x and y postive ; 2) x and y negative; 3) x positive and y negative ; 4) x negative and y postive
    i hope i explained my question better
     
  7. Jan 1, 2012 #6

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    Well, you can rearrange the cartesian integral so z is integrated last.
    In that case you would need to split the integral into 2 integrals.
    One integral for z from 0 to ##\frac R 2 \sqrt 2## and one from ##\frac R 2 \sqrt 2## to R.
    You would not need to split the x or y integrations though.

    Perhaps you can show what you think it might be?

    Cartesian means not-polar, but why do you think you would have to?
    There is no need to split x and y in the 4 domains you mention.
     
  8. Jan 2, 2012 #7
    because according to our proffesor when x is negative y has the possibility to be positive or negative right? so we have two cases
    now is x is positive then y can be either positive or negative right? so we also have two possibilities , thus we have 4 cases thats and that is only the case of the intersection of the plane z=constant with the cone
    if we do it with the sphere inside the domain we will obtain the same 4 parts thus the total will be 8 parts which is long
    that is why we choose polar here for a circle or to me be more specific a full disk because r is only positive and theta is between zero and two pie
     
  9. Jan 2, 2012 #8

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    I think you're mixing up a single integral with higher dimensional integrals.

    If you have a single integral to find the area of (part of) a circle disk, you have the problem that y has a negative and positive value and you have to choose between one of them.

    If you have a double integral over a circle disk, y runs from the negative value to the positive value, which covers the entire circle disk.
    So there is no need to split up the integral.
     
  10. Jan 2, 2012 #9
    but in class when i we projected the domain we got a circle , and it got 4 domains where x and y contain the domain of the circle
    okay if we have a disk on the xoy axis and i want to make double integration with respect to x and y dont i get 4 parts of the circles thus 4 domains?? then we use polar coordinate and becomes a simple integration right????????
    if this is true , MY question is that can r and theta be considered also as cartesian coordinates??????
     
  11. Jan 2, 2012 #10

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    Can you show me integrals of what you mean?

    As for your question whether r and theta can be considered cartesian coordinates, I believe we covered that in your other thread.
    All I can say is that the word "cartesian" means x and y, which is not r and theta.
     
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