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Iterated integrals over region w

  1. Jul 25, 2014 #1
    1. The problem statement, all variables and given/known data
    I am given

    [tex]W = \{ (x,z,z)| \frac{1}{2} \le z \le 1; x^2 + y^2 +z^2 \le 1\}[/tex]

    they want the iterated integrals to be of the form

    [tex]\iiint_W dzdydx[/tex]





    3. The attempt at a solution

    so I know z=1/2 will give me the larger bound for x

    [tex]x^2 + y^2 + (1/2)^2 =1 \rightarrow x^2 + y^2 = 3/4[/tex]


    [tex]x^2 + y^2 = 3/4[/tex] gives me [tex]- \sqrt{3/4} \le x \le \sqrt{3/4}[/tex] when y=0

    so y bounds are

    [tex]- \sqrt{3/4-x^2} \le y \le \sqrt{3/4-x^2}[/tex]

    and my z bounds

    [tex]\sqrt{1-y^2} + \frac{1}{2} \le z \le \sqrt{1-y^2} [/tex]


    so I get

    [tex]\int_{- \sqrt{3/4}}^{\sqrt{3/4}}\int_{- \sqrt{3/4-x^2}}^{\sqrt{3/4-x^2}}\int_{\sqrt{1-y^2} + \frac{1}{2}}^{\sqrt{1-y^2}}f(x,y,z)dzdydx[/tex]
     
  2. jcsd
  3. Jul 25, 2014 #2

    Orodruin

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    I suggest a change to cylinder coordinates to make one of the integrals trivial.

    The way you have written it, it is not clear what you intend to be the bounds for what integral.
     
  4. Jul 25, 2014 #3

    [tex]\int_{- \sqrt{3/4}}^{\sqrt{3/4}}\Bigg(\int_{- \sqrt{3/4-x^2}}^{\sqrt{3/4-x^2}}\Bigg(\int_{\sqrt{1-y^2} + \frac{1}{2}}^{\sqrt{1-y^2}}f(x,y,z)dz\Bigg)dy\Bigg)dx[/tex]

    maybe that will help

    I'll try it out with cylindrical coordinates as well
     
  5. Jul 25, 2014 #4

    LCKurtz

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    Your z bounds are incorrect. z should go from the lower surface to the upper surface, which is a function of both x and y.
     
  6. Jul 25, 2014 #5

    Orodruin

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    The integration bounds on z depend both on x and y. The lower bound on z is 1/2 regardless of x and y. The upper must be found for arbitrary x and y (within their bounds).

    I really really suggest the cylinder coordinates. :)
     
  7. Jul 25, 2014 #6
    hm, the way I did it was once I had x, I sliced with the x=0 plane so that I'd have a 2-D image then took the bounds of y and figured out how z changes in the plane from 1/2 to 1.
     
  8. Jul 25, 2014 #7
    I'll try cylindrical after my current problem

    are the z bounds [tex]\frac{1}{2} \le z \le \sqrt{1-x^2-y^2}[/tex]

    if it relies on both thats what makes sense
     
  9. Jul 25, 2014 #8

    LCKurtz

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    Doing z first, you must go from the lower surface to the upper surface. Then you look in the xy plane for the "shadow" of the volume which, in this problem, is an ellipse. That's where you get the x and y limits. And what is ##f(x,y,z)##?
     
  10. Jul 25, 2014 #9

    Orodruin

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    Yes. If you fix x and y to arbitrary values within their common bounds, those are the resulting bounds on z.
     
  11. Jul 25, 2014 #10
    f(x,y,z) is just a place older, theres no given function. it just asks for it to be set up as iterated integrals.

    in my course anytime we've been setting up integrals we use that as an arbitrary placeholder
     
  12. Jul 31, 2014 #11
    I think I converted that over correctly

    ##\displaystyle \int_{0}^{2\pi}\int_{0}^{1}\int_{1/2}^{1}r\cdot rdzdrd\theta = \frac{\pi}{3}##

    not positive why theres an extra 'r' in the integral but thats what seems to make it correct.




    I think this is is the integral I should have gotten

    ##\displaystyle \int_{0}^{2\pi}\int_{0}^{1}\int_{\frac{r}{2}}^{r}rdzdrd\theta = \frac{\pi}{3}##
     
    Last edited: Aug 1, 2014
  13. Aug 1, 2014 #12

    LCKurtz

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    You're getting worse, not better.

    Look in your post #7. The upper limit for ##z## is not ##1##. Nor is ##r=1## correct for its upper limit. And why the extra ##r##?

    And the upper limit for ##z## isn't ##r## either. And its lower limit isn't ##r/2##. And the upper limit for ##r## isn't ##1##. Perhaps you should just start over.
     
  14. Aug 1, 2014 #13
    the extra r came simply from a heuristic approach, it seemed correct.

    starting over:


    first thing converting over to cylindrical

    ##0 \le \theta \le 2\pi## this one I see no issues with

    I have ##x^2+y^2+z^2 \le 1##

    this is where I wasn't sure about converting, do I treat ##z^2## just as is or treat it as ##z^2=x^2+y^2##

    the first way of treating z

    ##r^2cos^2(\theta) + r^2sin^2(\theta) + z^2 =1## but graphing this is not the same sphere, it
    the second way

    ##r^2cos^2(\theta) + r^2sin^2(\theta) + (r^2cos^2(\theta) + r^2sin^2(\theta)) =1## but in order for this to be the same sphere it has to be equal to 2 or I take out ##(r^2cos^2(\theta) + r^2sin^2(\theta))## and keep it set as equal to 1

    some explanation on this would be appreciated immensely

    now for the r and z bounds

    ##\frac{1}{2} \le z \le \sqrt{1-x^2-y^2}## this does not work if put as ##\frac{1}{2} \le z \le \sqrt{1-r^2cos^2(\theta)-r^2sin^2(\theta)}##

    this can't be that difficult but I am missing something

    EDIT: also doing it in rectangular I graphed my x,y bounds and those aren't any good either, only way I could get the proper bounds was to have each of x,y,z in terms of the other two.
     
    Last edited: Aug 1, 2014
  15. Aug 11, 2014 #14
    I am returning to this problem even though my semester is over. I believe spherical would be better than cylindrical.

    ##\displaystyle \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{1/2}^{1}\rho^2sin\phi d\rho d\phi d\theta##
     
  16. Aug 11, 2014 #15

    LCKurtz

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    Did you pass you course?

    Your setup is still incorrect. ##\rho## is not constant on the lower surface ##z=1##. And the upper limit for ##\phi## is not ##\frac \pi 4##.
     
  17. Aug 11, 2014 #16

    I managed to get in A in the course, I am still having difficulties with triple integrals though so I am continuing to work on them

    z=1 is the upper surface isn't it? And I thought phi would go from 0 to ##\frac{\pi}{4}## because phi starts from the positive z axis.

    I am looking at it in grapher. I have a unit sphere and two plane, z=1 and z=1/2, slicing through it.

    ##0 \le \theta \le 2\pi## I find no issues with that

    fpr ##\phi## it makes sense looking at it that it goes form 0 to ##\pi/4##. I have a 2-D view of it graphed as well, like I sliced with the x=0 plane.
     
    Last edited: Aug 11, 2014
  18. Aug 11, 2014 #17

    LCKurtz

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    No. ##z=1## is the largest value of ##z## in the figure, but the upper surface is the sphere and the lower surface is the plane ##z=1/2##. Draw a cross sectional picture of the sphere and plane to figure out the max value of ##\phi##.
     
  19. Aug 11, 2014 #18
    isn't z=1 the largest value without the z=1 plane even in there? its a unit sphere so I don't even see why that plane is necessary.

    looking at a cross sectional picture in the yz plane: I have a unit circle z^2+y^2 = 1 sliced by a line z=1/2

    and phi starts at the positive z-axis and would stop where z=1/2 slices through so thats a movement of 0 to ##\pi/3##

    but ##\rho##, ##z=\rho cos(\phi)## so ##\rho = \frac{z}{cos(\phi)}##
     
    Last edited: Aug 11, 2014
  20. Aug 11, 2014 #19

    LCKurtz

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    Yes. That's better than just guessing that ##\rho = \pi/4## eh?

    But you want ##\rho## on the plane ##z=1/2##. So ##\rho## there is ..?
     
  21. Aug 11, 2014 #20

    LCKurtz

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    Look at your original statement. You want ##z## between ##1/2## and ##1##, but also inside the sphere ##x^2+y^2+z^2=1##. That region lies above the plane ##z=1/2## and below the upper hemisphere of ##x^2+y^2+z^2=1##. Those are your lower and upper surfaces.
     
    Last edited: Aug 11, 2014
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