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Checking method and answer of Triple Integral

  1. Apr 21, 2012 #1
    The question is:

    find the volume of the solid bounded above by x^2+y^2+z^2=9, below by z=0 and on the sides by the cylinder x^2+y^2=4.

    Now rho comes out to be 3. At the very top of the solid, phi is 0 so z=3.

    So limits of z are 0 (lower limit) and 3 (upper limit).

    As far as the limits of x and y are I found their lower and upper limits to be 0 and 2 respectively.

    When we do the triple integal ∫∫∫dxdydz (while pluggung the limits) I got 12 (units of volume) . Is this the correct way and answer for this question?
  2. jcsd
  3. Apr 21, 2012 #2


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    It looks to me like you are going to have to divide this into two integrals because the boundary does not change smoothly. The cylinder [itex]x^2+ y^2= 4[/itex] crosses the sphere [itex]x^2+ y^2+ z^2= 9[/itex] where [itex]4+ z^2= 9[/itex] or [itex]z= \sqrt{5}[/itex] (since z> 0).

    So do this as two separate integrals:

    First the cylinder. With [itex]x^2+ y^2= 4[/itex], [itex]y= \pm\sqrt{4- x^2}[/itex] while x goes, overall, from -2 to 2. The integral is
    [tex]\int_{z=0}^\sqrt{5}\int_{x=-2}^2\int_{y=-\sqrt{4- x^2}}^{\sqrt{4- x^2}} f(x,y,z) dydxdz[/tex]
    Because of the cylindrical symmetry, of course, it is simpler in cylindrical coordinates:
    [tex]\int_{z= 0}^\sqrt{5}\int_{r=0}^2\int_{\theta= 0}^{2\pi} f(r,\theta,z)d\theta dr dz[/tex]

    Now do the spherical "cap", x and y still cover the circle of radius \sqrt{5}[/itex] while z goes from [itex]\sqrt{4}[/itex] up to the sphere, [itex]x^2+y^2+ z^2= 9[/itex] which is the same as [itex]z^2= 9- x^2- y^2[/itex] or [itex]z= \sqrt{9- x^2- y^2}[/itex]. The integral is
    [tex]\int_{x= 0}^2\int_{y=-\sqrt{4- x^2}}^\sqrt{4- x^2}\int_{z=\sqrt{5}}^{\sqrt{9- x^2- y^2}} f(x,y,z) dzdydx[/tex]
    or, in cylindrical coordinates
    [tex]\int_{r= 0}^2\int_{\theta= 0}^{2\pi}\int_{z=\sqrt{5}}^\sqrt{9- r^2} f(r,\theta, z) rsin^2(\theta)dz d\theta dr[/tex]

    The original integral is the sum of those.
    Last edited by a moderator: Apr 21, 2012
  4. Apr 21, 2012 #3
    Here's what I get:

    1) I get why you split the integral and how the solid object is split into a simple cylinder and then a cap.

    Here's what I don't get:

    1) I was'nt given any function so can't I assume that f(x,y,z) of f(r,∅,θ) (just to be clear the symbol before ∅ is phi ) or f(r,θ,z) all are equal to 1.

    2) Did'nt you miss out an r in the integrand of the first cylindrical integration you did. ( The reason why I stated that is not to point out your mistake, but just in case you intentionally omitted the r, I want to know why, so that I don't make mistakes in the exam).

    3) When the cap's volume is being integrated you're taking the upper limit of z as sqrt(9-r^2) or sqrt(9-x^2-y^2). Is'nt it sort of pointless to include the r^2 or -x^2-y^2 terms in the upper limit as at the top of the cap the point will have only a single x,y,r value so why integrate it. I mean one integrates when one is summing over a variable. Now when you start at the bottom pf the cap ( say in Cartesian Coordinates) x and y are in the form of x^2+y^2=4. But at the top of the cap x^2+y^2=0. So why even bother to inlude the x^2+y^2 terms (or r^2) in the upper limit. If we don't write them it will be understood that they have gone to 0 and the top of the cap is a point.

    Hope you can clear the 3rd point as I'm most confused regarding it. I just hope that I have sufficiently conveyed what I'm confused about.
  5. Apr 21, 2012 #4


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    Yes, since the question is to find the volume, and [itex]V= \int\int\int dV[/itex], your integrand is 0.

    Yes. Sorry about that. I was focusing on the limit of integration.

    Because you have to integrate over the entire cap, not just the top of the cap. It is true that at the top of the cap x= y= 0 so z= 3 but when you are doing the "inside" z-integral, x and y can take any values from 0 to 2.

  6. Apr 22, 2012 #5
    Thanks a lot HallsofIvy.
  7. Apr 22, 2012 #6
    Sorry, last question:-

    I said that since the integrand is'nt given we should take it to be 1 ( or we can just write it as ∫∫∫ 1 dV or ∫∫∫ dv i.e. writing nothing as the integrand).

    You said we should take it as 0 ( ∫∫∫ 0 dV ). Won't the volume then always come out to be 0 since zero is being multiplied with everything.

    So should'nt we take the integrand as ∫∫∫ 1 dV or ∫∫∫ dv instead of ∫∫∫ 0 dV. In other words should'nt the integrand be 1 instead of 0.
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