Checking method and answer of Triple Integral

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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?
 

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  • #2
HallsofIvy
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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.
 
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  • #3
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Okay.
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.
 
  • #4
HallsofIvy
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Okay.
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.
Yes, since the question is to find the volume, and [itex]V= \int\int\int dV[/itex], your integrand is 0.

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


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.
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.

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
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Thanks a lot HallsofIvy.
 
  • #6
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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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