Volume integral of x^2 + (y-2)^2 +z^2 = 4 where x , y , z > 0

## by integrating the function over ## ( x-y ) ## and then taking the limit as ## x \rightarrow 0 ## and ## y \rightarrow \infty ##.f
  • #1

dyn

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Homework Statement
Consider the volume defined by the surface x^2 + (y-2)^2 +z^2 = 4 where x , y , z > 0 and the x=0 , y=0 , z=0 planes
(a) sketch this volume
(b) write down the definite integral which defines the volume in Cartesian coordinates
(c) integrate the function 1/ ( x^2 + (y-2)^2 +z^2 ) over this volume by making an appropriate coordinate transformation
Relevant Equations
Equation of sphere x^2+y^2+z^2 = constant , volume element in spherical polars is
dV = r^2 sin (theta) dr d(theta) d(phi)
(a) i sketched a quarter of a sphere centred at x=0 , y=2 , z=0

(b ) ∫ ∫ √ (4-x2 - (y-2)2) dx dy with limits 0 < x < 2 and 0 < y <4

(c ) i converted to spherical polars and took the integrand as 1/r2 . the volume element is r2sinθ drdθd∅
This leads to the triple integral of sinθ with limits 0< r <2 , 0 <θ <π/2 , 0 < ∅ < π
This does give the correct answer of 2π but i am not sure if i can just take the integrand as 1/r2 because the sphere is displaced from the origin. But it does give the correct answer !

How have i done with all 3 parts ?
Thanks
 
  • #2
Hi,

a) No picture ##\Rightarrow## can't comment on how you did on part a) :wink:

b) I would expect a triple integral ?

c) Actually, you first shifted the origin ##(y^* = y-2)## so that ##f ## became ##1/({x^*}^ 2+{y^*}^ 2+{z^*}^ 2) = 1/ {r^*}^ 2 \ ## and only after that you converted to spherical coordinates (polar coordinates are 2D!)
So, yes: the integrand has become ## 1/{r^*}^ 2 ##


##\ ##
 
  • #3
(b ) ∫ ∫ √ (4-x2 - (y-2)2) dx dy with limits 0 < x < 2 and 0 < y <4
Note that if you consider ##x##, ##y##, and ##z## to have units of length, your integral would give an answer with units of length^4.
 
  • #4
You said in the OP that the correct answer is ## 2 \pi ##. That's not what I get. This problem looks to me to be relatively simple without integral calculus, if you can use the formula for the volume of a sphere, etc.
 
  • #5
You said in the OP that the correct answer is ## 2 \pi ##. That's not what I get. This problem looks to me to be relatively simple without integral calculus, if you can use the formula for the volume of a sphere, etc.
The answer given is ## 2 \pi ## and the question asks for an integration.
 
  • #6
The answer given is ## 2 \pi ## and the question asks for an integration.
Did I miss something? This looks like a half of a hemisphere to me from the boundaries that were given.
 
  • #7
Part ( a) asks to sketch the region. I sketched a quarter of a sphere centred at x=0 , y=2 , z=0 then part ( c ) asks for the function in the OP to be integrated over that region
 
  • #8
Part ( a) asks to sketch the region. I sketched a quarter of a sphere centred at x=0 , y=2 , z=0 then part ( c ) asks for the function in the OP to be integrated over that region
That implies ## V=((4/3) \pi (2^3))/4 ## as the correct answer, and not ## 2 \pi ##.
 
  • #9
You said in the OP that the correct answer is ## 2 \pi ##. That's not what I get. This problem looks to me to be relatively simple without integral calculus, if you can use the formula for the volume of a sphere, etc.
The answer for part c) is ##2\pi##. The integrand is not ##\ 1\ ## but ##1/{r^*}^2\ ##.
For part b), ##\ ## just writing down the integral would have been sufficient.


Note that if you consider ##x##, ##y##, and ##z## to have units of length, your integral would give an answer with units of length^4.
I see a modest square root symbol ... and no ##dz##

Just writing down the integral would have been sufficient -- however, it looks as if the integration over ##z## has already been carried out (as a kind of bonus attempt :wink: ) ?

b) I would expect a triple integral ?
Did not coax @dyn to explain -- and perhaps discover room for improvement :frown: in the limits for ##y## and ##x##

##\ ##
 
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  • #10
That implies ## V=((4/3) \pi (2^3))/4 ## as the correct answer, and not ## 2 \pi ##.
The function to be integrated over that volume is

1 / ( x2 + (y-2)2 +z2 )
 
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  • #11
The function to be integrated over that volume is

1 / ( x2 + (y-2)2 +z2 )
ok=I hadn't read part (c), but they did pick a very simple form to the function in a transformed coordinate system. Part (c) is fairly straightforward.
 
  • #12
Part (c) is fairly straightforward.
And well executed ! Of course, nitpickers like the guy in #2 always find something to nag about :cool:.

Like this: @dyn: can you improve on the limits for part b) ?

##\ ##
 
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  • #14
a hint or two for the OP on the triple integral of part (b):

You chose to do ## dz ## first. For a given ## x ## and ##y ##, you get the limits on the ## dz ## integration. You got the functional form of the upper limit for ## dz ## correct, but it is perhaps preferable to do this part as a triple integral with limits, rather than as a double integral over the height. In any case this part is ok.

Next, the ## dx \, dy ## is done in the x-y plane. If you choose to do ## dx ## before ## dy ##, the ## x ## has zero as the lower limit independent of ## y ##, but what is its upper limit in the x-y plane for a given ## y ##?

(Note: It would probably be a little simpler to do the ## dy ## integration first, and then do the integration over the x-z plane).
 
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  • #15
The function to be integrated over that volume is

1 / ( x2 + (y-2)2 +z2 )
Even though it works i would like to know exactly why i can write the above function as 1/r2
The Cartesian origin is at (0, 0 , 0 ) but it seems like i can take the origin in spherical polars at the centre of the sphere wherever it appears to be. Is that right ?
 
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  • #16
Perhaps the simplest way to look at it is to take ## y'=y-2 ##, and work with ## (x,y',z) ##.
Then ## r ## is measured from the origin of ##(x,y',z) ##.
In any case, yes, what you said is correct.
 
  • #17
I will have a go at part ( b ) a bit later with the limits. As far as i know i can also calculate a volume as
∫∫ z(x , y) dxdy. Is that right ? In which case i will have a go at the limits for the double and triple integral
 
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  • #18
As far as i know i can also calculate a volume as $$\int z(x,y)\;dx\,dy$$ Is that right ?
A Cartesian volume element is ##\;dx\,dy\,dz##. The whole volume is ##V = \displaystyle \iiint\;dx\,dy\,dz## and you have established that if you want to do the ##dz## first, the bounds of ##z## are ##0## and,
dependent on ##x## and ##y##: ##\sqrt{4-x^2-(y-2)^2}\mathstrut ##:$$
V = \iint\;\Biggl (\int_0^\sqrt{4-x^2-(y-2)^2}\ dz \Biggr )\;dx\,dy$$Again: for part b) you don't have to do it, but if you exectute the integral over ##dz## you indeed get $$
V = \iint\;\sqrt{4-x^2-(y-2)^2}\;dx\,dy$$and, ##\ -## thanks to the lower bound ##0## ##-\ ## this is indeed, as you wrote$$
V = \iint\;z(x,y) \;dx\,dy$$in words: we are going to integrate the volumes of little columns ##\;\sqrt{4-x^2-(y-2)^2}\;dx\,dy\ ## over a suitable area in ##x,y##. And that is where you should look at
  • which one to do first
  • the bounds
Because your
with limits 0 < x < 2 and 0 < y <4
looks like a rectangle and that's not good at all ! The footprint on ##z=0## is not a rectangle; the ##x## and ##y## bounds are not independent.

##\ ##
 
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  • #19
Homework Statement:: Consider the volume defined by the surface x^2 + (y-2)^2 +z^2 = 4 where x , y , z > 0 and the x=0 , y=0 , z=0 planes
(a) sketch this volume
(b) write down the definite integral which defines the volume in Cartesian coordinates
Hi ; here goes with the limits. I shall offer 3 ways to evaluate this volume integral ; hopefully they are all valid and correct

1 - The triple integral , ∫ ∫ ∫ dz dx dy with limits 0 < z < √(4-x2 - (y-2)2 ) , 0 < x < √(4-(y-2)2) and 0<y<4

2 - The double integral , ∫ ∫ √(4-(y-2)2) dx dy with limits 0 < x < √(4-(y-2)2 ) and 0<y<4

3 - The triple integral in spherical polars ; ∫ ∫ ∫ r2sinθ dr dθ d∅ with limits 0<r<2 , 0 < θ < π/2 and 0 < ∅ < π
 
  • #20
I see a modest square root symbol ... and no ##dz##.
Yeah, I didn't notice the radical.
 
  • #21
The double integral needs a ## -x^2 ## inside the radical expression for the height.. Otherwise I think you might have everything correct.
 
  • #22
Thank you. Yes , i accidently missed out the -x2 in the double integral
 
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