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

• dyn
In summary: ## by integrating the function over ## ( x-y ) ## and then taking the limit as ## x \rightarrow 0 ## and ## y \rightarrow \infty ##.
dyn
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

Delta2
Hi,

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

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 ####\ ##

dyn said:
(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.

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.

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.

dyn said:
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.

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

dyn said:
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 ##.

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.
vela said:
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 ) ?

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

##\ ##

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 )

dyn said:
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.

Part (c) is fairly straightforward.
And well executed ! Of course, nitpickers like the guy in #2 always find something to nag about .

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

##\ ##

Yes, I agree=his limits on part (b) need some work.

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).

Last edited:
dyn said:
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 ?

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.

dyn
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

dyn said:
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
dyn said:
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.

##\ ##

dyn said:
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

BvU said:
I see a modest square root symbol ... and no ##dz##.
Yeah, I didn't notice the radical.

The double integral needs a ## -x^2 ## inside the radical expression for the height.. Otherwise I think you might have everything correct.

dyn
Thank you. Yes , i accidently missed out the -x2 in the double integral

## 1. What is the volume of the given integral?

The volume of the given integral can be calculated using the formula V = ∫∫∫ f(x,y,z) dV, where f(x,y,z) is the integrand and dV represents the infinitesimal volume element. In this case, the integrand is x^2 + (y-2)^2 + z^2 = 4 and the limits of integration are x, y, and z greater than 0. The resulting volume will be in cubic units.

## 2. How do you set up the integral for this problem?

To set up the integral, we first need to determine the limits of integration for each variable. Since x, y, and z are all greater than 0, the limits for each variable will be from 0 to the value that satisfies the given equation. In this case, we can rewrite the equation as z = √(4 - x^2 - (y-2)^2) and the limits for x and y will be from 0 to 2. Then, we can set up the integral as ∫∫∫ 4 - x^2 - (y-2)^2 dz dy dx.

## 3. What is the significance of the given equation in a scientific context?

The given equation represents a three-dimensional shape known as a sphere with a radius of 2 units. This type of equation is commonly used in physics and engineering to represent physical objects or phenomena that have spherical symmetry. The volume integral of this equation can be used to calculate the volume of the sphere or to solve various physical problems related to it.

## 4. Can this integral be solved analytically?

Yes, this integral can be solved analytically by using techniques such as substitution, integration by parts, or partial fractions. However, it may require multiple steps and calculations to arrive at the final solution. Alternatively, it can also be solved numerically using computational methods such as numerical integration or Monte Carlo simulation.

## 5. How can this integral be applied in real-world scenarios?

The volume integral of this equation can be applied in various real-world scenarios, such as calculating the volume of a spherical object or determining the amount of material needed to create a spherical structure. It can also be used in physics and engineering to analyze physical systems with spherical symmetry, such as electric fields, gravitational fields, or fluid flow. Additionally, this type of integral is commonly used in computer graphics to render three-dimensional objects and in medical imaging to reconstruct three-dimensional images from two-dimensional scans.

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