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

[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

 

[BvU]

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$$\$

 

[vela]

(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.

 

[Charles Link]

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.

 

[dyn]

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.

 

[Charles Link]

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.

 

[dyn]

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

 

[Charles Link]

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 $2 \pi$.

 

[BvU]

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 ) ?

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 $x$

$\$

 

[dyn]

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 )

 

[Charles Link]

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.

 

[BvU]

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) ?

$\$

 

[Charles Link]

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

 

[Charles Link]

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).

 

[dyn]

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 ?

 

[Charles Link]

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

 

[BvU]

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.

$\$

 

[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

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

 

[vela]

I see a modest square root symbol ... and no $dz$.

Yeah, I didn't notice the radical.

 

[Charles Link]

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