# Change of variables in multiple integrals

• WMDhamnekar

#### WMDhamnekar

MHB
Summary: Find the volume V of the solid inside both ## x^2 + y^2 + z^2 =4## and ## x^2 +y^2 =1##

My attempt to answer this question: given ## x^2 + y^2 +z^2 =4; x^2 + y^2 =1 \therefore z^2 =3 \Rightarrow z=\sqrt{3}##
## \displaystyle\iiint\limits_R 1dV = \displaystyle\int_0^{2\pi}\displaystyle\int_0^{\pi}\displaystyle\int_0^2 1 \rho^2 \sin{(\phi)} d\rho d\phi d\theta =\frac{32\pi}{3}##

Now using ##z=\sqrt{3}## we get ## \displaystyle\int_0^{2\pi}\displaystyle\int_0^\pi \displaystyle\int_0^{\sqrt{3}} 1 \rho^2\sin{\phi}d\rho d\phi d\theta = 4\sqrt{3}\pi##

Hence the area of the solid is ##\frac{32\pi}{3} -4\sqrt{3}\pi =\frac{4\pi}{3}*(8-3^{3/2})##
Actually, I couldn't draw out the graph of this solid.
I think the problem is solved now.

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I think you didn't understand correctly the statement of the problem. The key word for understanding (at least for my brain) is the word "inside". This means we have to convert equalities to inequalities and look for the region V where both inequalities hold, $$V={(x,y,z):x^2+y^2+z^2\leq 4, x^2+y^2\leq 1}$$.

But there is more, what region/shape do you think the inequality $$x^2+y^2\leq1$$ is? (hint: think in 3D)

Summary: Find the volume V of the solid inside both ## x^2 + y^2 + z^2 =4## and ## x^2 +y^2 =1##

Now using z=3 we get ∫02π∫0π∫031ρ2sin⁡ϕdρdϕdθ=43π
I am afraid you are wrong here, because all you do is to calculate the volume of a sphere with radius 2 and another with radius ##\sqrt{3}##. And then calculate the volume of the spherical shell with ##\sqrt{3}\leq r\leq 2## This is not what the problem asks for.

• WMDhamnekar
I am afraid you are wrong here, because all you do is to calculate the volume of a sphere with radius 2 and another with radius ##\sqrt{3}##. And then calculate the volume of the spherical shell with ##\sqrt{3}\leq r\leq 2## This is not what the problem asks for.
But the author has provided the same answer for this question. I think your logic is something different than author Prof.Michael Corral , schoolcraft college, United States of America. Name of the book "Vector Calculus Part 2"

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• Delta2
Actually, I couldn't draw out the graph of this solid.
Why not? The first equation represents a sphere of radius 2, centered at the origin. The second equation represents a circular cylinder of radius 1, whose central axis lies along the z-axis.

Being able to draw a sketch of the region of integration makes it less likely that you will set up the limits of integration incorrectly, as you did in your recent thread.

Edit: changed level of thread from A to I. This isn't post-graduate stuff.

• scottdave
Why not? The first equation represents a sphere of radius 2, centered at the origin. The second equation represents a circular cylinder of radius 1, whose central axis lies along the z-axis.

Being able to draw a sketch of the region of integration makes it less likely that you will set up the limits of integration incorrectly, as you did in your recent thread.

Edit: changed level of thread from A to I. This isn't post-graduate stuff.
Sorry. This stuff may be meant for under-graduate Mathematics student.

Hold on while I do the calculation (using the fact that we talk here about the intersection of a sphere with a cylinder) and see if I get the same answer.

Sorry. This stuff may be meant for under-graduate Mathematics student.
My main point was the importance of being able to sketch the region over which integration takes place. If you can't do this, it's very unlikely that you'll get the correct limits of integration, so you're also not likely to get the correct value for the integral.
I can't overemphasize the importance of getting a good visual representation of what you're trying to integrate.

• • PhDeezNutz and WMDhamnekar
$$2\sqrt{3}\pi+\frac{2}{3}\pi(2-\sqrt{3})^2(4+\sqrt{3})$$
That's what I get. The first term is the volume of that cylinder with radius 1 and height ##2\sqrt{3}## and the second term is the volume of the two spherical caps with height ##2-\sqrt{3}##.
After some algebra (which I am bored to tex, sorry) reduces to the same answer as yours. But still I don't think you formed the correct integrals sorry.

• WMDhamnekar
$$2\sqrt{3}\pi+\frac{2}{3}\pi(2-\sqrt{3})^2(4+\sqrt{3})$$
That's what I get. The first term is the volume of that cylinder with radius 1 and height ##2\sqrt{3}## and the second term is the volume of the two spherical caps with height ##2-\sqrt{3}##.
After some algebra (which I am bored to tex, sorry) reduces to the same answer as yours. But still I don't think you formed the correct integrals sorry.
Is it just coincidence that your answer and my answer correct? I think there are two different methods of answering this question.(My opinion may be wrong.). I shall check out your answer in my spare time.

Is it just coincidence that your answer and my answer correct? I think there are two different methods of answering this question.(My opinion may be wrong.). I shall check out your answer in my spare time.
In my opinion yes it is a coincidence that the two answers match, because all you calculated, as I already said, is the volume of a spherical shell , you did not calculate the volume that corresponds to the intersection of a sphere and a cylinder.

• • scottdave and WMDhamnekar
Is it just coincidence that your answer and my answer correct? I think there are two different methods of answering this question.(My opinion may be wrong.). I shall check out your answer in my spare time.
It's just a coincidence.

Can you critically assess your method and justify each step? Being able to do that is much more important than getting an answer that matches the one in the book. And when you can do that, you can have more confidence that the answer you obtained is correct.

Can you address Delta2's critique of your attempt? He pointed out specific problems with your solution, and so far, your only response has been, "but I got the answer in the book."

• • scottdave, WMDhamnekar and Delta2
Btw, I used directly the formula for spherical cap . One way you can calculate the volume of a spherical cap using triple integral is by using a mix of cartesian and polar coordinates. That is after you setup the triple integral in cartesian coordinates x,y,z, you can use polar coordinates to do the integration w.r.t x and y, and cartesian coordinate to do the integration with respect to z.

If you are wondering, what the volume of a spherical cap has to do with this problem ,then all I can say is that the intersection of a sphere and a cylinder who has radius smaller than the sphere, and height bigger than the diameter of the sphere and with its axis passing through the center of the sphere, is a smaller cylinder and two spherical caps.

There is a way to do this , without referring to geometrical shapes at all, just by processing correctly the inequalities that define the region V in post #2.

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It's just a coincidence.

Can you critically assess your method and justify each step? Being able to do that is much more important than getting an answer that matches the one in the book. And when you can do that, you can have more confidence that the answer you obtained is correct.
As it turns out, it is not merely a coincidence for that method to give the correct answer. However, the fact that it does work is not readily apparent, certainly not obvious.

@WMDhamnekar ,
I agree with vela that you should be able to defend this method, if you use this method. Likely the author covered the "why" of how it works.

I don't have access to the book by Prof. Michael Corral, but consider the following.

For any value of ##z_h## where ##|z_h|\le \sqrt{3}##, the following two areas are equal.

The area of a cross-section of the small sphere having radius of ##\sqrt{3}##.

The area of a cross-section of the large sphere having radius of ##2##, minus the area of a cross-section of the cylinder of radius ##1## .

• • WMDhamnekar and Delta2
Likely the author covered the "why" of how it works.

It turns out a PDF of the book is available for free from http://www.mecmath.net.

For any value of zh where |zh|≤3, the following two areas are equal.
Congrats, that's a genuine observation and explains why his approach gives the same result, however to me its a kind of coincidence that this observation holds, it has to do that the radius of the cylinder is exactly 1.

This approach does not depend upon the radius of the cylinder being 1.

This approach does not depend upon the radius of the cylinder being 1.
So you mean that the observation holds for any radius of the big sphere and the cylinder?

The body lives in a sphere, so converting to spherical coordinates might be fruitful. So put
$$x = r\cos\phi \sin\theta,\quad y = r\sin\phi\sin\theta, \quad z = r\cos\theta$$
where ##\phi \in [0,2\pi]## and ##\theta \in [0,\pi]##. Due to symmetry it suffices to let ##\theta \in [0,\pi /2]##. For ## \theta \in [0, \pi /6]## we have a piece of the volume given by
$$\int _0^{2\pi} \int _0^{\pi/6} \int _0^2 J \mathrm{d}r\mathrm{d}\theta \mathrm{d}\phi \overset{?}= \frac{8\pi}{3}\left (2-\sqrt{3}\right ),$$
where ##J## is the Jacobian. Now, for ##\theta \in [\pi/6,\pi /2]##, the maximum radius depends on ##\theta##. More precisely ##r(\theta) = \frac{1}{\sin\theta}##. Thus, the remaining volume is
$$\int _0^{2\pi} \int _{\pi /6} ^{\pi /2} \int_0^{\frac{1}{\sin \theta}} J \mathrm{d}r\mathrm{d}\theta \mathrm{d}\phi \overset{?}= \frac{2\sqrt{3}\pi}{3}.$$
The two sum up to the upper half of the body's volume, so multiply by ##2## to get the entire volume.

That's how I usually sketch these things. Finding the necessary limits is then straightforward. When you decide the limits, try to verify whether these limits allow you to cover every point in the body. It might be easier to compute separate pieces, because the limits could be parametrised with respect to some variables (and often are).

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• Delta2
Summary: Find the volume V of the solid inside both ## x^2 + y^2 + z^2 =4## and ## x^2 +y^2 =1##

My attempt to answer this question: given ## x^2 + y^2 +z^2 =4; x^2 + y^2 =1 \therefore z^2 =3 \Rightarrow z=\sqrt{3}##
What is the significance of ##z^2=3##?
## \displaystyle\iiint\limits_R 1dV = \displaystyle\int_0^{2\pi}\displaystyle\int_0^{\pi}\displaystyle\int_0^2 1 \rho^2 \sin{(\phi)} d\rho d\phi d\theta =\frac{32\pi}{3}##
This is the volume of the ball centered at origin with radius ##2##.
Now using ##z=\sqrt{3}## we get ## \displaystyle\int_0^{2\pi}\displaystyle\int_0^\pi \displaystyle\int_0^{\sqrt{3}} 1 \rho^2\sin{\phi}d\rho d\phi d\theta = 4\sqrt{3}\pi##
You have computed the volume of the ball centered at origin with radius ##\sqrt{3}##. That's not what you are tasked to find and neither does it help.
Hence the area of the solid is ##\frac{32\pi}{3} -4\sqrt{3}\pi =\frac{4\pi}{3}*(8-3^{3/2})##
Actually, I couldn't draw out the graph of this solid.
I think the problem is solved now.
We need the volume of the solid. You have computed the volume that is left over after the smaller ball (with radius ##\sqrt{3}##) is erased inside the bigger ball. That is definitely not what you are tasked to find.

My solution's strategy is that I break the triple volume integral into the volume of the cylinder from ##z=-\sqrt{3}## to ##z=\sqrt{3}## and the volume of two spherical caps from ##z=\pm\sqrt 3 ## to ##z=\pm 2##. So the integrals are $$\int_{-\sqrt{3}}^{\sqrt{3}}\int\int_{x^2+y^2\leq 1}1dxdydz+\int_{-2}^{-\sqrt{3}}\int\int_{x^2+y^2\leq 4-z^2}1dxdydz+\int_{\sqrt{3}}^{2}\int\int_{x^2+y^2\leq 4-z^2}1dxdydz$$

The integration with respect to x and y is done in polar coordinates and then follows the integration with respect to z. For the first term after we integrate in polar coordinates we get a function which is constant w.r.t z, but for the 2nd and 3rd term we get a function that depends on z.

• nuuskur
But the author has provided the same answer for this question. I think your logic is something different than author Prof.Michael Corral , schoolcraft college, United States of America. Name of the book "Vector Calculus Part 2"
Your reasoning is suboptimal. It is not the end result your grader is interested in but how you obtain it. You should prioritise technique. Provided your solution is correct you will get the correct answer.

Also, the answer section could contain mistakes. Not often, but it still happens from time to time.

• Delta2
Something not pointed out until many posts later in this thread:
My attempt to answer this question: given ## x^2 + y^2 +z^2 =4; x^2 + y^2 =1 \therefore z^2 =3 \Rightarrow z=\sqrt{3}##
The work above contains a very basic mistake: ##z^2 = 3## has two solutions; namely, ##z = \pm \sqrt 3##. In your subsequent work you mention only the positive square root.

Something not pointed out until many posts later in this thread:

The work above contains a very basic mistake: ##z^2 = 3## has two solutions; namely, ##z = \pm \sqrt 3##. In your subsequent work you mention only the positive square root.
It doesn't really matter cause all he does is to compute the volume of the "smaller" sphere with radius ##\rho=\sqrt{3}##.

However under the new light shed by @SammyS, his approach can be justified to be correct and not a simple coincidence. Let's wait to hear from the OP
@WMDhamnekar I suggest you read carefully post #14, it contains info that leads to justification of your method as correct (and me wrong that criticized you )...

• • WMDhamnekar and SammyS
My solution's strategy is that I break the triple volume integral into the volume of the cylinder from ##z=-\sqrt{3}## to ##z=\sqrt{3}## and the volume of two spherical caps from ##z=\pm\sqrt 3 ## to ##z=\pm 2##. So the integrals are $$\int_{-\sqrt{3}}^{\sqrt{3}}\int\int_{x^2+y^2\leq 1}1dxdydz+\int_{-2}^{-\sqrt{3}}\int\int_{x^2+y^2\leq 4-z^2}1dxdydz+\int_{\sqrt{3}}^{2}\int\int_{x^2+y^2\leq 4-z^2}1dxdydz$$

The integration with respect to x and y is done in polar coordinates and then follows the integration with respect to z. For the first term after we integrate in polar coordinates we get a function which is constant w.r.t z, but for the 2nd and 3rd term we get a function that depends on z.
Other two math experts have solved this question using two different methods:
Method 1

## V= \frac{32\pi}{3} - \displaystyle\int_0^{2\pi} \displaystyle\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \displaystyle\int_{\csc{\phi}}^2 \rho^2 \sin{\phi} d\rho d\phi d\theta = \frac{4\pi}{3} (8 -3\sqrt{3})##

Method 2

## V = \displaystyle\int_0^1 4\pi r (4-r^2) dr = 4\pi (\frac83 -\sqrt{3}) \sim 11.745##

Method 3

Using Monte Carlo method in any programming language also, we can answer this question.

• Delta2
All this is fine, just to utilize the hint from @SammyS to prove that your work does indeed calculate the correct result (and it is not a coincidence). That hint tell you essentially that the volume of the small sphere is equal to the volume of the big sphere minus the volume of the cylinder between the planes ##z=\sqrt{3} ## and ##z=-\sqrt{3}##. So when you subtract the volume of the small sphere from the volume of the big sphere, you calculate the volume of the two spherical caps, plus the volume of the aforementioned cylinder, which is the same as what I did.

Method 2

## V = \displaystyle\int_0^1 4\pi r (4-r^2) dr = 4\pi (\frac83 -\sqrt{3}) \sim 11.745##
$$\int _0^1 4\pi r(4-r^2)\mathrm{d}r = -2\pi \int _0^1 (4-r^2)\mathrm{d}(4-r^2) = 7\pi \not\approx 11.745$$
This integral makes no sense to me in the context of this problem.

Whether or not "other math experts" have got the right answer shouldn't be relevant and neither should you blindly quote something as a solution. Do you understand any of the solutions? That's the important part.

Method 1

## V= \frac{32\pi}{3} - \displaystyle\int_0^{2\pi} \displaystyle\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \displaystyle\int_{\csc{\phi}}^2 \rho^2 \sin{\phi} d\rho d\phi d\theta = \frac{4\pi}{3} (8 -3\sqrt{3})##
The integral is correct. I haven't checked if the evaluation is correct.

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\begin{align*} \int_0^{2\pi} \displaystyle\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \displaystyle\int_{\csc{\phi}}^2 \rho^2 \sin{\phi} d\rho d\phi d\theta &= 2 \int_0^{2\pi} \displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \displaystyle\int_{\csc{\phi}}^2 \rho^2 \sin{\phi} d\rho d\phi d\theta \\ &=\frac{2}{3} \int _0^{2\pi} \int _{\pi/6} ^{\pi /2} \left (8\sin\phi - \frac{1}{\sin ^2\phi}\right ) d\phi d\theta \\ &= \frac{4\pi}{3} \int _{\pi /6} ^{\pi /2} \left (8\sin\phi - \frac{1}{\sin ^2\phi}\right ) d\phi \\ &= \frac{4\pi}{3} \left (-8\cos\phi + \frac{1}{\tan\phi}\right )\Bigg\vert _{\pi /6}^{\pi /2} \\ &= \frac{12\sqrt{3}\pi}{3} \end{align*}
so
$$V = \frac{32\pi}{3} - \frac{12\sqrt{3}\pi}{3} = \frac{4\pi}{3} \left (8-3\sqrt{3} \right ).$$

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• Delta2
It turns out a PDF of the book is available for free from http://www.mecmath.net.
After looking at the PDF of the book, I would say that the author, never covered that scheme of subtracting the volume of a sphere of radius ##\sqrt{3}##. In fact, I would think that method would be used to avoid using Calculus altogether.

I suspect that OP looked at the Answer key: from which it's pretty straight forward to see the volume of a sphere of radius ##2## minus the volume of a sphere of radius ##\sqrt{3}##.

##\displaystyle \frac{4\pi}{3} 2^3 - \frac{4\pi}{3} \left(\sqrt{3}\right)^3 ##

• Delta2
$$\int _0^1 4\pi r(4-r^2)\mathrm{d}r = -2\pi \int _0^1 (4-r^2)\mathrm{d}(4-r^2) = 7\pi \not\approx 11.745$$
This integral makes no sense to me in the context of this problem.

Whether or not "other math experts" have got the right answer shouldn't be relevant and neither should you blindly quote something as a solution. Do you understand any of the solutions? That's the important part.

The integral is correct. I haven't checked if the evaluation is correct.

edit:
\begin{align*} \int_0^{2\pi} \displaystyle\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \displaystyle\int_{\csc{\phi}}^2 \rho^2 \sin{\phi} d\rho d\phi d\theta &= 2 \int_0^{2\pi} \displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \displaystyle\int_{\csc{\phi}}^2 \rho^2 \sin{\phi} d\rho d\phi d\theta \\ &=\frac{2}{3} \int _0^{2\pi} \int _{\pi/6} ^{\pi /2} \left (8\sin\phi - \frac{1}{\sin ^2\phi}\right ) d\phi d\theta \\ &= \frac{4\pi}{3} \int _{\pi /6} ^{\pi /2} \left (8\sin\phi - \frac{1}{\sin ^2\phi}\right ) d\phi \\ &= \frac{4\pi}{3} \left (-8\cos\phi + \frac{1}{\tan\phi}\right )\Bigg\vert _{\pi /6}^{\pi /2} \\ &= \frac{12\sqrt{3}\pi}{3} \end{align*}
so
$$V = \frac{32\pi}{3} - \frac{12\sqrt{3}\pi}{3} = \frac{4\pi}{3} \left (8-3\sqrt{3} \right ).$$

$$\int _0^1 4\pi r(4-r^2)\mathrm{d}r = -2\pi \int _0^1 (4-r^2)\mathrm{d}(4-r^2) = 7\pi \not\approx 11.745$$
This integral makes no sense to me in the context of this problem.

Whether or not "other math experts" have got the right answer shouldn't be relevant and neither should you blindly quote something as a solution. Do you understand any of the solutions? That's the important part.

The integral is correct. I haven't checked if the evaluation is correct.

edit:
\begin{align*} \int_0^{2\pi} \displaystyle\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \displaystyle\int_{\csc{\phi}}^2 \rho^2 \sin{\phi} d\rho d\phi d\theta &= 2 \int_0^{2\pi} \displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \displaystyle\int_{\csc{\phi}}^2 \rho^2 \sin{\phi} d\rho d\phi d\theta \\ &=\frac{2}{3} \int _0^{2\pi} \int _{\pi/6} ^{\pi /2} \left (8\sin\phi - \frac{1}{\sin ^2\phi}\right ) d\phi d\theta \\ &= \frac{4\pi}{3} \int _{\pi /6} ^{\pi /2} \left (8\sin\phi - \frac{1}{\sin ^2\phi}\right ) d\phi \\ &= \frac{4\pi}{3} \left (-8\cos\phi + \frac{1}{\tan\phi}\right )\Bigg\vert _{\pi /6}^{\pi /2} \\ &= \frac{12\sqrt{3}\pi}{3} \end{align*}
so
$$V = \frac{32\pi}{3} - \frac{12\sqrt{3}\pi}{3} = \frac{4\pi}{3} \left (8-3\sqrt{3} \right ).$$
Sorry. I stated wrongly the Method 2.
Corrected Method 2
is ## \displaystyle\int_{r=0}^1 4\pi r\sqrt{4-r^2} dr = \frac{\pi(32- 12\sqrt{3})}{3} \sim 11.745 ##

I know you didn't ask for my comment but it might be another method with "hidden" justification, just like yours is. You have to provide more details (at least to me) on why the integrand is ##4\pi r\sqrt{4-r^2}## and also why the limits of integration are from 0 to 1.

I know you didn't ask for my comment but it might be another method with "hidden" justification, just like yours is. You have to provide more details (at least to me) on why the integrand is ##4\pi r\sqrt{4-r^2}## and also why the limits of integration are from 0 to 1.
## z = \sqrt{4-r^2} ## and ## \theta = 2\pi r## where r= radius of sphere =2 ## \therefore \theta = 4\pi##

## z = \sqrt{4-r^2} ## and ## \theta = 2\pi r## where r= radius of sphere =2 ## \therefore \theta = 4\pi##
Sorry your answer is too laconic for my standards , in order to understand. Why we take that ##z##, why that ##\theta## and why we multiply them? Also r seems to vary from 0 to 1 not from 0 to 2.

## \displaystyle\int_{r=0}^1 4\pi r\sqrt{4-r^2} dr = \frac{\pi(32- 12\sqrt{3})}{3} \sim 11.745 ##
This does work.
## z = \sqrt{4-r^2} ## and ## \theta = 2\pi r## where r= radius of sphere =2 ## \therefore \theta = 4\pi##
##2\pi r## is the perimeter of a circle with radius ##r##. Multiplied by some "height", we get the surface area of the side of a cylinder. The intuition here is that the volume of the (upper/lower half of the) body is obtained by summing up infinitely many cylinders with appropriate heights.

Here's my question to you. Why does the integral, as given, account for the entire volume and not just one of the halves?

• SammyS and Delta2
This does work.

##2\pi r## is the perimeter of a circle with radius ##r##. Multiplied by some "height", we get the surface area of the side of a cylinder. The intuition here is that the volume of the (upper/lower half of the) body is obtained by summing up infinitely many cylinders with appropriate heights.

Here's my question to you. Why does the integral, as given, account for the entire volume and not just one of the halves?
Sorry. I don't know the answer to your question. If you answer it yourself, it would be better for me and other readers, viewers, mathematics students and members of this Physics Forum. By the way, I also want to know if the jacobian have been used in this Corrected Method 2

$$V = 2\int _0^1 2\pi r\sqrt{4-r^2}\mathrm{d}r.$$
• 