Triple integral problem

  • #1
42
3

Homework Statement


Given
[tex]
E = [(x,y,z) s. t. 0 \leq x \leq 2, 0 \leq y \leq \sqrt{2x - x^2}, 0 \leq z \leq 2]
[/tex]
Calculate
[tex]
\int_E z^3\sqrt{x^2+y^2}dxdydz
[/tex]

Homework Equations


In cylindrical coordinates:
[tex]
x=rcos(\theta)\\y=rsin(\theta)\\z=z\\dxdydz = \rho d\rho d\theta dz
[/tex]

The Attempt at a Solution


The integral is kind of weird to compute in Cartesian coordinates, so I tried to find what kind of curve are x and y bounded to vary on.
[tex]
y<\sqrt{2x-x^2}\\ y^2+x^2-2x<0
[/tex]
and by completing the square I got:
[tex]
y^2 + (x-1)^2 < 1
[/tex]
which is a circle of radius 1 centered in (1,0). So x and y vary in this circle and, given the upper bounds of x and z, I think this is the left half of a cylinder centered in ##(1,0)## with height 2.
Here comes my problem. I thought of using cylindrical coordinates to compute the integral and this leads to:
[tex]
x=\rho cos(\theta)+1\\y=\rho sin(\theta)\\z=z\\
0 \leq \rho \leq 1,-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2},0 \leq z \leq 2
[/tex]
After plugging in the integral:
[tex]
\int^1_0 \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\int^2_0 z^3 \sqrt{\rho^2+1+2\rho cos(\theta)}\rho d\rho d\theta dz
[/tex]
Computing this integral for z is easy, but when it comes to ##\theta## and ##\rho## I have no idea how to solve it. Can anyone give me any suggestions?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Given
[tex]
E = [(x,y,z) s. t. 0 \leq x \leq 2, 0 \leq y \leq \sqrt{2x - x^2}, 0 \leq z \leq 2]
[/tex]
Calculate
[tex]
\int_E z^3\sqrt{x^2+y^2}dxdydz
[/tex]

Homework Equations


In cylindrical coordinates:
[tex]
x=rcos(\theta)\\y=rsin(\theta)\\z=z\\dxdydz = \rho d\rho d\theta dz
[/tex]

I think it is easiest to just avoid cylindrical coordinates; stick to the original cartesian system. Then, after doing the ##z##-integration you are left with the problem of integrating
[tex] F(x) = \int_0^{\sqrt{2x - x^2}} \sqrt{x^2+y^2} \, dy [/tex]
from ##x = 0## to ##x = 2##.

It is messy and lengthy, but involves no more than typical changes of variables and/or integration by parts. Alternatively, you can consult a table of integrals or use a computer algebra package. The final answer is simple, but getting there takes work.
 
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  • #3
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I agree with Ray that it's easiest to do the integration in Cartesian coordinates.
I can see how cylindrical seems tempting with the volume of integration actually being half a cylinder. unfortunately the ##\sqrt{x^2+y^2}## term makes this hard. If we instead had ##\sqrt{(x-1)^2+y^2}## as our integrand cylindrical would've been a very easy way to solve the problem.
 
  • #4
SammyS
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The Attempt at a Solution


The integral is kind of weird to compute in Cartesian coordinates, so I tried to find what kind of curve are x and y bounded to vary on.[tex]
y<\sqrt{2x-x^2}\\ y^2+x^2-2x<0
[/tex]and by completing the square I got:[tex]
y^2 + (x-1)^2 < 1
[/tex]which is a circle of radius 1 centered in (1,0). So x and y vary in this circle and, given the upper bounds of x and z, I think this is the left half of a cylinder centered in ##(1,0)## with height 2.
Here comes my problem. I thought of using cylindrical coordinates to compute the integral and this leads to:[tex]
x=\rho cos(\theta)+1\\y=\rho sin(\theta)\\z=z\\
0 \leq \rho \leq 1,-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2},0 \leq z \leq 2
[/tex]After plugging in the integral:[tex]
\int^1_0 \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\int^2_0 z^3 \sqrt{\rho^2+1+2\rho cos(\theta)}\rho d\rho d\theta dz
[/tex]Computing this integral for z is easy, but when it comes to ##\theta## and ##\rho## I have no idea how to solve it. Can anyone give me any suggestions?
I don't see any problem with using cylindrical coordinates.

You have confused what's needed to change the integrand to cylindrical, with what's needed to change the limits of integration to cylindrical.

For the integrand, simply change ##\ x^2+y^2\ ## to ##\ \rho^2 \ ##.
 
  • #5
42
3
I tried solving the integral using Cartesian coordinates, it's quite a mess though. By setting ##x=ytan(u), dx=ysec^2(u)du## the integral is
##\int y\sqrt{1+tan^2(u)}sec^2(u)du## which leads to ##y\int sec^3(u) du##. To integrate this, I tried with the reduction formula
##\int sec^m(u)du = \frac{sin(u)sec^{m-1}(u)}{m-1} + \frac{m-2}{m-1}\int sec^{m-2}(u)du##. I could integrate that last secant but, since this is all in ##u##, I would have to change the boundaries of integration afterwards.. And that would lead me to have very complicated trigonometric functions that I'd have to integrate in y afterwards.
Sammy shouldn't I take into account that ##x=\rho cos(\theta)+1## while equating ##x^2+y^2## with ##\rho##?
 
  • #6
36
1
The problem seems to be designed for the use in Cartesian coordinates, I think. Because Both x and z are limited by planes in cartesian, and the upper limit for y (the only limit there that isn't constant) is written only in terms of x (notice how this this value for y fits well with that square root). Also, if you integrate first in y you will notice that the problem is in a form which suggests a certain kind of substitution.
 
  • #7
334
47
For the integrand, simply change ##\ x^2+y^2\ ## to ##\ \rho^2 \ ##.
That's not the same cylinder that we're supposed to integrate over. It says ##y < \sqrt{2x-x^2}## which give us the half of the cylinder ##(x-1)^2+y^2=1##. You can change it like you said but you then get integration limits that are no easier to deal with (and arguable harder).
 
  • #8
SammyS
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Sammy shouldn't I take into account that ##x=\rho cos(\theta)+1## while equating ##x^2+y^2## with ##\rho##?
In a word, NO.

On the x-y plane, the boundary is the half circle in the first quadrant.

Writing that as ##\ x^2 - 2x + y^2=0\ ## makes conversion to ##\ \rho , \theta \ ## fairly easy.

When it's done this way you may see that the limits are not difficult to obtain.
 
Last edited:
  • #9
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3
In a word, NO.

On the x-y plane, the boundary is the half circle in the first quadrant.

Writing that as ##\ x^2 - 2x + y^2=0\ ## makes conversion to ##\ \rho , \theta \ ## fairly easy.

When it's done this way you may see that the limits are not difficult to obtain.
Could you elaborate some more on this? I don't understand why I can set ##x^2+y^2 = \rho^2## if the circle is not centered at the origin. Thanks to everyone btw, been struggling for quite some time on this problem today.
EDIT: I've just realized that my first interpretation of the problem was wrong, as x varies from 0 to 2, the domain of integration is actually the whole cylinder centered in (1,0)
 
Last edited:
  • #10
SammyS
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Could you elaborate some more on this? I don't understand why I can set ##x^2+y^2 = \rho^2## if the circle is not centered at the origin. Thanks to everyone btw, been struggling for quite some time on this problem today.
Oh ! I think I see what you're attempting.

You seem to be attempting to translate the problem to a new coordinate system having it's origin at the center of that circle.


There is no need to do that.

Use standard cylindrical coordinates. The equation of the circle, ##\ x^2 -2x +y^2=0\ ## is easy enough to get in polar coordinates. (Same circle as ##\ (x-1)^2 +y^2=1\ ##)
 
  • #11
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3
Oh ! I think I see what you're attempting.

You seem to be attempting to translate the problem to a new coordinate system having it's origin at the center of that circle.


There is no need to do that.

Use standard cylindrical coordinates. The equation of the circle, ##\ x^2 -2x +y^2=0\ ## is easy enough to get in polar coordinates. (Same circle as ##\ (x-1)^2 +y^2=1\ ##)
I think I solved it. The boundaries for ##\rho## and ##\theta## are ##-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}##, because the circle is on the right w.r.t. our coordinate system, and ## 0 \leq \rho \leq 2cos(\theta)##, which I get from the equation of the circle. The integral then becomes:
[tex]
\int^\frac{\pi}{2}_{-\frac{\pi}{2}}d\theta \int^{2cos(\theta)}_0 d\rho \int^2_0 z^3\rho^2 dz = \frac{256}{9}
[/tex]
Was my reasoning correct Sammy? Thanks a lot for your hints by the way!
 
  • #12
SammyS
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I think I solved it. The boundaries for ##\rho## and ##\theta## are ##-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}##, because the circle is on the right w.r.t. our coordinate system, and ## 0 \leq \rho \leq 2cos(\theta)##, which I get from the equation of the circle. The integral then becomes:
[tex]
\int^\frac{\pi}{2}_{-\frac{\pi}{2}}d\theta \int^{2cos(\theta)}_0 d\rho \int^2_0 z^3\rho^2 dz = \frac{256}{9}
[/tex]
Was my reasoning correct Sammy? Thanks a lot for your hints by the way!
It's only a half circle.

Looks like it should be 0 to π/2 to me.


(Added in Edit:) Ray's answer in the next post is correct.
 
Last edited:
  • #13
Ray Vickson
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I think I solved it. The boundaries for ##\rho## and ##\theta## are ##-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}##, because the circle is on the right w.r.t. our coordinate system, and ## 0 \leq \rho \leq 2cos(\theta)##, which I get from the equation of the circle. The integral then becomes:
[tex]
\int^\frac{\pi}{2}_{-\frac{\pi}{2}}d\theta \int^{2cos(\theta)}_0 d\rho \int^2_0 z^3\rho^2 dz = \frac{256}{9}
[/tex]
Was my reasoning correct Sammy? Thanks a lot for your hints by the way!

I get the answer 64/9 ≈ 7.1111 .

We can argue easily that your answer 256/9 ≈ 28.4444 is quite a bit too large. After doing the z-integration (to get 4), the integrand is ##f = 4\sqrt{x^2+y^2}##, integrated over a semicircle with center (1,0) and radius 1. We have ##f \leq 4 \sqrt{2^2+0^2} = 8## over the whole region, so the integral is ##\leq 8 \pi/2 = 4 \pi \approx 12.566##, and that is an upper bound on the integral.
 
  • #14
42
3
I get the answer 64/9 ≈ 7.1111 .

We can argue easily that your answer 256/9 ≈ 28.4444 is quite a bit too large. After doing the z-integration (to get 4), the integrand is ##f = 4\sqrt{x^2+y^2}##, integrated over a semicircle with center (1,0) and radius 1. We have ##f \leq 4 \sqrt{2^2+0^2} = 8## over the whole region, so the integral is ##\leq 8 \pi/2 = 4 \pi \approx 12.566##, and that is an upper bound on the integral.
Just finished redoing it (like Sammy pointed out I was integrating over the wrong region) and I got ##\frac{64}{9}## as well :D
 

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