Volume Calculation of Region w/ Triple Integral

In summary, the expert thinks that the cylindrical coordinate is more complicated than the spherical coordinate, and that the difference in volumes is due to a mistake in calculating the divergence for the vector field.
  • #1
Telemachus
835
30
Hi, I have to compute de volume for the region:

[tex]\{(x,y,z):x^2+y^2+z^2\leq{16};z\geq{1} \}[/tex]

I've tried to do by two different parametrizations, in spherics and cylindrical coordinates

For cylindrical coordinates I've made:

[tex]3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}dz d \theta dr\approx{162.26}[/tex]

For spherics I've considered [tex]4\sin \phi=\sqrt[ ]{15}\Rightarrow{\phi=Arcsen\left( \displaystyle\frac{\sqrt[ ]{15}}{4} \right)}=u[/tex]

[tex]3 \displaystyle\int_{0}^{u}\int_{0}^{2\pi}\int_{\cos \phi}^{4}d\rho d \theta d\phi\approx{81.13 }[/tex]

I've used wolfram to make the calculus, and it gives that for cylindrical coord. I'm having twice the volume that I have for spherical coord. I don't know where I'm committing the mistake. Bye, thanks for helping :)

Note: I'm multiplying by 3 because I was going to use for Gauss theorem and is the divergence for the vector field given by the problem
 
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  • #2
wheres your Jacobian for spherical and cylindrical coordinates.
 
  • #3
Thanks :P I forgot about that, I thought I wasn't making any change of variables, but now I see. It's because I've been doing some surface integrals and the jacobian doesn't appear on those kind of integrals, I'm not sure why is that, but I think that role is played by the normal vector, which is pretty much like the jacobian.
 
Last edited:
  • #4
Well, I added the jacobians, but still giving different, I don't know why. The difference now is more sutil: In cylindrical coord. gives (clic here)

In spherical coord gives


I tried to proceed with cylindrical, but the exercise gets more complicated, so I want to be sure that what I'm doing with spherical is ok, so I can proceed that way, but still giving different answers.
 
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  • #5
I think you bound for phi in spherical should be [itex] \phi=arctan(\sqrt{15}) [/itex]
and not your arcsin that you have.
 
  • #6
why is that? I've reasoned the bound this way:

attachment.php?attachmentid=37870&stc=1&d=1312912197.png


Then [tex]4 \sin \phi=\sqrt[]{15}[/tex]
 

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  • #7
That gives the same limits because you have, equivalently,
\begin{align*}
\cos \phi &= \frac{1}{4} \\
\sin \phi &= \sqrt{1-\left(\frac{1}{4}\right)^2} = \frac{\sqrt{15}}{4} \\
\tan \phi &= \frac{\sin\phi}{\cos\phi} = \sqrt{15}
\end{align*}
I think the lower bound of [itex]\rho[/itex] is wrong.
 
  • #8
Oh, yes, it says cos phi, but I used secant in wolfram. I'll correct the typo, thanks vela.

I can´t edit now the first post, but this is the integration I've computed with wolfram:[tex]3 \displaystyle\int_{0}^{u}\int_{0}^{2\pi}\int_{\sec \phi}^{4}\rho^2 \sin \phi d\rho d \theta d\phi=81\pi[/tex]

And for cylindrical:[tex]3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}r dz d \theta dr\approx{80\pi}[/tex]
 
  • #9
their equivalent i just did it with tangent. Yours is fine . It looks like your bounds are right but i did both integral and got 80 and 100 . so i got different answers but maybe i made a mistake.
 
  • #10
I got [itex]81\pi[/itex] for both using Mathematica.
 
  • #11
Thanks vela, perhaps it's only a bug in wolfram alpha :)
 
  • #12
[tex]3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}r dz d \theta dr=80\pi[/tex]

This is actually right, I just corroborated it doing it by hand:

[tex]3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}r dz d \theta dr=3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}(r\sqrt[ ]{16-r^2}-r)d \theta dr[/tex]
Using:
[tex]u=16-r^2\rightarrow du=-2r[/tex]
[tex]3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}(r\sqrt[ ]{16-r^2}-r)d \theta dr=-3\pi\displaystyle\int_{16}^{0}\sqrt[ ]{u}du-6\pi \left( \frac{r^2}{2} \right ) _0^4=2\pi 16^{3/2}-48\pi=80\pi[/tex]
 
  • #13
You got the limits for u wrong and plugged in the wrong limits for the other term.
 
  • #14
Uups you're right. That was the mistake, thank you vela :) I made a big mess with this involving spherical and cylindrical coordinates :P but the thing is that for the volume it's easier to compute it using cylindrical coordinates (unless I think so). But for the surface integrals spheric coordinates seems to be better (gives simpler expressions).

I wanted to corroborate the Gauss theorem for that region, the problem gives a vector field. I have some doubts about that too, I should make another topic for that, right?
 
  • #15
Probably a good idea to start a new thread for that.
 

What is the concept of triple integral?

Triple integral is a mathematical concept used to calculate the volume of a three-dimensional region. It involves integrating a function over a region in three-dimensional space, with respect to three different variables.

How do you set up a triple integral?

To set up a triple integral, you first need to determine the limits of integration for each variable based on the boundaries of the region. Then, you need to choose the order of integration, which can be either x, y, z or any other combination. Finally, you need to write the function to be integrated in terms of the chosen variables and set up the integral using the limits and the function.

What is the difference between a definite and indefinite triple integral?

A definite triple integral has specific limits of integration, which means it calculates the exact volume of a region. An indefinite triple integral has no specified limits and is used to find the general volume of a region or to evaluate a function.

How does the shape of the region affect the calculation of volume using triple integral?

The shape of the region affects the limits of integration and the order of integration chosen for the triple integral. Some regions may require multiple integrals to accurately calculate the volume, while others may only need a single triple integral.

Can triple integrals be used to calculate volumes of irregularly shaped regions?

Yes, triple integrals can be used to calculate the volume of any three-dimensional region, regardless of its shape. However, the limits of integration and the order of integration may be more complex for irregularly shaped regions.

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