Triple integral

1. Aug 8, 2011

Telemachus

Hi, I have to compute de volume for the region:

$$\{(x,y,z):x^2+y^2+z^2\leq{16};z\geq{1} \}$$

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

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

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

$$3 \displaystyle\int_{0}^{u}\int_{0}^{2\pi}\int_{\cos \phi}^{4}d\rho d \theta d\phi\approx{81.13 }$$

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

2. Aug 8, 2011

cragar

wheres your Jacobian for spherical and cylindrical coordinates.

3. Aug 8, 2011

Telemachus

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: Aug 8, 2011
4. Aug 9, 2011

Telemachus

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.

Last edited: Aug 9, 2011
5. Aug 9, 2011

cragar

I think you bound for phi in spherical should be $\phi=arctan(\sqrt{15})$
and not your arcsin that you have.

6. Aug 9, 2011

Telemachus

why is that? I've reasoned the bound this way:

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

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7. Aug 9, 2011

vela

Staff Emeritus
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 $\rho$ is wrong.

8. Aug 9, 2011

Telemachus

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:

$$3 \displaystyle\int_{0}^{u}\int_{0}^{2\pi}\int_{\sec \phi}^{4}\rho^2 \sin \phi d\rho d \theta d\phi=81\pi$$

And for cylindrical:

$$3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}r dz d \theta dr\approx{80\pi}$$

9. Aug 9, 2011

cragar

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. Aug 9, 2011

vela

Staff Emeritus
I got $81\pi$ for both using Mathematica.

11. Aug 9, 2011

Telemachus

Thanks vela, perhaps it's only a bug in wolfram alpha :)

12. Aug 9, 2011

Telemachus

$$3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}r dz d \theta dr=80\pi$$

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

$$3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}r dz d \theta dr=3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}(r\sqrt[ ]{16-r^2}-r)d \theta dr$$
Using:
$$u=16-r^2\rightarrow du=-2r$$
$$3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}(r\sqrt[ ]{16-r^2}-r)d \theta dr=-3\pi\displaystyle\int_{16}^{0}\sqrt[ ]{u}du-6\pi \left( \frac{r^2}{2} \right ) _0^4=2\pi 16^{3/2}-48\pi=80\pi$$

13. Aug 9, 2011

vela

Staff Emeritus
You got the limits for u wrong and plugged in the wrong limits for the other term.

14. Aug 9, 2011

Telemachus

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. Aug 9, 2011

vela

Staff Emeritus
Probably a good idea to start a new thread for that.