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Question about surface and volume integrals

http://workspace.imperial.ac.uk/mathematics/public/students/ug/exampapers/2008/M2AA2-2008.PDF

2(b)

So we calculate the divergence and it's just 2. That's nice. But then how do we do the volume integrals? What is this volume and how do we deal with it? I tried to use cylindrical polar coordinates but that doesn't work because both cylinders are not oriented the same way. And spherical polar coordinates don't help much either.
Reply 1
Avatar for gangsta316
gangsta316
OP
13
Does anyone have any ideas?
Reply 2
Avatar for Mathletics
Mathletics
I thought about this for about ten minutes and now my head is hurting!
(edited 13 years ago)
Reply 3
Avatar for Mathletics
Mathletics
How about this.

Unparseable latex formula:

\alpha(r,\theta,\phi)=(r cos\theta cos\phi, \frac{2r\theta}{\pi}sin\phi, \frac{2r\phi}{\pi}sin\theta)\ for\ [br] 0 \leq r \leq a,\[br] 0 \leq \theta \leq \frac{\pi}{2},[br]\ 0\leq \phi \leq \frac{\pi}{2}



I *THINK* this gives the parameterisation for the region restricted to the strictly positive octant.
(edited 13 years ago)
Reply 4
Avatar for Mathletics
Mathletics
I had a go with that parametrisation but had serious trouble finding the jacobian.
I had another thought, isn't the volume

8∫0a∫0a2−x2∫0a2−x21dydzdx=8∫0aa2−x2dx=16a33?8 \displaystyle \int^a_0 \displaystyle \int^{\sqrt{a^2-x^2}}_0 \displaystyle \int^{\sqrt{a^2-x^2}}_0 1 dydzdx=8\displaystyle \int^a_0 a^2-x^2 dx=\frac{16a^3}{3}?



And so your answer should be twice this.
(edited 13 years ago)
Reply 5
Avatar for gangsta316
gangsta316
OP
13
Original post by Mathletics
I had a go with that parametrisation but had serious trouble finding the jacobian.
I had another thought, isn't the volume

8∫0a∫0a2−x2∫0a2−x21dydzdx=8∫0aa2−x2dx=16a33?8 \displaystyle \int^a_0 \displaystyle \int^{\sqrt{a^2-x^2}}_0 \displaystyle \int^{\sqrt{a^2-x^2}}_0 1 dydzdx=8\displaystyle \int^a_0 a^2-x^2 dx=\frac{16a^3}{3}?



And so your answer should be twice this.


Awesome. That looks right. So I was trying to be smart and use some kind of polars when the "inferior" Cartesian coordinates worked just as well!

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