I need to find the volume enclosed by $\displaystyle x^2+y^2+z^2=a^2$ and $\displaystyle x^2+y^2=ax$ where $\displaystyle a>0.$

How do I find the bounds? Do I apply spherical coordinates as written?

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- Mar 11th 2011, 06:51 PMKillerFinding volume by using triple integral
I need to find the volume enclosed by $\displaystyle x^2+y^2+z^2=a^2$ and $\displaystyle x^2+y^2=ax$ where $\displaystyle a>0.$

How do I find the bounds? Do I apply spherical coordinates as written? - Mar 11th 2011, 06:55 PMProve It
To start with, I expect that you have written the second bound wrongly, since that is the equation of a plane figure, not a solid...

- Mar 11th 2011, 07:00 PMKiller
I edited the first equation, but what's wrong with the second one?

- Mar 11th 2011, 11:30 PMProve It
It's a 2 Dimensional object, i.e. a plane figure. How are we supposed to know where along the $\displaystyle \displaystyle z$ axis it's supposed to lie?

- Mar 12th 2011, 01:57 AMtom@ballooncalculus
Assuming the cylinder extends indefinitely up and down the z dimension, we have Viviani's Curve. Doing just the top half, we have z going from 0 (where it 'starts', on the (x,y) plane) up to $\displaystyle \sqrt{a^2 - r^2}$ (where it hits the hemisphere). And we have r going from 0 at the centre (z axis), up to a cos theta i.e. everywhere inside the cylinder. And theta is turning through the x-positive half of the space, i.e. from minus pi/2 to pi/2. So...

$\displaystyle \displaystyle{V = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\ \int_{0}^{a \cos \theta} \int_{0}^{\sqrt{a^2 - r^2}} r\ dz\ dr\ d\theta}$

Just in case a picture helps to follow through from the inside out, we can start bottom left here, integrating r with respect to z...

http://www.ballooncalculus.org/draw/intMulti/two.png

... where (key in spoiler) ...

__Spoiler__:

Which leaves a couple of blanks to fill. Hope this helps.

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Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - Mar 12th 2011, 03:21 AMHallsofIvy
- Mar 12th 2011, 03:24 AMProve It
- Mar 12th 2011, 09:41 AMHallsofIvy
No, it

**was**stated that this problem was in three dimensions. The fact that there was no restriction put on z meant that it could be anything.