# Multiple variable integration using polar coordinates: Quick question

• Mar 28th 2009, 04:42 AM
Pierre-Alexandre
Multiple variable integration using polar coordinates: Quick question
Hello folks,

I have a quick - and probably stupid - question regarding this particular method of integrating multiple variable functions. The volume of the region given by the formula goes until and stops at the oXY plane, right?

Let's say I want the volume of a unit sphere. I would then have to multiply the $\displaystyle \sqrt{x^2+y^2+C}$ by 2, correct?

Thanks!
• Mar 28th 2009, 06:08 AM
Scott H
Yes, you are correct. :)

The upper half of a unit sphere is defined by the function

\displaystyle \begin{aligned} z&=\sqrt{(x-a)^2+(y-b)^2}\\ &=\sqrt{x^2-2ax+a^2+y^2-2by+b^2}\\ &=\sqrt{x^2+y^2-2(ax+by)+a^2+b^2}. \end{aligned}

In finding the volume of the upper half of a unit sphere (say, for $\displaystyle a=b=0$), we would integrate this function in its circular domain and multiply by two.
• Mar 28th 2009, 06:27 AM
Jester
Quote:

Originally Posted by Scott H
Yes, you are correct. :)

The upper half of a unit sphere is defined by the function

\displaystyle \begin{aligned} z&=\sqrt{(x-a)^2+(y-b)^2}\\ &=\sqrt{x^2-2ax+a^2+y^2-2by+b^2}\\ &=\sqrt{x^2+y^2-2(ax+by)+a^2+b^2}. \end{aligned}

In finding the volume of the upper half of a unit sphere (say, for $\displaystyle a=b=0$), we would integrate this function in its circular domain and multiply by two.

I believe that it would be (with the center at the origin)

$\displaystyle x^2+y^2+z^2=1$ or for the top half

$\displaystyle z = \sqrt{1 - x^2 - y^2}$
• Mar 28th 2009, 06:27 AM
HallsofIvy
Quote:

Originally Posted by Pierre-Alexandre
Hello folks,

I have a quick - and probably stupid - question regarding this particular method of integrating multiple variable functions. The volume of the region given by the formula goes until and stops at the oXY plane, right?

I don't know what you mean by that.

Quote:

Let's say I want the volume of a unit sphere. I would then have to multiply the $\displaystyle \sqrt{x^2+y^2+C}$ by 2, correct?

Thanks!
The unit sphere is given by $\displaystyle x^2+ y^2+ x^2= 1$ or $\displaystyle z= \pm\sqrt{1- x^2- y^2}$. $\displaystyle z= \sqrt{1-x^2-y^2}$ is the upper half of the sphere so, by symmetry, you could get the volume of the sphere by integrating 2 times that. I don't know where you got $\displaystyle \sqrt{x^2+ y^2+ C}$ or what "C" is.