# Calculate the volume.

• Sep 13th 2008, 11:24 AM
NicoleB
Calculate the volume.
Hi. Was hoping someone could help em out.

I have to calculate the volume between the two functions.

$x^2+y^2 \leq z \leq sqrt(3-x^2-y^2)$

So i figured I try to find the area D:

$z^2-3=-x^2-y^2$
$z=x^2+x^2$

Wich gives: $3-z^2=z$ => ${(sqrt(13)-1} )/{ 2}$

I get a integral:

$\int_{0}^{2\pi}\int_{0}^{((sqrt(13)-1)/2}sqrt(3-r^2)-r^2) rdrd\theta$

However solving the integral I get a rather nasty algebra. Is there a more clever way to solve the equation?

Very thankful for a little help.
• Sep 13th 2008, 11:54 AM
Chris L T521
Quote:

Originally Posted by NicoleB
Hi. Was hoping someone could help em out.

I have to calculate the volume between the two functions.

$x^2+y^2 \leq z \leq sqrt(3-x^2-y^2)$

So i figured I try to find the area D:

$z^2-3=-x^2-y^2$
$z=x^2+x^2$

Wich gives: $3-z^2=z$ => ${(sqrt(13)-1} )/{ 2}$

I get a integral:

$\int_{0}^{2\pi}\int_{0}^{((sqrt(13)-1)/2}sqrt(3-r^2)-r^2) rdrd\theta$

However solving the integral I get a rather nasty algebra. Is there a more clever way to solve the equation?

Very thankful for a little help.

I'd say that you should convert to polar first:

$r^2\leq z\leq \sqrt{3-r^2}$

The value of r where they are equal would be:

$r^4=3-r^2\implies r^4+r^2-3=0\implies r^2=\frac{-1\pm\sqrt{13}}{2}$

So we have $r^2=\frac{\sqrt{13}-1}{2}\implies r=\sqrt{\frac{\sqrt{13}-1}{2}}$ (noting that $r\geq0$)

Also note that $0\leq\vartheta\leq2\pi$

Thus, the integral would be $\int_0^{2\pi}\int_0^{\sqrt{\frac{\sqrt{13}-1}{2}}}(\sqrt{3-r^2}-r^2)r\,dr\,d\vartheta=\int_0^{2\pi}\int_0^{\sqrt{\ frac{\sqrt{13}-1}{2}}}r\sqrt{3-r^2}-r^3\,dr\,d\vartheta$

Mathematica gives me a messy solution. The integration shouldn't be hard, but the simplifying can get a little long.

Here is my suggestion on tackling the integral:

$\int_0^{2\pi}\int_0^{\sqrt{\frac{\sqrt{13}-1}{2}}}r\sqrt{3-r^2}-r^3\,dr\,d\vartheta=\int_0^{2\pi}\left[\int_0^{\sqrt{\frac{\sqrt{13}-1}{2}}}r\sqrt{3-r^2}\,dr-\int_0^{\sqrt{\frac{\sqrt{13}-1}{2}}}r^3\,dr\right]\,d\vartheta$

--Chris
• Sep 13th 2008, 12:01 PM
NicoleB
Thank you! I apologize I wasn't being clear.

Thank you for the integral suggestion, but I did solve the integral, It's just that the solution is very algebraic, i can't get anything nice out of it when calculating it.

So I think it might be wrong. I was looking into help more in the lines of if I even started out right before the integral.

More in the lines of, Is this how you would solve a volume problem? Am I on the right track at all? thanks again!
• Sep 13th 2008, 12:11 PM
Chris L T521
Quote:

Originally Posted by NicoleB
Thank you! I apologize I wasn't being clear.

Thank you for the integral suggestion, but I did solve the integral, It's just that the solution is very algebraic, i can't get anything nice out of it when calculating it.

Trust me, I've seen worse calculations in other types of double and triple integrals. (Rofl)

Quote:

So I think it might be wrong. I was looking into help more in the lines of if I even started out right before the integral.
Don't let a nasty looking solution discourage you. Sometimes it may seem wrong, but it is right. However, I'd make sure you copied down the problem right, or just go through your work and see if you made any simple mistakes (like missing a minus sign somewhere, you accidentally differentiate instead of integrate, etc.)

Quote:

More in the lines of, Is this how you would solve a volume problem? Am I on the right track at all? thanks again!
This is one way you would tackle a problem as such. If the only information that is given to you are the functions, then you will have to find some common region $R$ that they have in common. Usually, this ends up in having to find where they have the same z value.

Sometimes, they give you the bounds and that makes life easier.

It varies from problem to problem, but after a while, you will better understand how to tackle various problems.

--Chris