# Using integral to find volume problem

• May 7th 2011, 12:04 PM
iva
Using integral to find volume problem
Hi there,

I have an exercise I couldn't solve, have the answer which i couldn't get to, any guidance will be appreciated:

Calculate volume of D where D is the region in $\displaystyle R^3$enclosed by the surfaces
z=3-x^2-y^2 and z=x^2+y^2 -3 (Sorry Latex was giving me errors)

So this looks like 2 identical parabaloids one startign at z=3 going down and one starting at z=-3 going up.

I used a triple integral with cylindrical coordinates ( if theres another better way let me know)
So I integrated over

theta = 0 to 2pi ( ie all the way around)
r= sqrt(3) by equating the 2 parabaoids to find the intesection
z integrated over 3-r^2 and r^2-3.

The first issue here is that when i do the z integration i come out with zero (what did I do wrong here)
So i decided to integrate in 2 identical parts split at z=0.
I followed this through but didn't get the books answer of 9pi. i got 24root(3)/5 for one half, how does the book get to 9 pi? At which step did I go wrong?

(if this is impossible to read I'll reattempt the Latex)

Thank you
• May 7th 2011, 12:27 PM
TheEmptySet
Quote:

Originally Posted by iva
Hi there,

I have an exercise I couldn't solve, have the answer which i couldn't get to, any guidance will be appreciated:

Calculate volume of D where D is the region in $\displaystyle R^3$enclosed by the surfaces
z=3-x^2-y^2 and z=x^2+y^2 -3 (Sorry Latex was giving me errors)

So this looks like 2 identical parabaloids one startign at z=3 going down and one starting at z=-3 going up.

I used a triple integral with cylindrical coordinates ( if theres another better way let me know)
So I integrated over

theta = 0 to 2pi ( ie all the way around)
r= sqrt(3) by equating the 2 parabaoids to find the intesection
z integrated over 3-r^2 and r^2-3.

The first issue here is that when i do the z integration i come out with zero (what did I do wrong here)
So i decided to integrate in 2 identical parts split at z=0.
I followed this through but didn't get the books answer of 9pi. i got 24root(3)/5 for one half, how does the book get to 9 pi? At which step did I go wrong?

(if this is impossible to read I'll reattempt the Latex)

Thank you

$\displaystyle \int_{0}^{2\pi}\int_{0}^{\sqrt{3}}\int_{r^2-3}^{3-r^2}rdzdrd\theta=2\int_{0}^{2\pi}\int_{0}^{\sqrt{3 }}(3r-r^3)drd\theta=2\left( \int_{0}^{2\pi}d\theta\right)\left(\int_{0}^{\sqrt {3}} (3r-r^3)dr\right)$
• May 7th 2011, 12:32 PM
alexmahone
Quote:

Originally Posted by iva
Hi there,

I have an exercise I couldn't solve, have the answer which i couldn't get to, any guidance will be appreciated:

Calculate volume of D where D is the region in $\displaystyle R^3$enclosed by the surfaces
z=3-x^2-y^2 and z=x^2+y^2 -3 (Sorry Latex was giving me errors)

So this looks like 2 identical parabaloids one startign at z=3 going down and one starting at z=-3 going up.

I used a triple integral with cylindrical coordinates ( if theres another better way let me know)
So I integrated over

theta = 0 to 2pi ( ie all the way around)
r= sqrt(3) by equating the 2 parabaoids to find the intesection
z integrated over 3-r^2 and r^2-3.

The first issue here is that when i do the z integration i come out with zero (what did I do wrong here)
So i decided to integrate in 2 identical parts split at z=0.
I followed this through but didn't get the books answer of 9pi. i got 24root(3)/5 for one half, how does the book get to 9 pi? At which step did I go wrong?

(if this is impossible to read I'll reattempt the Latex)

Thank you

$\displaystyle V=\int_0^{2\pi}\int_0^{\sqrt{3}}(z_{top}-z_{bottom})rdrd\theta$

= $\displaystyle \int_0^{2\pi}\int_0^{\sqrt{3}}(6-2r^2)rdrd\theta$

= $\displaystyle \int_0^{2\pi}\int_0^{\sqrt{3}}(6r-2r^3)drd\theta$

= $\displaystyle \int_0^{2\pi}\left[3r^2-\frac{r^4}{2}\right]_0^{\sqrt{3}}d\theta$

= $\displaystyle \int_0^{2\pi}(3\times 3-\frac{9}{2})d\theta$

= $\displaystyle \int_0^{2\pi}\frac{9}{2}d\theta$

= $\displaystyle \frac{9}{2}\times 2\pi$

= $\displaystyle 9\pi$
• May 7th 2011, 12:48 PM
Deveno
the "z integration" is NOT 0, you should wind up with $\displaystyle r((3-r^2) - (-3+r^2)) = 6r - 2r^3$.

integrating that with respect to r from 0 to √3, will yield $\displaystyle 3r^2 - \frac{r^4}{2}$, evaluated at √3 which gives you 9/2.

integrating the constant function 9/2 over θ = 0 to θ = $\displaystyle 2\pi$, gives $\displaystyle 9\pi$, as your textbook indicates.