Using integral to find volume problem

**iva** · May 7th 2011, 12:04 PM

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 $R^3$ 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

**TheEmptySet** · May 7th 2011, 12:27 PM

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 $R^3$ 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

Your idea is sound.

$\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}\int_{r^2-3}^{3-r^2}rdzdrd\theta=2\int_{0}^{2\pi}\int_{0}^{\sqrt{3 }}(3r-r^3)drd\theta=2\left( \int_{0}^{2\pi}d\theta\right)\left(\int_{0}^{\sqrt {3}} (3r-r^3)dr\right)$

**alexmahone** · May 7th 2011, 12:32 PM

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 $R^3$ 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

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

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

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

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

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

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

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

= $9\pi$

**Deveno** · May 7th 2011, 12:48 PM

the "z integration" is NOT 0, you should wind up with $r((3-r^2) - (-3+r^2)) = 6r - 2r^3$ .

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

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

Thread: Using integral to find volume problem

LinkBack

Thread Tools

Display

Using integral to find volume problem

Similar Math Help Forum Discussions

Triple integral to find volume

Volume integral, how do I find the limits for my integral?

Find the Volume, Triple Integral

Find the volume of a region using an iterated integral

Use triple integral to find volume

Search Tags