# Find the volume of the region bounded by y=sqrt(z-x^2) and x^2+y^2+2z=12?

## find the volume of the region bounded by y=sqrt(z-x^2) and x^2+y^2+2z=12?

Aug 15, 2016

$= 12 \pi$

#### Explanation:

this is symmetric about z axis so we use polar co-ords

first paraboloid

$y = \sqrt{z - {x}^{2}}$
${y}^{2} = z - {x}^{2}$
${z}_{1} = {y}^{2} + {x}^{2} = {r}^{2}$

second

${x}^{2} + {y}^{2} + 2 z = 12$
${z}_{2} = \frac{12 - \left({x}^{2} + {y}^{2}\right)}{2} = 6 - {r}^{2} / 2$

${z}_{1} = {z}_{2} \implies 6 - {r}^{2} / 2 = {r}^{2}$
$\implies r = 2$
(and ${z}_{1} = {z}_{2} = 4$)

here is a 2-D plot:

We need to find the volume that is made by revolving the area between the red and blue curves around the z-axis. That is the area under the purple curve, which is entered as ${z}_{3} = {z}_{2} - {z}_{1}$ in the box on the left.

Ie, we need to find volume

$V = {\int}_{A} \setminus \left({z}_{2} - {z}_{1}\right) \mathrm{dA}$

where volume element in polar is $\mathrm{dA} = r \setminus \mathrm{dr} \setminus d \theta$

$\implies {\int}_{0}^{2 \pi} {\int}_{0}^{2} \left(6 - \frac{3}{2} {r}^{2}\right) r \setminus \mathrm{dr} \setminus d \theta$

$= {\int}_{0}^{2 \pi} \setminus d \theta \cdot \setminus {\int}_{0}^{2} 6 r - \frac{3}{2} {r}^{3} \setminus \mathrm{dr}$

$= 2 \pi {\left[3 {r}^{2} - \frac{3}{8} {r}^{4}\right]}_{0}^{2}$

$= 2 \pi {\left[3 {r}^{2} \left(1 - \frac{1}{8} {r}^{2}\right)\right]}_{0}^{2}$

$= 2 \pi \left(12 \left(1 - \frac{1}{2}\right)\right)$

$= 12 \pi$

As an aside, @abubakar wanted this expressed as a triple integral . I originally skipped that step because it just adds extra notation but if we start with the general volume element and switching straight to cylindrical co-ordinates, we can say that

${\int}_{V} \mathrm{dV} = {\int}_{{\theta}_{0}}^{{\theta}_{1}} \setminus {\int}_{{r}_{0}}^{{r}_{1}} \setminus {\int}_{{z}_{0}}^{{z}_{1}} \setminus r \mathrm{dr} d \theta \mathrm{dz}$

now neither r or z depend upon $\theta$ so we can immediately, using Fubini's theorem, lift $\theta$ outside the integration as it is independent so we have

${\int}_{V} \mathrm{dV} = \left({\int}_{0}^{2 \pi} d \theta\right) \cdot \left(\setminus {\int}_{{r}_{0}}^{{r}_{1}} \setminus {\int}_{{z}_{0}}^{{z}_{1}} \setminus r \mathrm{dr} \mathrm{dz}\right)$

$= 2 \pi \setminus {\int}_{{r}_{0}}^{{r}_{1}} \setminus {\int}_{{z}_{0}}^{{z}_{1}} \setminus r \mathrm{dr} \mathrm{dz}$

I did that to simplify the next bit, because we have $r = r \left(z\right)$ and (z = z(r)#. z and r are inter-dependent. clearly!

In general terms, to do this we need either do

$= 2 \pi \setminus {\int}_{{z}_{0}}^{{z}_{1}} \setminus {\int}_{r = {r}_{0} \left(z\right)}^{{r}_{1} \left(z\right)} \setminus r \setminus \mathrm{dr} \setminus \mathrm{dz}$

OR

$= 2 \pi \setminus {\int}_{{r}_{0}}^{{r}_{1}} \setminus {\int}_{z = {z}_{0} \left(r\right)}^{{z}_{1} \left(r\right)} \setminus r \setminus \mathrm{dz} \setminus \mathrm{dr}$

The explanation and drawings here are helpful

The key to this is that these are iterated integrals. The $\mathrm{dA} = \mathrm{dx} \setminus \mathrm{dy} = r \setminus \mathrm{dr} \setminus d \theta$ element notation is very useful but perhaps misleading.

Aug 19, 2016

A graphic contribution.

#### Explanation:

A graphic contribution.