How do you find the volume bounded by y=3-x^2 and y=2 revolved about the y=2?

Dec 28, 2017

Volume $= \frac{16 \pi}{15}$ cu.units

Explanation:

(Sorry if this is in excessive detail)
The given regions in the Cartesian plane look like:

If this is rotated about $y = 2$,
we have:

For ease of calculation, it is convenient if we shift this solid down so the axis of rotation falls along the X-axis:

Note that the radius (relative to the X-axis) of this shifted volume is
equal to ${x}^{2} - 1$ for $x \in \left[- 1 , + 1\right]$
and we can slice this solid into thin disks, each with a thickness of $\delta$,
so that each disk has a volume of $\delta \cdot p r {r}^{2} = \pi \cdot \delta \cdot {\left({x}^{2} - 1\right)}^{2}$

With very small values of $\delta$ the sum of the volumes of all such disks will be the volume of the rotated solid.

We can evaluate this sum with $\delta \rightarrow 0$ using the integral:
$\textcolor{w h i t e}{\text{XXX}} {\int}_{- 1}^{+ 1} \pi \cdot {\left({x}^{2} - 1\right)}^{2} \mathrm{dx}$

$\textcolor{w h i t e}{\text{XXX}} = \pi \cdot {\int}_{- 1}^{+ 1} {x}^{4} - 2 {x}^{2} + 1 \mathrm{dx}$

$\textcolor{w h i t e}{\text{XXX}} = \pi \cdot \left(\frac{{x}^{5}}{5} - \frac{2 {x}^{3}}{3} + x {|}_{- 1}^{+ 1}\right)$

$\textcolor{w h i t e}{\text{XXX}} = \pi \cdot \left(\frac{3 - 10 + 15}{15} - \frac{\left(- 3\right) - \left(- 10\right) + \left(- 15\right)}{15}\right)$

$\textcolor{w h i t e}{\text{XXX}} = \pi \cdot \left(\frac{8}{15} - \frac{- 8}{15}\right)$

$\textcolor{w h i t e}{\text{XXX}} = \frac{16}{15} \pi$