# The area bounded by the curve y=3+2x-x^2 and line y=3 is rotated completely about the line y=3. Find the volume of the solid of revolution obtained?

Jun 10, 2015

$V = \frac{16}{15} \pi \approx 3.35103$

#### Explanation:

The area are the solution of this system:
$\left\{\begin{matrix}y \le - {x}^{2} + 2 x + 3 \\ y \ge 3\end{matrix}\right.$
And it is sketched in this plot:

The formula for the volume of a x-axis rotation solid is:
$V = \pi \cdot {\int}_{a}^{b} {f}^{2} \left(z\right) \mathrm{dz}$.

To apply the formula we should translate the half moon on the x-axis, the area won't change, and so it won't change also the volume:
$y = - {x}^{2} + 2 x + 3 \textcolor{red}{- 3} = - {x}^{2} + 2 x$
$y = 3 \textcolor{red}{- 3} = 0$
In this way we obtain $f \left(z\right) = - {z}^{2} + 2 z$.
The translated area now is plotted here:

But which are the a and b of the integral? The solutions of the system:
$\left\{\begin{matrix}y = - {x}^{2} + 2 x \\ y = 0\end{matrix}\right.$
So $a = 0 \mathmr{and} b = 2$.

Let's rewrite and solve the integral:
$V = \pi \cdot {\int}_{0}^{2} {\left(- {z}^{2} + 2 z\right)}^{2} \mathrm{dz}$
$V = \pi \cdot {\int}_{0}^{2} {z}^{4} - 4 {z}^{3} + 4 {z}^{2} \mathrm{dz}$
$V = \pi \cdot {\left[{z}^{5} / 5 - \frac{4 {z}^{4}}{4} + \frac{4 {z}^{3}}{3}\right]}_{0}^{2}$
$V = \pi \cdot {\left[{z}^{5} / 5 - {z}^{4} + \frac{4 {z}^{3}}{3}\right]}_{0}^{2}$
$V = \pi \cdot \left({2}^{5} / 5 - {2}^{4} + \frac{4 \cdot {2}^{3}}{3} - {0}^{5} / 5 + {0}^{4} - \frac{4 \cdot {0}^{3}}{3}\right)$
$V = \pi \cdot \left(\frac{32}{5} - 16 + \frac{32}{3} + 0\right)$
$V = \pi \cdot \left(\frac{96}{15} - \frac{240}{15} + \frac{160}{15}\right)$
$V = \pi \cdot \left(\frac{96}{15} - \frac{240}{15} + \frac{160}{15}\right)$
$V = \frac{16}{15} \pi \approx 3.35103$
And this "lemon" is the solid obtained: