# How do you find the volume of the solid obtained by rotating the region bounded by the curves y=2x^2+5, and y=x+3 and the y-axis, and x=3 rotated around the x axis?

Jun 21, 2015

Use disks/washers.

#### Explanation:

Sketch the region. Note that $2 {x}^{2} + 5$ is above (greater than) $x + 3$, so the parabola is farther from the axis of rotation.

Therefore:
At a particular $x$, the large radius is, $R = 2 {x}^{2} + 5$, and the small radius is $r = x + 3$. The thickness of the disks is $\mathrm{dx}$

The volume of each representative disk would be $\pi \cdot \text{radius"^2 * "thickness}$. So the large disk has volume: $\pi {\left(2 {x}^{2} + 5\right)}^{2} \mathrm{dx}$ and the small one has volume $\pi {\left(x + 3\right)}^{2} \mathrm{dx}$

The volume of the washer is the difference, or $\pi {R}^{2} \mathrm{dx} - \pi {r}^{2} \mathrm{dx}$and the resulting solid has volume:

$V = \pi {\int}_{0}^{3} \left({\left(2 {x}^{2} + 5\right)}^{2} - {\left(x + 3\right)}^{2}\right) \mathrm{dx}$

$= \pi {\int}_{0}^{3} \left(4 {x}^{4} + 19 {x}^{2} - 6 x + 16\right) \mathrm{dx}$

You can finish the integral to get $\frac{1932 \pi}{5}$