gaiaecologicaq2

2022-07-01

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

Kayley Jackson

Sketch the region. Note that $y=2{x}^{2}+5$ 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 dx
The volume of each representative disk would be $\pi \ast radiu{s}^{2}\ast thikness$. So the large disk has volume: $\pi \left(2{x}^{2}+5{\right)}^{2}dx$ and the small one has volume $\pi \left(x+3{\right)}^{2}dx$
The volume of the washer is the difference, or $\pi {R}^{2}dx-\pi {r}^{2}dx$ and the resulting solid has volume:
$V=\pi {\int }_{0}^{3}\left(\left(2{x}^{2}+5{\right)}^{2}-\left(x+3{\right)}^{2}\right)dx$
$=\pi {\int }_{0}^{3}\left(4{x}^{4}+19{x}^{2}-6x+16\right)dx$
You can finish the integral to get $\frac{1932\pi }{5}$

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