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

Beginner2022-07-02Added 16 answers

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: $\pi (2{x}^{2}+5{)}^{2}dx$ and the small one has volume $\pi (x+3{)}^{2}dx$

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

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

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

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

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: $\pi (2{x}^{2}+5{)}^{2}dx$ and the small one has volume $\pi (x+3{)}^{2}dx$

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

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

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

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

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