Sunshine's questions at Yahoo Answers regarding solids of revolution

[MarkFL]

Here are the questions:

Calculus Question-disk/washer method?

a)Let R be the region bounded by the given curves. Sketch R on a graph. If R is revolved around the x-axis, find the volume of the solid revolution by the disk/washer method.
y=x^3
y=2x

b)Let R be the region bounded by the given curves. Sketch R on a graph. IF R is revolved around the y-axis find the volume of the solid revolution by the disk/washer method.
y=x^2+5
y=3
x=-2
x=2

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Re: Subshine's questions at Yahoo! Answers regarding solids of revolution

Hello sunshine,

a) Here is a graph of the region $R$ to be revolved about the $x$-axis:

View attachment 899

We may revolve the area shaded in red and then double the volume to find the requested volume.

The volume of an arbitrary washer is:

\(\displaystyle dV=\pi\left(R^2-r^2 \right)\,dx\)

where:

\(\displaystyle R=2x\)

\(\displaystyle r=x^3\)

and so we have:

\(\displaystyle dV=\pi\left((2x)^2-\left(x^3 \right)^2 \right)\,dx=\pi\left(4x^2-x^6 \right)\,dx\)

To determine the limits of integration, we may equate the two functions:

\(\displaystyle x^3=2x\)

\(\displaystyle x\left(x^2-2 \right)=0\)

\(\displaystyle x=0,\,\pm2\)

Hence, summing the disks by integrating, we find:

\(\displaystyle V=2\pi\int_0^{\sqrt{2}}4x^2-x^6\,dx=2\pi\left[\frac{4}{3}x^3-\frac{1}{7}x^7 \right]_0^{\sqrt{2}}=\frac{16\sqrt{2}\pi}{21}\left(7-3 \right)=\frac{64\sqrt{2}\pi}{21}\)

Now, whenever possible, I like to use the shell method to check my work.

The volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dy\)

where:

\(\displaystyle r=y\)

\(\displaystyle h=y^{\frac{1}{3}}-\frac{y}{2}\)

and so we have:

\(\displaystyle dV=2\pi y\left(y^{\frac{1}{3}}-\frac{y}{2} \right)\,dy=2\pi\left(y^{\frac{4}{3}}-\frac{y^2}{2} \right)\,dy\)

To determine the limits of integration, we may equation the two functions:

\(\displaystyle y^{\frac{1}{3}}=\frac{y}{2}\)

\(\displaystyle y-2y^{\frac{1}{3}}=0\)

\(\displaystyle y^{\frac{1}{3}}\left(y^{\frac{2}{3}}-2 \right)\)

\(\displaystyle y=0,\,\pm2\sqrt{2}\)

Thus, summing the shells by integration we find:

\(\displaystyle V=4\pi\int_0^{2\sqrt{2}}y^{\frac{4}{3}}-\frac{y^2}{2}\,dy=4\pi\left[\frac{3}{7}y^{\frac{7}{3}}-\frac{1}{6}y^3 \right]_0^{2\sqrt{2}}=\frac{16\sqrt{2}\pi}{21}\left(18-14 \right)=\frac{64\sqrt{2}\pi}{21}\)

This checks with the result we obtained using the washer method.

b) Here is a graph of the region $R$ to be revolved about the $y$-axis:

View attachment 900

We need only revolve the area shaded in red to find the requested volume.

Using the disk/washer method, we should observe that for $3\le y\le5$ we simply have a right circular cylinder of radius 2 and height 2, hence:

\(\displaystyle V_1=\pi(2)^22=8\pi\)

To find the remaining volume, we may use the washer method. The volume of an arbitrary washer is:

\(\displaystyle dV_2=\pi\left(R^2-r^2 \right)\,dy\)

where:

\(\displaystyle R=2\)

\(\displaystyle r=\sqrt{y-5}\)

and so we have:

\(\displaystyle dV_2=\pi\left((2)^2-\left(\sqrt{y-5} \right)^2 \right)\,dy=\pi\left(9-y \right)\,dy\)

Summing the washers by integration, we find:

\(\displaystyle V_2=\pi\int_5^9 9-y\,dy=\pi\left[9y-\frac{1}{2}y^2 \right]_5^9=\frac{\pi}{2}\left(\left(2\cdot9^2-9^2 \right)-(90-25) \right)=8\pi\)

Thus, the total volume is:

\(\displaystyle V=V_1+V_2=8\pi+8\pi=16\pi\)

Now, using the shell method to check our work, we find the volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dx\)

where:

\(\displaystyle r=x\)

\(\displaystyle h=x^2+5-3=x^2+2\)

and so we have:

\(\displaystyle dV=2\pi x\left(x^2+2 \right)\,dx=2\pi\left(x^3+2x \right)\,dx\)

Summing the shells by integrating, we find:

\(\displaystyle V=2\pi\int_0^2 x^3+2x\,dx=2\pi\left[\frac{1}{4}x^4+x^2 \right]_0^2=2\frac{\pi}{2}\left(2^4+2^4 \right)=16\pi\)

This checks with our previous result.