Panda Bear's question at Yahoo Answers regarding solid of revolution

[MarkFL]

Here is the question:

Need calculus help Volumes using Disks/Washers?

find the volume of the solid that is obtained by revolving the region about the indicated axis or line
y=square root of (1-x )
x= -3, 1 y=1

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

 

[MarkFL]

Hello Panda Bear,

First, let's take a look at the region to be revolved:

View attachment 1100

Using the disk method, we see we will have two regions to revolve, the region on the left and the region on the right. However, since the radius is to be squared, and the radius on the right is the negative of the radius on the right, we may simply use one integral.

The volume of an arbitrary disk is:

\(\displaystyle dV=\pi r^2\,dx\)

where:

\(\displaystyle r=\pm\left(\sqrt{1-x}-1 \right)\)

and so we have:

\(\displaystyle dV=\pi\left(\sqrt{1-x}-1 \right)r^2\,dx=\pi\left(2-x-2\sqrt{1-x} \right)\,dx\)

Summing the disks by integrating, we have:

\(\displaystyle V=\pi\int_{-3}^1 2-x-2\sqrt{1-x}\,dx\)

For the third term in the integrand, let's use the substitution:

\(\displaystyle u=1-x\,\therefore\,du=-dx\) and we have:

\(\displaystyle V=\pi\int_{-3}^1 2-x\,dx-2\pi\int_0^4u^{\frac{1}{2}}\,du\)

Applying the anti-derivative form of the FTOC, we find:

\(\displaystyle V=\pi\left(\left[2x-\frac{1}{2}x^2 \right]_{-3}^1-\frac{4}{3}\left[u^{\frac{3}{2}} \right]_0^4 \right)=\)

\(\displaystyle \pi\left(\left(\left(2(1)-\frac{1}{2}(1)^2 \right)-\left(2(-3)-\frac{1}{2}(-3)^2 \right) \right)-\frac{4}{3}\left(\left(4^{\frac{3}{2}} \right)-\left(0^{\frac{3}{2}} \right) \right) \right)=\)

\(\displaystyle \pi\left(12-\frac{32}{3} \right)=\frac{4\pi}{3}\)

We can check our work using the shell method. For the area on the left, we have:

The volume of an arbitrary shell is:

\(\displaystyle dV_1=2\pi rh\,dy\)

where:

\(\displaystyle r=y-1\)

\(\displaystyle h=\left(1-y^2 \right)-(-3)=4-y^2\)

and so we have:

\(\displaystyle dV_1=2\pi(y-1)\left(4-y^2 \right)\,dy=2\pi\left(-y^3+y^2+4y-4 \right)\,dy\)

Summing the shells, we find:

\(\displaystyle V_1=2\pi\int_1^2 -y^3+y^2+4y-4\,dy=2\pi\left[-\frac{1}{4}y^4+\frac{1}{3}y^3+2y^2-4y \right]_1^2=\)

\(\displaystyle 2\pi\left(\left(-\frac{1}{4}(2)^4+\frac{1}{3}(2)^3+2(2)^2-4(2) \right)-\left(-\frac{1}{4}(1)^4+\frac{1}{3}(1)^3+2(1)^2-4(1) \right) \right)=\)

\(\displaystyle 2\pi\left(\left(-4+\frac{8}{3}+8-8 \right)-\left(-\frac{1}{4}+\frac{1}{3}+2-4 \right) \right)=\)

\(\displaystyle 2\pi\left(-\frac{4}{3}+\frac{23}{12} \right)=\frac{7\pi}{6}\)

Now, for the area on the right, we find the volume of an arbitrary shell is:

\(\displaystyle dV_2=2\pi rh\,dy\)

where:

\(\displaystyle r=1-y\)

\(\displaystyle h=1-\left(1-y^2 \right)=y^2\)

Hence:

\(\displaystyle dV_2=2\pi (1-y)\left(y^2 \right)\,dy=2\pi\left(y^2-y^3 \right)\,dy\)

Summing the shells, we find:

\(\displaystyle V_2=2\pi\int_0^1 y^2-y^3\,dy=2\pi\left[\frac{1}{3}y^3-\frac{1}{4}y^4 \right]_0^1=2\pi\left(\frac{1}{3}-\frac{1}{4} \right)=\frac{\pi}{6}\)

Adding the two volumes, we find the total is:

\(\displaystyle V=V_1+V_2=\frac{7\pi}{6}+\frac{\pi}{6}=\frac{4\pi}{3}\)