MHB Sam m's question at Yahoo Answers regarding a solid of revolution

[MarkFL]

Here is the question:

Find the volume of the solid generated by revolving around the line x = 4?

given the region bounded by y=x^2+2, y=1, x=0, x=1?

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello sam m,

Let's try the shell method first. The volume of an arbitrary shell is:

$$dV=2\pi rh\,dx$$

where:

$$r=4-x$$

$$h=x^2+2-1=x^2+1$$

And so we have:

$$dV=2\pi(4-x)\left(x^2+1 \right)\,dx=2\pi\left(-x^3+4x^2-x+4 \right)\,dx$$

Hence, summing the shells, we find:

$$V=2\pi\int_0^1 -x^3+4x^2-x+4\,dx$$

Application of the FTOC yields:

$$V=2\pi\left[-\frac{1}{4}x^4+\frac{4}{3}x^3-\frac{1}{2}x^2+4x \right]_0^1=2\pi\left(-\frac{1}{4}+\frac{4}{3}-\frac{1}{2}+4 \right)=\frac{55}{6}\pi$$

Now, we can check our answer by using the washer method. And we can ease our calculations by observing that for $1\le y\le2$ the inner and outer radii are constant, so we have a washer whose thickness is 1 unit and its ourter radius is 4and its inner radius is 3, so this part of the volume is:

$$V_1=\pi\left(4^2-3^2 \right)(1)=7\pi$$

Now, for $2\le y\le3$ we find the outer radius varies, and so the volume of an arbitrary washer is:

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

where:

$$R=4-x=4-\sqrt{y-2}$$

$$r=3$$

And so we have:

$$dV_2=\pi\left(\left(4-\sqrt{y-2} \right)^2-3^2 \right)\,dy=\pi\left(y-8\sqrt{y-2}+5 \right)\,dy$$

Adding the washers, we obtain:

$$V_2=\pi\int_2^3 y-8\sqrt{y-2}+5\,dy$$

Application of the FTOC yields:

$$V_2=\pi\left[\frac{1}{2}y^2-\frac{16}{3}(y-2)^{\frac{3}{2}}+5y \right]_2^3=\pi\left(\left(\frac{9}{2}-\frac{16}{3}+15 \right)-\left(2+10 \right) \right)=\pi\left(\frac{85}{6}-12 \right)=\frac{13}{6}\pi$$

Adding the two volumes to get the total:

$$V=V_1+V_2=7\pi+\frac{13}{6}\pi=\frac{55}{6}\pi$$

And so, this checks with the shell method. Thus we find the volume of the solid of revolution in units cubed is:

$$V=\frac{55}{6}\pi$$