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

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The volume of the solid generated by revolving the region bounded by y=x^2+2, y=1, x=0, and x=1 around the line x=4 is calculated using both the shell and washer methods. The shell method yields a volume of V=55/6π, derived from integrating the expression for the volume of an arbitrary shell. The washer method confirms this result by calculating the volumes for two segments of the region, ultimately summing to the same volume of V=55/6π. Both methods validate the accuracy of the calculations. The final volume of the solid of revolution is V=55/6π cubic units.
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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.
 
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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:

$$V=\frac{55}{6}\pi$$
 
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