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sam m's question at Yahoo! Answers regarding a solid of revolution
- Thread starter MarkFL
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Hello sam m,
Let's try the shell method first. The volume of an arbitrary shell is:
\(\displaystyle dV=2\pi rh\,dx\)
where:
\(\displaystyle r=4-x\)
\(\displaystyle h=x^2+2-1=x^2+1\)
And so we have:
\(\displaystyle dV=2\pi(4-x)\left(x^2+1 \right)\,dx=2\pi\left(-x^3+4x^2-x+4 \right)\,dx\)
Hence, summing the shells, we find:
\(\displaystyle V=2\pi\int_0^1 -x^3+4x^2-x+4\,dx\)
Application of the FTOC yields:
\(\displaystyle V=2\pi\left[-\frac{1}{4}x^4+\frac{4}{3}x^3-\frac{1}{2}x^2+4x \right]_0^1=2\pi\left(-\frac{1}{4}+\frac{4}{3}-\frac{1}{2}+4 \right)=\frac{55}{6}\pi\)
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:
\(\displaystyle V_1=\pi\left(4^2-3^2 \right)(1)=7\pi\)
Now, for $2\le y\le3$ we find the outer radius varies, and so the volume of an arbitrary washer is:
\(\displaystyle dV_2=\pi\left(R^2-r^2 \right)\,dy\)
where:
\(\displaystyle R=4-x=4-\sqrt{y-2}\)
\(\displaystyle r=3\)
And so we have:
\(\displaystyle dV_2=\pi\left(\left(4-\sqrt{y-2} \right)^2-3^2 \right)\,dy=\pi\left(y-8\sqrt{y-2}+5 \right)\,dy\)
Adding the washers, we obtain:
\(\displaystyle V_2=\pi\int_2^3 y-8\sqrt{y-2}+5\,dy\)
Application of the FTOC yields:
\(\displaystyle V_2=\pi\left[\frac{1}{2}y^2-\frac{16}{3}(y-2)^{\frac{3}{2}}+5y \right]_2^3=\pi\left(\left(\frac{9}{2}-\frac{16}{3}+15 \right)-\left(2+10 \right) \right)=\pi\left(\frac{85}{6}-12 \right)=\frac{13}{6}\pi\)
Adding the two volumes to get the total:
\(\displaystyle V=V_1+V_2=7\pi+\frac{13}{6}\pi=\frac{55}{6}\pi\)
And so, this checks with the shell method. Thus we find the volume of the solid of revolution in units cubed is:
\(\displaystyle V=\frac{55}{6}\pi\)
Let's try the shell method first. The volume of an arbitrary shell is:
\(\displaystyle dV=2\pi rh\,dx\)
where:
\(\displaystyle r=4-x\)
\(\displaystyle h=x^2+2-1=x^2+1\)
And so we have:
\(\displaystyle dV=2\pi(4-x)\left(x^2+1 \right)\,dx=2\pi\left(-x^3+4x^2-x+4 \right)\,dx\)
Hence, summing the shells, we find:
\(\displaystyle V=2\pi\int_0^1 -x^3+4x^2-x+4\,dx\)
Application of the FTOC yields:
\(\displaystyle V=2\pi\left[-\frac{1}{4}x^4+\frac{4}{3}x^3-\frac{1}{2}x^2+4x \right]_0^1=2\pi\left(-\frac{1}{4}+\frac{4}{3}-\frac{1}{2}+4 \right)=\frac{55}{6}\pi\)
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:
\(\displaystyle V_1=\pi\left(4^2-3^2 \right)(1)=7\pi\)
Now, for $2\le y\le3$ we find the outer radius varies, and so the volume of an arbitrary washer is:
\(\displaystyle dV_2=\pi\left(R^2-r^2 \right)\,dy\)
where:
\(\displaystyle R=4-x=4-\sqrt{y-2}\)
\(\displaystyle r=3\)
And so we have:
\(\displaystyle dV_2=\pi\left(\left(4-\sqrt{y-2} \right)^2-3^2 \right)\,dy=\pi\left(y-8\sqrt{y-2}+5 \right)\,dy\)
Adding the washers, we obtain:
\(\displaystyle V_2=\pi\int_2^3 y-8\sqrt{y-2}+5\,dy\)
Application of the FTOC yields:
\(\displaystyle V_2=\pi\left[\frac{1}{2}y^2-\frac{16}{3}(y-2)^{\frac{3}{2}}+5y \right]_2^3=\pi\left(\left(\frac{9}{2}-\frac{16}{3}+15 \right)-\left(2+10 \right) \right)=\pi\left(\frac{85}{6}-12 \right)=\frac{13}{6}\pi\)
Adding the two volumes to get the total:
\(\displaystyle V=V_1+V_2=7\pi+\frac{13}{6}\pi=\frac{55}{6}\pi\)
And so, this checks with the shell method. Thus we find the volume of the solid of revolution in units cubed is:
\(\displaystyle V=\frac{55}{6}\pi\)