- #1
MarkFL
Gold Member
MHB
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Hello MHB,
As students of calculus, we are taught to find the volumes of solids of rotation obtained by revolving given regions about horizontal and vertical axes of rotation. But, what if the axis of rotation is neither horizontal nor vertical? Please consider the following diagram:
View attachment 1398
We wish to revolve the region shaded in green about the line $y=mx+b$. Using the disk method, where the radius of a disk is $r$ and its thickness is $du$, we may write:
\(\displaystyle dV=\pi r^2\,du\)
Using the formula where the point is $(x,f(x))$ and the line is $y=mx+b$, we have:
\(\displaystyle r=\frac{\left|f(x)-mx-b \right|}{\sqrt{m^2+1}}\)
To find $du$ in terms of $dx$, let $dD$ represent the arc-length along $f(x)$ corresponding to $dx$, as in the following diagram:
https://www.physicsforums.com/attachments/1399
From this, we obtain:
\(\displaystyle dD=\frac{dx}{\cos(\theta)}\)
Now, to obtain the projection of $dD$ onto $y=mx+b$, we find that by rotating everything by \(\displaystyle -\tan^{-1}(m)\), we may write:
\(\displaystyle du=dD\cos\left(\theta-\tan^{-1}(m) \right)\)
Hence, we obtain:
\(\displaystyle du=\frac{\cos\left(\theta-\tan^{-1}(m) \right)}{\cos(\theta)}\,dx\)
Using the angle difference identity for cosine, we find:
\(\displaystyle du=\frac{\cos\left(\theta\right)\cos\left(\tan^{-1}(m) \right)+\sin\left(\theta\right)\sin\left(\tan^{-1}(m) \right)}{\cos(\theta)}\,dx\)
\(\displaystyle du=\left(\cos\left(\tan^{-1}(m) \right)+\tan(\theta)\cos\left(\tan^{-1}(m) \right) \right)\,dx\)
Using:
\(\displaystyle \cos\left(\tan^{-1}(m) \right)=\frac{1}{\sqrt{m^2+1}}\)
\(\displaystyle \sin\left(\tan^{-1}(m) \right)=\frac{m}{\sqrt{m^2+1}}\)
\(\displaystyle \tan(\theta)=f'(x)\)
we then obtain:
\(\displaystyle du=\left(\frac{1}{\sqrt{m^2+1}}+\frac{m}{\sqrt{m^2+1}}f'(x) \right)\,dx\)
\(\displaystyle du=\frac{1}{\sqrt{m^2+1}}\left(1+mf'(x) \right)\,dx\)
And so we may now give the area $A$ of the shaded region as:
\(\displaystyle A=\int_{x_i}^{x_f} r\,du=\int_{x_i}^{x_f}\left(\frac{\left|f(x)-mx-b \right|}{\sqrt{m^2+1}} \right)\left(\frac{1}{\sqrt{m^2+1}}\left(1+mf'(x) \right)dx \right)\)
\(\displaystyle A=\frac{1}{m^2+1}\int_{x_i}^{x_f}\left|f(x)-mx-b \right|\left(1+mf'(x) \right)\,dx\)
And the volume $V$ of the solid of revolution is:
\(\displaystyle V=\pi\int_{x_i}^{x_f} r^2\,du=\pi\int_{x_i}^{x_f}\left(\frac{\left|f(x)-mx-b \right|}{\sqrt{m^2+1}} \right)^2\left(\frac{1}{\sqrt{m^2+1}}\left(1+mf'(x) \right)\,dx \right)\)
\(\displaystyle V=\frac{\pi}{\left(m^2+1 \right)^{\frac{3}{2}}}\int_{x_i}^{x_f} \left(f(x)-mx-b \right)^2\left(1+mf'(x) \right)\,dx\)
Let's apply this to a simple example. Let's rotate $f(x)=x^2$ about the line $y=x-1$ on $[0,1]$.
The area of the region is:
\(\displaystyle A=\frac{1}{2}\int_0^1\left|x^2-x+1 \right|(1+2x)\,dx\)
Because \(\displaystyle x^2-x+1>0\) for all real $x$, we may write:
\(\displaystyle A=\frac{1}{2}\int_0^1\left(x^2-x+1 \right)(1+2x)\,dx\)
Expanding the integrand, we have:
\(\displaystyle A=\frac{1}{2}\int_0^1 2x^3-x^2+x+1\,dx\)
Applying the FTOC, we find:
\(\displaystyle A=\frac{1}{2}\left[\frac{1}{2}x^4-\frac{1}{3}x^3+\frac{1}{2}x^2+x \right]_0^1=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{2}+1 \right)=\frac{1}{2}\cdot\frac{5}{3}=\frac{5}{6}\)
The volume of the solid of rotation is:
\(\displaystyle V=\frac{\pi}{2\sqrt{2}}\int_0^1\left(x^2-x+1 \right)^2(1+2x)\,dx\)
Expanding the integrand, we obtain:
\(\displaystyle V=\frac{\pi}{2\sqrt{2}}\int_0^1 2x^5-3x^4+4x^3-x^2+1\,dx\)
Applying the FTOC, we find:
\(\displaystyle V=\frac{\pi}{2\sqrt{2}}\left[\frac{1}{3}x^6-\frac{3}{5}x^5+x^4-\frac{1}{3}x^3+x \right]_0^1=\frac{\pi}{2\sqrt{2}}\left(\frac{1}{3}-\frac{3}{5}+1-\frac{1}{3}+1 \right)=\frac{\pi}{2\sqrt{2}}\cdot\frac{7}{5}= \frac{7\pi}{10\sqrt{2}}\)
As students of calculus, we are taught to find the volumes of solids of rotation obtained by revolving given regions about horizontal and vertical axes of rotation. But, what if the axis of rotation is neither horizontal nor vertical? Please consider the following diagram:
View attachment 1398
We wish to revolve the region shaded in green about the line $y=mx+b$. Using the disk method, where the radius of a disk is $r$ and its thickness is $du$, we may write:
\(\displaystyle dV=\pi r^2\,du\)
Using the formula where the point is $(x,f(x))$ and the line is $y=mx+b$, we have:
\(\displaystyle r=\frac{\left|f(x)-mx-b \right|}{\sqrt{m^2+1}}\)
To find $du$ in terms of $dx$, let $dD$ represent the arc-length along $f(x)$ corresponding to $dx$, as in the following diagram:
https://www.physicsforums.com/attachments/1399
From this, we obtain:
\(\displaystyle dD=\frac{dx}{\cos(\theta)}\)
Now, to obtain the projection of $dD$ onto $y=mx+b$, we find that by rotating everything by \(\displaystyle -\tan^{-1}(m)\), we may write:
\(\displaystyle du=dD\cos\left(\theta-\tan^{-1}(m) \right)\)
Hence, we obtain:
\(\displaystyle du=\frac{\cos\left(\theta-\tan^{-1}(m) \right)}{\cos(\theta)}\,dx\)
Using the angle difference identity for cosine, we find:
\(\displaystyle du=\frac{\cos\left(\theta\right)\cos\left(\tan^{-1}(m) \right)+\sin\left(\theta\right)\sin\left(\tan^{-1}(m) \right)}{\cos(\theta)}\,dx\)
\(\displaystyle du=\left(\cos\left(\tan^{-1}(m) \right)+\tan(\theta)\cos\left(\tan^{-1}(m) \right) \right)\,dx\)
Using:
\(\displaystyle \cos\left(\tan^{-1}(m) \right)=\frac{1}{\sqrt{m^2+1}}\)
\(\displaystyle \sin\left(\tan^{-1}(m) \right)=\frac{m}{\sqrt{m^2+1}}\)
\(\displaystyle \tan(\theta)=f'(x)\)
we then obtain:
\(\displaystyle du=\left(\frac{1}{\sqrt{m^2+1}}+\frac{m}{\sqrt{m^2+1}}f'(x) \right)\,dx\)
\(\displaystyle du=\frac{1}{\sqrt{m^2+1}}\left(1+mf'(x) \right)\,dx\)
And so we may now give the area $A$ of the shaded region as:
\(\displaystyle A=\int_{x_i}^{x_f} r\,du=\int_{x_i}^{x_f}\left(\frac{\left|f(x)-mx-b \right|}{\sqrt{m^2+1}} \right)\left(\frac{1}{\sqrt{m^2+1}}\left(1+mf'(x) \right)dx \right)\)
\(\displaystyle A=\frac{1}{m^2+1}\int_{x_i}^{x_f}\left|f(x)-mx-b \right|\left(1+mf'(x) \right)\,dx\)
And the volume $V$ of the solid of revolution is:
\(\displaystyle V=\pi\int_{x_i}^{x_f} r^2\,du=\pi\int_{x_i}^{x_f}\left(\frac{\left|f(x)-mx-b \right|}{\sqrt{m^2+1}} \right)^2\left(\frac{1}{\sqrt{m^2+1}}\left(1+mf'(x) \right)\,dx \right)\)
\(\displaystyle V=\frac{\pi}{\left(m^2+1 \right)^{\frac{3}{2}}}\int_{x_i}^{x_f} \left(f(x)-mx-b \right)^2\left(1+mf'(x) \right)\,dx\)
Let's apply this to a simple example. Let's rotate $f(x)=x^2$ about the line $y=x-1$ on $[0,1]$.
The area of the region is:
\(\displaystyle A=\frac{1}{2}\int_0^1\left|x^2-x+1 \right|(1+2x)\,dx\)
Because \(\displaystyle x^2-x+1>0\) for all real $x$, we may write:
\(\displaystyle A=\frac{1}{2}\int_0^1\left(x^2-x+1 \right)(1+2x)\,dx\)
Expanding the integrand, we have:
\(\displaystyle A=\frac{1}{2}\int_0^1 2x^3-x^2+x+1\,dx\)
Applying the FTOC, we find:
\(\displaystyle A=\frac{1}{2}\left[\frac{1}{2}x^4-\frac{1}{3}x^3+\frac{1}{2}x^2+x \right]_0^1=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{2}+1 \right)=\frac{1}{2}\cdot\frac{5}{3}=\frac{5}{6}\)
The volume of the solid of rotation is:
\(\displaystyle V=\frac{\pi}{2\sqrt{2}}\int_0^1\left(x^2-x+1 \right)^2(1+2x)\,dx\)
Expanding the integrand, we obtain:
\(\displaystyle V=\frac{\pi}{2\sqrt{2}}\int_0^1 2x^5-3x^4+4x^3-x^2+1\,dx\)
Applying the FTOC, we find:
\(\displaystyle V=\frac{\pi}{2\sqrt{2}}\left[\frac{1}{3}x^6-\frac{3}{5}x^5+x^4-\frac{1}{3}x^3+x \right]_0^1=\frac{\pi}{2\sqrt{2}}\left(\frac{1}{3}-\frac{3}{5}+1-\frac{1}{3}+1 \right)=\frac{\pi}{2\sqrt{2}}\cdot\frac{7}{5}= \frac{7\pi}{10\sqrt{2}}\)
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