Volume of Solid of Revolution about Slant Line

We know how to find the volume of a solid of revolution obtained by rotating a region about a horizontal or vertical line (see Method of Disks/Rings and Method of Cylindrical Shells). But what if we rotate about a slanted line, that is, a line that is neither horizontal nor vertical?

Let ${C}$ be the arc of the curve ${y}={f{{\left({x}\right)}}}$ between the points ${P}={\left({p},{f{{\left({p}\right)}}}\right)}$ and ${B}={\left({q},{f{{\left({q}\right)}}}\right)}$ and let ${R}$ be the region bounded by ${C}$, by the line ${y}={m}{x}+{b}$ (which lies entirely below ${C}$), and by the perpendiculars to the line from ${A}$ and ${B}$.

We first will try to find area of region ${R}$. For this we will need another sketch.

We again approximate area by the sum or rectangles, but this time rectangles are perpendicular to the line ${y}={m}{x}+{b}$. Area of i-th rectangle is ${L}_{{i}}\Delta{u}$. Thus, approximate area of ${R}$ is ${A}\approx{\sum_{{{i}={1}}}^{{n}}}{L}_{{i}}\Delta{u}$.

Taking ${n}$ very large we have that ${A}=\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{n}}}{L}_{{i}}\Delta{u}$.

We need to express ${L}_{{i}}$ and $\Delta{u}$ in terms of $\Delta{x}$.

From the sketch we immediately have that ${\tan{{\left(\alpha\right)}}}={f{'}}{\left({x}_{{i}}\right)}$ (slope of tangent line) and ${\tan{{\left(\beta\right)}}}={m}$.

Also, ${D}_{{i}}=\frac{{\Delta{x}}}{{\cos{{\left(\alpha\right)}}}}$, and $\Delta{u}={D}_{{i}}{\cos{{\left(\beta-\alpha\right)}}}={D}_{{i}}{\left({\cos{{\left(\beta\right)}}}{\cos{{\left(\alpha\right)}}}+{\sin{{\left(\alpha\right)}}}{\sin{{\left(\beta\right)}}}\right)}$.

From this we have that $\Delta{u}=\frac{{\Delta{x}}}{{{\cos{{\left(\alpha\right)}}}}}{\left({\cos{{\left(\beta\right)}}}{\cos{{\left(\alpha\right)}}}+{\sin{{\left(\alpha\right)}}}{\sin{{\left(\beta\right)}}}\right)}=\Delta{x}{\left({\cos{{\left(\beta\right)}}}+{\tan{{\left(\alpha\right)}}}{\sin{{\left(\beta\right)}}}\right)}$.

Now, we need to express ${\sin{{\left(\beta\right)}}}$ and ${\cos{{\left(\beta\right)}}}$ in terms of ${\tan{{\left(\beta\right)}}}$.

From identity ${{\tan}}^{{2}}{\left(\beta\right)}+{1}=\frac{{1}}{{{{\cos}}^{{2}}{\left(\beta\right)}}}$ we have that ${\cos{{\left(\beta\right)}}}=\sqrt{{\frac{{1}}{{{1}+{{\tan}}^{{2}}{\left(\beta\right)}}}}}=\sqrt{{\frac{{1}}{{{1}+{{m}}^{{2}}}}}}$.

Since ${{\sin}}^{{2}}{\left(\beta\right)}+{{\cos}}^{{2}}{\left(\beta\right)}={1}$ then ${\sin{{\left(\beta\right)}}}=\frac{{m}}{\sqrt{{{{m}}^{{2}}+{1}}}}$.

Therefore, $\Delta{u}=\Delta{x}{\left(\frac{{1}}{{\sqrt{{{{m}}^{{2}}+{1}}}}}+{f{'}}{\left({x}_{{i}}\right)}\frac{{m}}{\sqrt{{{{m}}^{{2}}+{1}}}}\right)}=\Delta{x}\frac{{{1}+{m}{f{'}}{\left({x}_{{i}}\right)}}}{{\sqrt{{{{m}}^{{2}}+{1}}}}}$.

Now, we need to determine ${L}_{{i}}$.

It is fairly easy to do: ${L}_{{i}}={\left({f{{\left({x}_{{i}}\right)}}}-{\left({m}{x}_{{i}}+{b}\right)}\right)}{\cos{{\left(\beta\right)}}}=\frac{{{f{{\left({x}_{{i}}\right)}}}-{m}{x}_{{i}}-{b}}}{{\sqrt{{{{m}}^{{2}}+{1}}}}}$.

That's all. ${A}=\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{n}}}{L}_{{i}}\Delta{u}=\frac{{{f{{\left({x}_{{i}}\right)}}}-{m}{x}_{{i}}-{b}}}{\sqrt{{{{m}}^{{2}}+{1}}}}\cdot\frac{{{1}+{m}{f{'}}{\left({x}_{{i}}\right)}}}{{\sqrt{{{{m}}^{{2}}+{1}}}}}\Delta{x}=\frac{{1}}{{{{m}}^{{2}}+{1}}}\lim_{{{n}\to\infty}}{\sum_{{{i}={1}}}^{{n}}}{\left({f{{\left({x}_{{i}}\right)}}}-{m}{x}_{{i}}-{b}\right)}{\left({1}+{m}{f{'}}{\left({x}_{{i}}\right)}\right)}$.

This is limit of the Riemann sum, i.e. definite integral, so

Now, to find volume we use method of disks: we slice perpendicular to the line ${y}={m}{x}+{b}$: radius of disk is ${L}=\frac{{{f{{\left({x}\right)}}}-{m}{x}-{b}}}{\sqrt{{{{m}}^{{2}}+{1}}}}$.

Thus, ${A}{\left({x}\right)}=\pi{{L}}^{{2}}=\pi{{\left(\frac{{{f{{\left({x}\right)}}}-{m}{x}-{b}}}{{\sqrt{{{{m}}^{{2}}+{1}}}}}\right)}}^{{2}}$.

Therefore, ${V}={\int_{{p}}^{{q}}}\pi{{L}}^{{2}}{d}{u}={\int_{{p}}^{{q}}}\pi{{\left(\frac{{{f{{\left({x}\right)}}}-{m}{x}-{b}}}{\sqrt{{{{m}}^{{2}}+{1}}}}\right)}}^{{2}}\frac{{{1}+{m}{f{'}}{\left({x}\right)}}}{\sqrt{{{{m}}^{{2}}+{1}}}}{d}{x}$.