Solutions - 0090-02

(1) Integral formula for surface area, revolution around $x$-axis:

$$S={\int }_a^{b}2\pi f(x)\sqrt{1+{(f^{\prime }(x))}^2}dx$$

---

(2) Work out integrand:

$$\begin{array}{c}f(x)=mx\gg \gg f^{\prime }(x)=m\\ \\ S\gg \gg {\int }_0^{h}2\pi x\sqrt{1+m^2}dx\end{array}$$

---

(4) Evaluate integral:

$$\begin{array}{c}{\int }_0^{h}2\pi mx\sqrt{1+m^2}dx\\ \gg \gg 2\pi m\sqrt{1+m^2}{\int }_0^{h}xdx\\ \gg \gg 2\pi m\sqrt{1+m^2}{\left(\frac{x^2}{2}\right)|}_0^h\\ \gg \gg \pi h^{2}m\sqrt{1+m^2}\end{array}$$

---

(5) Verify with geometry:

Note that unrolling the cone forms a sector with radius $h\sqrt{1+m^2}$ and arc length $2\pi mh$. The total circumference is $2\pi h\sqrt{1+m^2}$. So the area of the sector is:

$$\begin{array}{c}A=\frac{\text{sector arc}}{\text{total arc}}\cdot \text{area circle}\\ \\ \gg \gg \frac{2\pi mh}{2\pi h\sqrt{1+m^2}}\cdot \pi {\left(h\sqrt{1+m^2}\right)}^2\\ \\ \gg \gg \pi h^{2}m\sqrt{1+m^2}\end{array}$$

Notes:

• Sector radius is the lateral length of the cone: hypotenuse of right triangle with legs $h$ (on $x$-axis) and $mh$ (on $y$-axis).
• Sector arc is $2\pi mh$ because $mh$ is radius of the base.