Question

Find by integration the volume and surface of the right circular
formed by the revolution of a right-angled triangle about a side which contains the right angle.​

Answer 1

Correct Question:-

Find, by integration, the volume and surface area of the right circular cone formed by the revolution of a right angled triangle about a side which contains the right angle.​

Solution:-

The base of the right triangle is the same as the base radius of the cone. Let it be r.

Both the heights of the right triangle and the cone are same. Let it be h.

Consider the base of triangle $\sf{r}$ is rotated through a very small circular displacement $\sf{dx}$ along forming the cone, in the form of arc which subtends a small angle $\sf{d\theta}$ at the center, i.e., at the right angled corner of the triangle.

Hence we have,

$\longrightarrow\sf{Arc=Angle\times Radius}$

$\longrightarrow\sf{dx=r\ d\theta\quad\quad\dots(1)}$

A small portion of cone in the form of a triangular pyramid is obtained here (the blue solid in the fig.), whose base area can be the area of the small triangle below it (ΔACD in the fig.), so that $\sf{h}$ is taken as the height.

Hence the volume of this small triangular pyramid is,

$\displaystyle\longrightarrow\sf{dV=\dfrac{1}{3}\cdot\dfrac{1}{2}\,r\ dx\cdot h}$

$\displaystyle\longrightarrow\sf{dV=\dfrac{1}{6}\,rh\ dx}$

From (1),

$\displaystyle\longrightarrow\sf{dV=\dfrac{1}{6}\,r^2h\ d\theta}$

The total volume is found out as the right triangle is rotating for one revolution, i.e., by rotating through an angle from $\sf{0}$ to $\sf{2\pi.}$

$\displaystyle\longrightarrow\sf{V=\int\limits_0^{2\pi}\dfrac{1}{6}\,r^2h\ d\theta}$

$\displaystyle\longrightarrow\sf{V=\dfrac{1}{6}\,r^2h\big[\theta\big]_0^{2\pi}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{V=\dfrac{1}{3}\pi r^2h}}}$

So this is the volume of a right circular cone!

Let $\sf{BC=BD=l}$ be the slant height of the cone which is perpendicular to $\sf{dx.}$ So the area of the slanted triangle (ΔBCD in the fig.) is,

$\displaystyle\longrightarrow\sf{dA_c=\dfrac{1}{2}\,l\ dx}$

From (1),

$\displaystyle\longrightarrow\sf{dA_c=\dfrac{1}{2}\,rl\ d\theta}$

So the total curved surface area of the cone is,

$\displaystyle\longrightarrow\sf{A_c=\int\limits_0^{2\pi}\dfrac{1}{2}\,rl\ d\theta}$

$\displaystyle\longrightarrow\sf{A_c=\dfrac{1}{2}\,rl\big[\theta\big]_0^{2\pi}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{A_c=\pi rl}}}$

So this is the curved surface area of a right circular cone!

The area of the ΔACD of base r with altitude $\sf{dx}$ is,

$\displaystyle\longrightarrow\sf{dA_b=\dfrac{1}{2}\,r\ dx}$

From (1),

$\displaystyle\longrightarrow\sf{dA_b=\dfrac{1}{2}\,r^2\ d\theta}$

because CD is perpendicular to AC even it's an arc, since an infinitesimally small arc can be considered as a straight line.

Hence the total area of the base circle is,

$\displaystyle\longrightarrow\sf{A_b=\int\limits_0^{2\pi}\dfrac{1}{2}\,r^2\ d\theta}$

$\displaystyle\longrightarrow\sf{A_b=\dfrac{1}{2}\,r^2\big[\theta\big]_0^{2\pi}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{A_b=\pi r^2}}}$

So this is the base area of a right circular cone, or simply the area of a circle!

And the total surface area of the cone is,

$\displaystyle\longrightarrow\sf{A=A_b+A_c}$

$\displaystyle\longrightarrow\sf{A=\pi r^2+\pi rl}$

$\displaystyle\longrightarrow\sf{\underline{\underline{A=\pi r\left(r+l\right)}}}$

So this is the total surface area of a right circular cone!