Question

How to get the surface area of a cone using integral calculus $?$

Answer 1

This question describes the operation of finding the area of a circle and lateral surface of the cone. We need to know the basic integration and differentiation. Also, we need to know the process of finding$\sin \theta$, $\cos \theta$and $\tan \theta$from the triangle.

We need to know trigonometric conditions to make an easy calculation.

Complete step by step solution:

In the given question we have to find the total surface area of a cone using integral calculus. For finding the surface area the following figure will help us,

In the above figure,

$C$ is the center of the bottom of the cone,

$l$is the length of slant edge in the cone,

$h$is the vertical height of the cone,

$\alpha$ is the angle between $OC$and $OB$,

$dl$ is the small part of the slant edge,

$dx$ is the small part of vertical height,

$r$is the radius of the bottom cone.

We know that the bottom of the cone is a circle. So, we have to find the area of the circle. For finding the area of the circle we assume the following figure,

The above figure was drawn with the reference of$fig:1$. From the$fig:2$, we get the following equation

${x^2} + {y^2} = {r^2} \to \left( 1 \right)$ (By Pythagoras theorem)

The above equation can also be written as

${y^2} = {r^2} - {x^2}$

$y = \sqrt {{r^2} - {x^2}}$

The area of the circle,

${A_{circle}}$=$\int {\sqrt {{r^2} - {x^2}} } dx$

Let’s take${r^2}$it as a common term, so we get

${A_{circle}} = \int {\sqrt {{r^2}\left( {1 - \dfrac{{{x^2}}}{{{r^2}}}} \right)} dx}$

We know that$\sqrt {{r^2}} = r$

${A_{circle}} = r\int {\sqrt {\left( {1 - \dfrac{{{x^2}}}{{{r^2}}}} \right)} dx} \to \left( 2 \right)$

If we consider the limit between$- r$and$r$ we can evaluate the area of half of the circle.
From$fig:2$we get the following values,

$$\cos \theta = \dfrac{{adj}}{{opposite}} = \dfrac{x}{r} \\\ \dfrac{x}{r} = {\cos ^2}\theta = 1 - {\sin ^2}\theta \\\$$

$\theta = arc\operatorname{Cos} \left( {\dfrac{x}{r}} \right)$

Let’s take$x = - r$

$\theta = arc\operatorname{Cos} \left( {\dfrac{{ - r}}{r}} \right)$
$\theta = arc\operatorname{Cos} \left( { - 1} \right)$

We know that, $arc\operatorname{Cos} ( - 1) = \pi$

So, $\theta = \pi$

Let’s take $x = r$

$\theta = arc\operatorname{Cos} \left( {\dfrac{r}{r}} \right)$
$\theta = arc\operatorname{Cos} \left( 1 \right)$

We know that, $arc\operatorname{Cos} (1) = 0$

So, $\theta = 0$

We know that,

$\cos \theta = \dfrac{x}{r}$

Let’s differentiate the above equation, we get

$- \sin \theta d\theta = \dfrac{1}{r}dx$
So, we get

$$\dfrac{{dx}}{r} = - \sin \theta d\theta \\\ dx = - r\sin \theta d\theta \\\$$

So, the equation$\left( 2 \right)$is,

${A_{circle}} = r\int {\sqrt {\left( {1 - \dfrac{{{x^2}}}{{{r^2}}}} \right)} dx}$

(For finding the area of half of the circle the limit will be$\pi to0$.)

Let’s substitute $\dfrac{{{x^2}}}{{{r^2}}} = 1 - {\sin ^2}\theta ,dx = - r\sin \theta d\theta$in the above equation, we get

$$\dfrac{1}{2}{A_{circle}} = r\int\limits_\pi ^0 {\sqrt {1 - (1 - {{\sin }^2}\theta )} ( - r\sin \theta d\theta )} \\\ \dfrac{1}{2}{A_{circle}} = r\int\limits_\pi ^0 {\sqrt {1 - 1 + {{\sin }^2}\theta } ( - r\sin \theta d\theta )} \\\$$

Here, $r$is constant, so it can be taken to the outside of the interval.

)$$ $$\dfrac{1}{2}{A_{circle}} = - {r^2}\int\limits_\pi ^0 {\sin \theta .\sin \theta d\theta } $$ $$\dfrac{1}{2}{A_{circle}} = - {r^2}\int\limits_\pi ^0 {{{\sin }^2}\theta d\theta } $$ We know that, $${\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$$. So, we get $$\dfrac{1}{2}{A_{circle}} = - {r^2}\int\limits_\pi ^0 {\dfrac{{1 - \cos 2\theta }}{2}d\theta } = - {r^2}\int\limits_\pi ^0 {\left( {\dfrac{1}{2} - \dfrac{{\cos 2\theta }}{2}} \right)d\theta } $$ From equation$$\left( 1 \right)$$ $${x^2} + {y^2} = {r^2} \to \left( 3 \right)$$ Let’s differentiate the above equation, we get $$2xdx + 2ydy = 2rdr$$ In the above equation$$r$$is constant so, $$dr = 0$$. So, the above equation becomes, $$2xdx + 2ydy = 0$$

2xdx = - 2ydy \\
\dfrac{{dy}}{{dx}} = \dfrac{{ - 2x}}{{2y}} = \dfrac{{ - x}}{y} \to \left( 4 \right) \\

From equation$$\left( 1 \right)$$

{y^2} = {r^2} - {x^2} \\
y = \sqrt {{r^2} - {x^2}} \\

By substituting the above-mentioned value in the equation$$\left( 4 \right)$$we get, $$\dfrac{{dy}}{{dx}} = \dfrac{{ - x}}{{\sqrt {{r^2} - {x^2}} }}$$ The arc length L is $$L = \int {ds} $$ Here, $$ds = \sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} dx$$ Let’s substitute the$$\dfrac{{dy}}{{dx}}$$value in the above equation, we get

ds = \sqrt {1 + {{\left( {\dfrac{{ - x}}{{\sqrt {{r^2} - {x^2}} }}} \right)}^2}} dx \\

ds = \sqrt {1 + \dfrac{{{x^2}}}{{{r^2} - {x^2}}}} dx \\

$$$$

ds = \sqrt {\dfrac{{{r^2} - {x^2} + {x^2}}}{{{r^2} - {x^2}}}} dx \\

ds = \sqrt {\dfrac{{{r^2}}}{{{r^2} - {x^2}}}} dx \\

Half of the circumference of the circle is given below, $$L = \int\limits_{ - r}^r {ds} $$ By substituting the value of$$ds$$ we get, $$L = \int\limits_{ - r}^r {\sqrt {\dfrac{{{r^2}}}{{{r^2} - {x^2}}}} } dx$$ We know that, $$x = r\cos \theta $$, $$dx = - r\sin \theta d\theta $$so, the above equation becomes $$L = \int\limits_\pi ^0 {\sqrt {\dfrac{{{r^2}}}{{{r^2} - {x^2}}}} } \left( { - r\sin \theta } \right)d\theta

$L = \int\limits_\pi ^0 {\sqrt {\dfrac{{{r^2}}}{{{r^2} - {{\left( {r\cos \theta } \right)}^2}}}} } \left( { - r\sin \theta } \right)d\theta$

$L = \int\limits_\pi ^0 {\sqrt {\dfrac{{{r^2}}}{{{r^2}\left( {1 - {{\cos }^2}\theta } \right)}}} } \left( { - r\sin \theta } \right)d\theta$

We know that, $1 - {\cos ^2}\theta = {\sin ^2}\theta$so, the above equation become,

$L = \int\limits_\pi ^0 {\sqrt {\dfrac{{{r^2}}}{{{r^2}\left( {{{\sin }^2}\theta } \right)}}} } \left( { - r\sin \theta } \right)d\theta$

$L = - r\int\limits_\pi ^0 {\dfrac{1}{{\sin \theta }}} .\sin \theta d\theta$
$L = - r\int\limits_\pi ^0 1 d\theta$

$$L = - r\left[ \theta \right]_\pi ^0 = - r\left[ {0 - \pi } \right] \\\ L = \pi r \\\$$

So, the full circumference of the circle is$2L = 2\pi r$

So, the lateral surface area of the cone
$S = \int\limits_0^h {2\pi f\left( x \right)} dl$
Here,$y = f\left( x \right)$

From the given figure

$$\cos \alpha = \dfrac{h}{l} \\\ l = \dfrac{h}{{\cos \alpha }} \\\ dl = \dfrac{{dh}}{{\cos \alpha }} = \dfrac{{dx}}{{\cos \alpha }} \\\ \tan \alpha = \dfrac{r}{h} = \dfrac{{f\left( x \right)}}{x} \\\$$

$f\left( x \right) = x\tan \alpha$

By using the above values we get,

$S = \int\limits_0^h {2\pi x\tan \alpha \dfrac{{dx}}{{\cos \alpha }}}$

$S = 2\pi \dfrac{{\tan \alpha }}{{\cos \alpha }}\int\limits_0^h {xdx}$

$$S = 2\pi \dfrac{{\left( {\dfrac{r}{h}} \right)}}{{\left( {\dfrac{h}{l}} \right)}}\left[ {\dfrac{{{x^2}}}{2}} \right]_0^h \\\ S = 2\pi \left( {\dfrac{{rl}}{{{h^2}}}} \right)\left[ {\dfrac{{{h^2}}}{2} - \dfrac{{{0^2}}}{2}} \right] \\\ S = 2\pi r \\\ \\\$$

So, the total surface area of the cone$=$total surface area of the circle$+$total surface area of the lateral surface.

The total surface area of the cone=$\pi {r^2} + 2\pi r$

Note: We have to consider the bottom of the cone as a circle. At first, we want to find the area of a circle, and next, we want to find the area of the lateral surface. Also, note that the integral of $\cos \theta$is$\sin \theta$ and the differentiation of $\cos \theta$is$- \sin \theta$. If a constant term is present inside the integral, we can take the term outside of the integral.