Area of Bounded Regions
2.0 Working Rule for finding the Area
2.0 Working Rule for finding the Area
1. Sign Convention
If the curve is above $x$-axis then the value of integration will be positive but if the curve is below the $x$-axis the value will be negative but we know that area can't be negative and so we always take the modulus of integration. For this there are three possible cases:
Case 1: If the curve is above the $x$-axis $\forall x \in (a,b)$ then the value of integration will be positive and there is no modification in $A = \int\limits_a^b {f(x)dx} $
Case 2: If the curve is below the $x$-axis $\forall x \in (a,b)$ then the value of integration will be negative and so we take the modulus of value$$i.e,\ A = \left| {\int\limits_a^b {f(x)dx} } \right|$$
Case 3: If the curve is on both sides of $x$-axis i.e, above as well as below the $x$-axis, then calculate both areas separately and add their modulii to get the total area.
In general, if the curve $y=f(x)$ crosses x-axis 'n' times when $x$ varies from $a$ to $b$, then the area between the curve $y=f(x)$, the $x$-axis and the lines $x=a$ and $x=b$ is given by $$A = \left| {{A_1}} \right| + \left| {{A_2}} \right| + ... + \left| {{A_n}} \right| + \left| {{A_{n + 1}}} \right|$$
Question 2. Find the area bounded by $y = \operatorname{cosx} $, $x = \frac{{ - \pi }}{2}$, $x={2}{\pi}$ and the $x$-axis.
Solution:
The required area is $A = \left| {{A_1}} \right| + \left| {{A_2}} \right| + \left| {{A_3}} \right|$
$$\begin{equation} \begin{aligned} {A_1} = \int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\cos xdx} = [\sin x]_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} = 2 \\ {A_2} = \int\limits_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\cos xdx} = [\sin x]_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} = 2 \\ {A_3} = \int\limits_{\frac{{3\pi }}{2}}^{2\pi } {\cos xdx} = [\sin x]_{\frac{{3\pi }}{2}}^{2\pi } = 1 \\\end{aligned} \end{equation} $$
Therefore, the total area bounded is $5$ sq.units($A=2+2+1$)
2. If the curve is symmetrical about the $x$-axis , or the $y$-axis, or both, then calculate the area of one symmetrical part and multiply it by the number of symmetrical parts to get whole area.
Question 3. Find the area interior of the circle ${x^2} + {y^2} = {r^2}$
Solution: The curve is symmetrical about both the $x$-axis and the $y$-axis.
Therefore, the total area bounded by the circle is, $$A = 4 \times a$$
From the equation of the circle,$$\begin{equation} \begin{aligned} {x^2} + {y^2} = {r^2} \\ \Rightarrow {y^2} = {r^2} - {x^2} \\ \Rightarrow y = \pm \sqrt {{r^2} - {x^2}} \\\end{aligned} \end{equation} $$Therefore,the equation of the upper part of the circle is $y = + \sqrt {{r^2} - {x^2}} $ (since, $y$ is positive above $x$-axis).
Hence,$$a = \int\limits_0^r {\sqrt {{r^2} - {x^2}} dx} $$
Put, $$x = r\sin \theta$$$$\Rightarrow dx = r\cos \theta d\theta $$
Hence, the limits of $\theta $ will be $0$ and $\frac{\pi }{2}$, $$\begin{equation} \begin{aligned} \Rightarrow a = \int\limits_0^{\frac{\pi }{2}} {\left\{ {\sqrt {{r^2} - {r^2}{{\sin }^2}\theta } } \right\}r\sin \theta d\theta } \\ \Rightarrow a = \int\limits_0^{\frac{\pi }{2}} {\left\{ {r\cos \theta } \right\}r\cos \theta d\theta } \\ \Rightarrow a = {r^2}\int\limits_0^{\frac{\pi }{2}} {{{\left\{ {\cos \theta } \right\}}^2}d\theta } \\ \Rightarrow a = {r^2}\int\limits_0^{\frac{\pi }{2}} {\left\{ {\frac{{1 + \cos 2\theta }}{2}} \right\}d\theta } \\ \Rightarrow a = {r^2}\left\{ {\frac{\pi }{4}} \right\} \\\end{aligned} \end{equation} $$
Hence, the value of $a = \frac{{\pi {r^2}}}{4}$ and the total area of circle is four times of $a$ i.e, ${\pi {r^2}}$.
