 ## Area Bounded By Closed Curve

In this tutorial we will learn how to find the area of a region bounded by a closed curve. To do so consider the figure (1.1) given in the thumbnail.

Let PM = y1 and P’M = y2. The area of the verical strip (given in the figure in thumbnail) is (y1 – y2) dx. And the area of the region bounded by the closed curve will be $Area=\int ^{b}_{a}\left( y_{1}-y_{2}\right) dx$. Here OB = a and OB’ = b.

Example: Find the area bounded by closed curve y = (4x2 – x4)1/2.
Solution: The given function represents a right side of a loop which intersects the x-axis at (0, 0) and (2, 0). Now the area of the reqired region will be

\begin{aligned}Area=2\int ^{2}_{0}\left( x\sqrt{4-x^{2}}\right) dx\Rightarrow Let,4-x^{2}=t\\ \Rightarrow -2xdx=dt\Rightarrow -\int ^{0}_{4}t^{1/2}dt\\ \Rightarrow \int ^{4}_{0}\left( t\right) ^{\dfrac{1}{2}}dt=\dfrac{2}{3}\left( t^{\dfrac{3}{2}}\right) _{0}^{4}=\dfrac{16}{3}\end{aligned}

Example: Find the area bounded by closed curve given by the equation |y| = 1 – x2.
Solution: The given function represents two parabola y = x2 – 1 and y = 1 – x2 with vertices (0, -1) and (0, 1). Both parabola together forms a closed curve and intesect the x-axis at (-1, 0) and (1, 0). Given functionscollectively will represent a shape similar to a circle (fig. 2.1 in thumbnail). Hence the required area of the desired region will be

\begin{aligned}Area=4\int ^{1}_{0}\left( 1-x^{2}\right) dx=4\int ^{1}_{0}dx-4\int ^{1}_{0}x^{2}dx\\ \Rightarrow 4-\dfrac{4}{3}\left( x^{3}\right) _{0}^{1}=4-\dfrac{4}{3}=\dfrac{8}{3}\end{aligned}

Example: Find the area of the closed region bounded by the curve given by the equation (y – Sin-1x)2 = x – x2.
Solution: Consider the figure 3.1 given in the thumbnail. Since x – x2 ≥ 0, 0 ≤ x ≤ 1 and y = Sin-1x ± (x – x2)1/2. Hence the given equation represents two branches of the given function. Thease branches intersect at (0, 0) and (1, π/2). Hence the required area will be given by

\begin{aligned}\int ^{1}_{0}\left( \left( \sin ^{-1}\left( x\right) +\sqrt{x-x^{2}}\right) -\left( \sin ^{-1}\left( x\right) -\sqrt{x-x^{2}}\right) \right) dx\\ =2\int ^{1}_{0}\left( \sqrt{x-x^{2}}\right) dx=\dfrac{1}{4}\left( \dfrac{\pi }{2}+\dfrac{\pi }{2}\right) \\ \ =\dfrac{\pi }{4}\end{aligned}

Now attempt at least five questions based on this topic. So this is it from this tutorial. Hoping you people will attempt a few problems based on the topic discussed in this tutorial. In the next tutorial, we will discuss the topic “Area Of The Region Bounded By the Curves Given In their Parametric Form”.