Question

Using integration, find the area of the region enclosed between the circles ${{x}^{2}}+{{y}^{2}}=4$ and ${{\left( x-2 \right)}^{2}}+{{y}^{2}}=4$ .

Answer 1

Hint: For solving this question, first we will plot the given curves on the same $x-y$ plane. Then, we will find the desired region whose area is asked in the question. After that, we will solve the given equations and find the coordinates of the intersection points. Then, we will divide the given region into 4 equal parts and take an elementary horizontal strip of width $dy$ in one of its parts and try to write its length in terms of the variable $y$ . Then, we will write the area of the elementary in terms of $y$ and $dy$ by multiplying its length and width. And finally, we will multiply it by 4 and use the formula $\int{\sqrt{{{a}^{2}}-{{y}^{2}}}dy=\dfrac{y}{2}\sqrt{{{a}^{2}}-{{y}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \dfrac{y}{a} \right)}+c$ to integrate the area of the elementary strip with suitable limits to get the total area of the given region.

Complete step-by-step answer:
Given:
We have to find the area of the region enclosed between the circles ${{x}^{2}}+{{y}^{2}}=4$ and ${{\left( x-2 \right)}^{2}}+{{y}^{2}}=4$ .
Now, before we proceed we should plot the circles ${{x}^{2}}+{{y}^{2}}=4$ and the ${{\left( x-2 \right)}^{2}}+{{y}^{2}}=4$ on the same $x-y$ plane. For more clarity, look at the figure given below:

In the above figure, we have to find the area of the region OACB.
Now, from the above figure, the point $O\equiv \left( 0,0 \right)$ is the centre of the circle ${{x}^{2}}+{{y}^{2}}=4$ and point $C\equiv \left( 2,0 \right)$ is the centre of the circle ${{\left( x-2 \right)}^{2}}+{{y}^{2}}=4$ . And we can say that for the coordinates of points A and B, we should equate the equations ${{x}^{2}}+{{y}^{2}}=4$ and ${{\left( x-2 \right)}^{2}}+{{y}^{2}}=4$ . Then,
$\begin{aligned} & {{x}^{2}}+{{y}^{2}}={{\left( x-2 \right)}^{2}}+{{y}^{2}}=4 \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}={{x}^{2}}+4-4x+{{y}^{2}} \\\ & \Rightarrow 4x=4 \\\ & \Rightarrow x=1 \\\ \end{aligned}$
Now, from the above result and the figure, to get the coordinates of points A and B, we should put $x=1$ in the equation ${{x}^{2}}+{{y}^{2}}=4$ . Then,
$\begin{aligned} & {{x}^{2}}+{{y}^{2}}=4 \\\ & \Rightarrow {{1}^{2}}+{{y}^{2}}=4 \\\ & \Rightarrow {{y}^{2}}=3 \\\ & \Rightarrow y=\pm \sqrt{3} \\\ \end{aligned}$
Now, from the above result and figure, we conclude that, coordinates of points $A\equiv (1,\sqrt{3})$ and $B\equiv (1,-\sqrt{3})$ .
Now, we will join the points A and B, O and C. For more clarity, look at the figure given below:

In the above figure, segment AB and OC intersect at a point $D\equiv (1,0)$ on the $x-axis$ . And by the concept of symmetry, we can say that area of region ODA, region CDA, region ODB, region CDB will be equal to one-fourth of the area of the region OACB. Then,
Area of the region OACB $=4\times$ (Area of the region CDA).
Now, we will integrate for the area of the region CDA and then, we will multiply it by 4 to get the area for the region OACB.
Now, we take an elementary horizontal strip at $y$ of width $dy$ in region CDA. For more clarity, look at the figure given below:

Now, here we should not take ${{x}_{right}}=-\sqrt{4-{{y}^{2}}}$ , because in the first and fourth quadrant $x>0$ .
Now, to find the length of the elementary strip, we should subtract the ${{x}_{right}}=\sqrt{4-{{y}^{2}}}$ and ${{x}_{left}}=1$ . Then,
Length of the elementary strip $={{x}_{right}}-{{x}_{left}}=\sqrt{4-{{y}^{2}}}-1$ .
Now, as we know that width of the elementary strip is $dy$ . So, the area of the elementary strip will be length multiplied by width. Then,
Area of the elementary strip $=dA=\left( \sqrt{4-{{y}^{2}}}-1 \right)dy$ .
Now, to get the total area of the region, we should add the area of such elementary strips from $y=0$ to $y=\sqrt{3}$ so, to get the desired area we should integrate the expression $\left( \sqrt{4-{{y}^{2}}}-1 \right)dy$ from $y=0$ to $y=\sqrt{3}$ . Then,
Area of the region CDA $=\int\limits_{0}^{\sqrt{3}}{\left( \sqrt{4-{{y}^{2}}}-1 \right)dy}$ .
So, Area of the region OACB $=4\times$ (Area of the region CDA) $=4\left( \int\limits_{0}^{\sqrt{3}}{\left( \sqrt{4-{{y}^{2}}}-1 \right)dy} \right)$
Now, we will use the formula $\int{{{y}^{n}}dy=\dfrac{{{y}^{n+1}}}{n+1}+c}$ and $\int{\sqrt{{{a}^{2}}-{{y}^{2}}}dy=\dfrac{y}{2}\sqrt{{{a}^{2}}-{{y}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \dfrac{y}{a} \right)}+c$ to integrate the above integral. Then,

& 4\left( \int\limits_{0}^{\sqrt{3}}{\left( \sqrt{4-{{y}^{2}}}-1 \right)dy} \right) \\\ & \Rightarrow 4\left[ \dfrac{y}{2}\sqrt{4-{{y}^{2}}}+\dfrac{4}{2}{{\sin }^{-1}}\left( \dfrac{y}{2} \right)-y \right]_{0}^{\sqrt{3}} \\\ & \Rightarrow 4\left[ \left( \dfrac{\sqrt{3}}{2}\times \sqrt{4-{{\left( \sqrt{3} \right)}^{2}}}+2{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)-\sqrt{3} \right)-\left( 0 \right) \right] \\\ & \Rightarrow 4\left[ \dfrac{\sqrt{3}}{2}\times \sqrt{4-3}+2{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)-\sqrt{3} \right] \\\ & \Rightarrow 4\left[ \dfrac{\sqrt{3}}{2}-\sqrt{3}+2{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \right] \\\ & \Rightarrow 4\left[ 2{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)-\dfrac{\sqrt{3}}{2} \right] \\\ \end{aligned}$$ Now, we will put ${{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{3}$ in the above. Then, $$\begin{aligned} & 4\left[ 2{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)-\dfrac{\sqrt{3}}{2} \right] \\\ & \Rightarrow 4\left[ 2\times \dfrac{\pi }{3}-\dfrac{\sqrt{3}}{2} \right] \\\ & \Rightarrow 4\left[ \dfrac{2\pi }{3}-\dfrac{\sqrt{3}}{2} \right] \\\ & \Rightarrow \dfrac{8\pi }{3}-2\sqrt{3} \\\ \end{aligned}$$ Now, from the above result, we conclude that area of the region OACB will be $$\dfrac{8\pi }{3}-2\sqrt{3}\text{ sq}\text{.units}$$ . Thus, the area of the region enclosed between the circles ${{x}^{2}}+{{y}^{2}}=4$ and ${{\left( x-2 \right)}^{2}}+{{y}^{2}}=4$ will be equal to $$\dfrac{8\pi }{3}-2\sqrt{3}\text{ sq}\text{.units}$$ . Note: Here, the student should first plot the given curves carefully and then find the desired region whose area is asked in the question and proceed in a stepwise manner. Then, for easy calculation, we should make use of the symmetricity of the region OACB, and we should be careful while writing the dimensions of the elementary strip and for that, we should take help from the plot of the given curves. Then, we should apply formulas like $\int{\sqrt{{{a}^{2}}-{{y}^{2}}}dy=\dfrac{y}{2}\sqrt{{{a}^{2}}-{{y}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \dfrac{y}{a} \right)}$ correctly. Moreover, we could have used substituted $y=2\sin \theta $ , $dy=2\cos \theta d\theta $ in the integral $4\left( \int\limits_{0}^{\sqrt{3}}{\left( \sqrt{4-{{y}^{2}}}-1 \right)dy} \right)$ and transformed it to $4\left( \int\limits_{0}^{\dfrac{\pi }{3}}{\left( 2\cos \theta -1 \right)2\cos \theta d\theta } \right)$ and solved it further to get the area of the desired region.