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Question: Find the area of the region into which the circle \({x^2} + {y^2} = 4\) as divided by the line \(x +...

Find the area of the region into which the circle x2+y2=4{x^2} + {y^2} = 4 as divided by the line x+3y=2x + \sqrt 3 y = 2.

Explanation

Solution

In the above question, first we have to draw a circle of radius 22 and then we have to draw the required line which intersects the given circle at two points. Then we will draw the area between the x-axis and the line x+3y=2x + \sqrt 3 y = 2 and then we have to find the area of that region using integration.

Complete answer:
In the above question, first we will draw a circle x2+y2=4{x^2} + {y^2} = 4 and a line x+3y=2x + \sqrt 3 y = 2and then we will find the points of intersection of the line and the circle.
So, to find the point of intersection we have to solve both the equations simultaneously.
We have,
x+3y=2x + \sqrt 3 y = 2
⇒y=2−x3.............(1)\Rightarrow y = \dfrac{{2 - x}}{{\sqrt 3 }}.............\left( 1 \right)
Now, put this value in the equation of the circle.
⇒(x)2+(2−x3)2=4\Rightarrow {\left( x \right)^2} + {\left( {\dfrac{{2 - x}}{{\sqrt 3 }}} \right)^2} = 4
⇒(x)2+(2−x)23=4\Rightarrow {\left( x \right)^2} + \dfrac{{{{\left( {2 - x} \right)}^2}}}{3} = 4
⇒3(x)2+(2−x)2=4×3\Rightarrow 3{\left( x \right)^2} + {\left( {2 - x} \right)^2} = 4 \times 3
⇒3x2+4+x2−4x=12\Rightarrow 3{x^2} + 4 + {x^2} - 4x = 12
⇒4x2+4−4x−12=0\Rightarrow 4{x^2} + 4 - 4x - 12 = 0
⇒4x2−4x−8=0\Rightarrow 4{x^2} - 4x - 8 = 0
⇒x2−x−2=0\Rightarrow {x^2} - x - 2 = 0
⇒x2−2x+x−2=0\Rightarrow {x^2} - 2x + x - 2 = 0
⇒x(x−2)+1(x−2)=0\Rightarrow x\left( {x - 2} \right) + 1\left( {x - 2} \right) = 0
⇒(x−2)(x+1)=0\Rightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0
Therefore, the x coordinates of two points of intersection is (2,−1)\left( {2, - 1} \right).
Now, we will put both these values in equation one to find the value of y.
⇒y=2−23\Rightarrow y = \dfrac{{2 - 2}}{{\sqrt 3 }} and y=2−(−1)3y = \dfrac{{2 - \left( { - 1} \right)}}{{\sqrt 3 }}

⇒y=0 \Rightarrow y = 0 and y=33y = \dfrac{3}{{\sqrt 3 }}
⇒y=0\Rightarrow y = 0 and y=3y = \sqrt 3
Therefore, the y-coordinates of two points of intersection is (0,3)\left( {0,\sqrt 3 } \right)
So, the two points of intersection are (2,0)  and  (−1,3)\left( {2,0} \right)\,\,and\,\,\left( { - 1,\sqrt 3 } \right).
So the figure will be,


So, in the above figure first we will draw the area of region AOB and then the region of BOC using integration and then we have to add both the values.
Total area == Area of region AOB ++ Area of region BOC
⇒∫AxBxy dx + ∫BxCxy dx\Rightarrow \int\limits_{{A_x}}^{{B_x}} {y\,dx} \, + \,\int\limits_{{B_x}}^{{C_x}} {y\,dx}
In the region of AOB we will integrate the region under the circle and in the region of BOC we will integrate the region under the line.
⇒∫−2−14−x2 dx + ∫−122−x3 dx\Rightarrow \int\limits_{ - 2}^{ - 1} {\sqrt {4 - {x^2}} \,dx} \, + \,\int\limits_{ - 1}^2 {\dfrac{{2 - x}}{{\sqrt 3 }}\,dx}
⇒∫−2−1(2)2−x2 dx + ∫−1223−x3 dx\Rightarrow \int\limits_{ - 2}^{ - 1} {\sqrt {{{\left( 2 \right)}^2} - {x^2}} \,dx} \, + \,\int\limits_{ - 1}^2 {\dfrac{2}{{\sqrt 3 }} - \dfrac{x}{{\sqrt 3 }}\,dx}
Now, using (∫a2−x2 dx= a22sin⁡−1x2+x2a2−x2)\left( {\int {\sqrt {{a^2} - {x^2}} \,dx = \dfrac{{\,{a^2}}}{2}{{\sin }^{ - 1}}\dfrac{x}{2} + \dfrac{x}{2}} \sqrt {{a^2} - {x^2}} } \right)
On integration, we get
⇒[42sin⁡−1x2+x24−x2]−2−1+[2x3]−12−[x223]−12\Rightarrow \left[ {\dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{x}{2} + \dfrac{x}{2}\sqrt {4 - {x^2}} } \right]_{ - 2}^{ - 1} + \left[ {\dfrac{{2x}}{{\sqrt 3 }}} \right]_{ - 1}^2 - \left[ {\dfrac{{{x^2}}}{{2\sqrt 3 }}} \right]_{ - 1}^2
⇒[42sin⁡−1−12+−124−(−1)2−42sin⁡−1−22+−224−(−2)2]+[2×23−(2(−1)3)]−[22−(−1)223]\Rightarrow \left[ {\dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{{ - 1}}{2} + \dfrac{{ - 1}}{2}\sqrt {4 - {{\left( { - 1} \right)}^2}} - \dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{{ - 2}}{2} + \dfrac{{ - 2}}{2}\sqrt {4 - {{\left( { - 2} \right)}^2}} } \right] + \left[ {\dfrac{{2 \times 2}}{{\sqrt 3 }} - \left( {\dfrac{{2\left( { - 1} \right)}}{{\sqrt 3 }}} \right)} \right] - \left[ {\dfrac{{{2^2} - {{\left( { - 1} \right)}^2}}}{{2\sqrt 3 }}} \right]
⇒[−2sin⁡−112−124−1+2sin⁡−11−4−4]+[43+23]−[4−123]\Rightarrow \left[ { - 2{{\sin }^{ - 1}}\dfrac{1}{2} - \dfrac{1}{2}\sqrt {4 - 1} + 2{{\sin }^{ - 1}}1 - \sqrt {4 - 4} } \right] + \left[ {\dfrac{4}{{\sqrt 3 }} + \dfrac{2}{{\sqrt 3 }}} \right] - \left[ {\dfrac{{4 - 1}}{{2\sqrt 3 }}} \right]
⇒[−2π6−32+2π2]+[63]−[323]\Rightarrow \left[ { - \dfrac{{2\pi }}{6} - \dfrac{{\sqrt 3 }}{2} + \dfrac{{2\pi }}{2}} \right] + \left[ {\dfrac{6}{{\sqrt 3 }}} \right] - \left[ {\dfrac{3}{{2\sqrt 3 }}} \right]
⇒[π−π3−32]+[23]−[32]\Rightarrow \left[ {\pi - \dfrac{\pi }{3} - \dfrac{{\sqrt 3 }}{2}} \right] + \left[ {2\sqrt 3 } \right] - \left[ {\dfrac{{\sqrt 3 }}{2}} \right]
⇒2π3+3\Rightarrow \dfrac{{2\pi }}{3} + \sqrt 3
Hence, the required area is 2π3+3\dfrac{{2\pi }}{3} + \sqrt 3 sq. units.

Note: We can also do this question by integrating the y coordinate with respect to x. In this method we will change the limits of x-coordinates with y-coordinates and dx with dy and y with x and find the area between the curve and the y axis.