Question 196 of 230HardMCQ
The area (in sq. units) of the smaller portion enclosed between the curves, x2 + y2 = 4 and y2 = 3x, is :
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Official solution
The given equation x2+y2=4 is equation of circle of radius 2 centred at origin and equation y2=3x is the equation of parabola.
x2+y2=4 ..... (1)
y2=3x ..... (2)
Substituting Eq. (2) in Eq. (1), we get
x2+3x−4=0
\[ \Rightarrow {x^2} + 4x - x - 4 = 0\]
\[ \Rightarrow x(x + 4) - 1(x + 4) = 0\]
\[ \Rightarrow (x - 1)(x + 4) = 0\]
\[ \Rightarrow (x - 1) = 0\] and (x+4)=0
Therefore, x = 1, -4. Considering x = 1, then from Eq. (2), we get \[y = \sqrt 3 , - \sqrt 3 \].
Therefore \[(1,\sqrt 3 )\] and \[(1, - \sqrt 3 )\] are the points of intersection of parabola and circle.
The required area (A) is the area of the shaded region shown in the figure. Therefore,
\[A = 2\left[ {\int\limits_0^1 {{y_2}dx + \int\limits_1^2 {{y_1}dx} } } \right]\]
From Eq. (1), we get \[{y_1} = \sqrt {4 - {x^2}} \]
From Eq. (2), we get \[{y_2} = \sqrt {3x} \]
Therefore, \[A = 2\left[ {\int\limits_0^1 {\sqrt {3x} dx + \int\limits_1^2 {\sqrt {4 - {x^2}} dx} } } \right]\]
\[ = 2\left[ {\int\limits_0^1 {\sqrt 3 {x^{1/2}}dx + \int\limits_1^2 {\sqrt {{2^2} - {x^2}} dx} } } \right]\]
Using standard integral : \[\int {{x^n}dx = {{{x^{n - 1}}} \over {n + 1}}} \], we have
\[\int {\sqrt {{a^2} - {x^2}} dx = {x \over 2}\sqrt {{a^2} - {x^2}} + {{{a^2}} \over 2}{{\tan }^{ - 1}}{x \over {\sqrt {{a^2} - {x^2}} }}} \]
Therefore,
\[A = 2\left[ {\left. {\left( {\sqrt 3 {{{x^{3/2}}} \over {3/2}}} \right)} \right|_0^1 + \left. {\left( {{x \over 2}\sqrt {4 - {x^2}} + {4 \over 2}{{\tan }^{ - 1}}{x \over {\sqrt {4 - {x^2}} }}} \right)} \right|_1^2} \right]\]
\[ = 2\left[ {\sqrt 3 .{1 \over {3/2}} - 0 + {2 \over 2}\sqrt {4 - 4} + 2{{\tan }^{ - 1}}{2 \over {\sqrt {4 - 4} }} - {1 \over 2}\sqrt {4 - 1} - 2{{\tan }^{ - 1}}{1 \over {\sqrt {4 - 1} }}} \right]\]
\[ = 2\left[ {{{2\sqrt 3 } \over 3} + {{\tan }^{ - 1}}(\infty ) - {{\sqrt 3 } \over 2} - 2{{\tan }^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)} \right]\]
Now, \[\tan {\pi \over 2} = \infty \] and \[\tan {\pi \over 6} = {1 \over {\sqrt 3 }}\]. Therefore, the area of the smaller portion enclosed between the two curves is obtained as follows:
\[A = 2\left[ {{{2\sqrt 3 } \over 3} - {{\sqrt 3 } \over 2} + 2{{\tan }^{ - 1}}\left( {\tan {\pi \over 2}} \right) - 2{{\tan }^{ - 1}}\left( {\tan {\pi \over 6}} \right)} \right]\]
\[ = 2\left[ {\sqrt 3 {1 \over 6} + 2{\pi \over 2} - 2{\pi \over 6}} \right]\]
\[ = 2\left[ {\sqrt 3 + {1 \over 6} + 2{{2\pi } \over 6}} \right] = 2\left[ {{{\sqrt 3 } \over 6} + {{4\pi } \over 6}} \right]\]
\[ = {{\sqrt 3 } \over 3} + {{4\pi } \over 3} = \left( {{1 \over {\sqrt 3 }} + {{4\pi } \over 3}} \right)\] sq. units
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