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Question

The area (in $s q$ units) of the smaller portion enclosed between the curves, $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 4$ and $y to the power of 2 end exponent equals 3 x$ , is

$\frac{1}{2 \sqrt{3}} + \frac{π}{3}$
$\frac{1}{2 \sqrt{3}} + \frac{2 π}{3}$
$\frac{1}{\sqrt{3}} + \frac{2 π}{3}$
$\frac{1}{\sqrt{3}} + \frac{4 π}{3}$

The correct answer is: $fraction numerator 1 over denominator square root of 3 end fraction plus fraction numerator 4 pi over denominator 3 end fraction$

The given equation $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 4$ is equation of circle of rradius 2 centred at origin and equation $y to the power of 2 end exponent equals 3 x$ is the equation of parabola.
$x to the power of 2 end exponent plus y to the power of 2 end exponent equals 4$ (1)
$y to the power of 2 end exponent equals 3 x$ (2)
$left parenthesis 2 comma blank 1 right parenthesis$
Substituting Eq (2) in Eq (1), we get
$x to the power of 2 end exponent plus 3 x minus 4 equals 0$
$rightwards double arrow x to the power of 2 end exponent plus 4 x minus x minus 4 equals 0$
$rightwards double arrow x left parenthesis x plus 4 right parenthesis minus 1 left parenthesis x plus 4 right parenthesis equals 0$
$rightwards double arrow left parenthesis x minus 1 right parenthesis left parenthesis x plus 4 right parenthesis equals 0$
$rightwards double arrow left parenthesis x minus 1 right parenthesis equals 0$ and $left parenthesis x plus 4 right parenthesis equals 0$
Therefore, $equals 1 comma$ $negative 4$ Considering $x equals 1$ , then from Eq (2), we get $y equals square root of 3 ´ minus square root of 3.$
Thus, $left parenthesis right square bracket comma square root of 3 right parenthesis$ and $left parenthesis 1 comma blank minus square root of 3 right parenthesis$ 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,
$A equals 2 left square bracket not stretchy integral subscript 0 end subscript superscript 1 end superscript y subscript 2 end subscript d x plus not stretchy integral subscript 1 end subscript superscript 2 end superscript y subscript 1 end subscript d x right square bracket$
From Eq (1), we get
$y subscript 1 to the power of equals end exponent end subscript superscript square root of 4 minus x to the power of 2 end exponent end root end superscript$
From Eq (2), we get
$y subscript 2 end subscript equals square root of 3 x end root$
Therefore,
$A equals 2 left square bracket not stretchy integral subscript 0 end subscript superscript 1 end superscript square root of 3 x end root d x plus not stretchy integral subscript 1 end subscript superscript 2 end superscript square root of 4 minus x to the power of 2 end exponent end root d x right square bracket$
$equals 2 left square bracket not stretchy integral subscript 0 end subscript superscript 1 end superscript square root of 3 x to the power of 1 divided by 2 end exponent d x plus not stretchy integral subscript 1 end subscript superscript 2 end superscript square root of 2 to the power of 2 end exponent minus x to the power of 2 end exponent end root d x right square bracket$
Using standard integral, $not stretchy integral subscript blank superscript blank x to the power of n end exponent d x equals fraction numerator x to the power of n minus 1 end exponent over denominator n plus 1 end fraction$ , we have
$not stretchy integral subscript blank superscript blank square root of 0 to the power of 2 end exponent minus x to the power of 2 end exponent end root d x equals fraction numerator x over denominator 2 end fraction square root of 0 to the power of 2 end exponent minus x to the power of 2 end exponent end root plus fraction numerator o to the power of 2 end exponent over denominator 2 end fraction t a n to the power of negative 1 end exponent fraction numerator x over denominator square root of 0 to the power of 2 end exponent minus x to the power of 2 end exponent end root end fraction$
Therefore,
$A equals 2 left square bracket left parenthesis square root of 3 fraction numerator x to the power of 3 divided by 2 end exponent over denominator 3 divided by 2 end fraction right parenthesis vertical line subscript 0 end subscript superscript 1 end superscript plus left parenthesis fraction numerator x over denominator 2 end fraction square root of 4 minus x to the power of 2 end exponent end root plus fraction numerator 4 over denominator 2 end fraction t a n to the power of negative 1 end exponent fraction numerator x over denominator square root of 4 minus x to the power of 2 end exponent end root end fraction 1 vertical line subscript 1 end subscript superscript 2 end superscript right square bracket$
$equals 2 left square bracket square root of 3. fraction numerator 1 over denominator 3 divided by 2 end fraction minus 0 plus fraction numerator 2 over denominator 2 end fraction square root of 4 minus 4 end root plus 2 t a n to the power of negative 1 end exponent fraction numerator 2 over denominator square root of 4 minus 4 end root end fraction minus fraction numerator 1 over denominator 2 end fraction square root of 4 minus 1 end root minus 2 t a n to the power of negative 1 end exponent fraction numerator 1 over denominator square root of 4 minus 1 end root end fraction right square bracket$
$equals 2 left square bracket fraction numerator 2 square root of 3 over denominator 3 end fraction plus 2 t a n to the power of negative 1 end exponent left parenthesis infinity right parenthesis minus fraction numerator square root of 3 over denominator 2 end fraction minus 2 t a n to the power of negative 1 end exponent left parenthesis fraction numerator 1 over denominator square root of 3 end fraction right parenthesis right square bracket$
Now, $t a n blank fraction numerator pi over denominator 2 end fraction equals infinity$ and $t a n blank fraction numerator pi over denominator 6 end fraction equals fraction numerator 1 over denominator square root of 3 end fraction$ Therefore, the area of the smaller portion enclosed between the two curves is obtained as follows:
$A equals 2 left square bracket fraction numerator 2 square root of 3 over denominator 3 end fraction minus fraction numerator square root of 3 over denominator 2 end fraction plus 2 t a n to the power of negative 1 end exponent left parenthesis blank t a n blank fraction numerator pi over denominator 2 end fraction right parenthesis minus 2 t a n to the power of negative 1 end exponent left parenthesis blank t a n blank fraction numerator pi over denominator 6 end fraction right parenthesis right square bracket$
$equals 2 left square bracket square root of 3 fraction numerator 1 over denominator 6 end fraction plus 2 fraction numerator pi over denominator 2 end fraction minus 2 fraction numerator pi over denominator 6 end fraction right square bracket$
$equals 2 left square bracket square root of 3 fraction numerator 1 over denominator 6 end fraction plus 2 fraction numerator 2 pi over denominator 6 end fraction right square bracket equals 2 left square bracket fraction numerator square root of 3 over denominator 6 end fraction plus fraction numerator 4 pi over denominator 6 end fraction right square bracket equals fraction numerator square root of 3 over denominator 3 end fraction plus fraction numerator 4 pi over denominator 3 end fraction$
$equals left parenthesis fraction numerator 1 over denominator square root of 3 end fraction plus fraction numerator 4 pi over denominator 3 end fraction right parenthesis s q.$ unit
Hence, the correct answer is option (D).

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