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Vector Algebra

Question
CBSEENMA12032709

Find the area of the region in the first quadrant enclosed by the x-axis, the line straight x equals square root of 3 space straight y and the circle x2 + y2 = 4.

Solution

The equation of circle is
                   straight x squared plus straight y squared space equals space 4                      ...(1)
The equation of line is
                      straight x space equals square root of 3 space straight y                       ...(2)
From (1) and (2), we get,
                3 straight y squared plus straight y squared space equals space 4 space space space space or space space space space 4 straight y squared space equals space 4
therefore space space space space space space space space straight y squared space equals space 1
rightwards double arrow space space space space space space straight y space equals space 1 space space space space space space space space space rightwards double arrow space space space space space straight x space equals space square root of 3

∴     point of intersection of circle (1) and line (2) is P left parenthesis square root of 3 comma space 1 right parenthesis.
From P, draw PM ⊥ x-axis.
Also OA = radius of circle = 2
∴ A is (2, 0)
Required area = Area OMAPO = Area of ∆OMP + area MAP
= A1 + A2    ..(1 )
where straight A subscript 1 space equals space area space of space increment OMP space equals space fraction numerator 1 over denominator square root of 3 end fraction integral subscript 0 superscript square root of 3 end superscript space straight x space dx

                equals space fraction numerator 1 over denominator square root of 3 end fraction open square brackets straight x squared over 2 close square brackets subscript 0 superscript square root of 3 end superscript space equals space left parenthesis 1 right parenthesis space fraction numerator 1 over denominator 2 square root of 3 end fraction left parenthesis 3 minus 0 right parenthesis space equals space fraction numerator square root of 3 over denominator 2 end fraction
and straight A subscript 2 space equals space integral subscript square root of 3 end subscript superscript 2 space straight y space dx space equals space integral subscript square root of 3 end subscript superscript 2 square root of 4 minus straight x squared end root space dx space equals space integral subscript square root of 3 end subscript superscript 2 square root of left parenthesis 2 right parenthesis squared minus straight x squared end root space dx
             equals space open square brackets fraction numerator straight x square root of 4 minus straight x squared end root over denominator 2 end fraction plus fraction numerator left parenthesis 2 right parenthesis squared over denominator 2 end fraction sin to the power of negative 1 end exponent open parentheses straight x over 2 close parentheses close square brackets subscript square root of 3 end subscript superscript 2
equals space open parentheses fraction numerator 2 square root of 4 minus 4 end root over denominator 2 end fraction plus 2 sin to the power of negative 1 end exponent 1 close parentheses space minus space open parentheses fraction numerator square root of 3 space square root of 4 minus 3 end root over denominator 2 end fraction plus 2 space sin to the power of negative 1 end exponent fraction numerator square root of 3 over denominator 2 end fraction close parentheses
equals space open parentheses 0 plus 2 space cross times straight pi over 2 close parentheses space minus space open parentheses fraction numerator square root of 3 over denominator 2 end fraction plus 2 cross times straight pi over 3 close parentheses space equals space straight pi minus fraction numerator square root of 3 over denominator 2 end fraction minus fraction numerator 2 straight pi over denominator 3 end fraction space equals straight pi over 3 minus fraction numerator square root of 3 over denominator 2 end fraction
therefore space space from space left parenthesis 1 right parenthesis comma space required space area space equals space fraction numerator square root of 3 over denominator 2 end fraction plus straight pi over 3 minus fraction numerator square root of 3 over denominator 2 end fraction equals space straight pi over 3 space sq. space units
              

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