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Vector Algebra

Question
CBSEENMA12032710

Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x, and the circle x2 + y2 = 32.  

Solution

The equation of circle is
                      straight x squared plus straight y squared space equals space 32                   ...(1)
The equation of line is
                      y = x                                  ...(2)
From (1) and (2), we get
                     straight x squared plus straight x squared space equals space 32 space space or space space 2 space straight x squared space equals space 16
therefore space space space straight x squared space equals space 16 space space rightwards double arrow space space space space straight x space equals space 4 space space rightwards double arrow space space space straight y space equals space 4
therefore     circle (1) and line (2) meet in P(4, 4)
Radius OA of circle  = square root of 32 space equals space square root of 16 cross times 2 end root space equals space 4 square root of 2
therefore space space space straight A space is space left parenthesis 4 square root of 2 comma space 0 right parenthesis
Required area = Area OAP
                       = Area of increment OMP + area MAP
                       equals space straight A subscript 1 plus straight A subscript 2                                   ...(1)
where straight A subscript 1 space equals space integral subscript 0 superscript 4 space straight y space dx space equals space integral subscript 0 superscript 4 straight x space dx space equals space open vertical bar straight x squared over 0 close vertical bar subscript 0 superscript 4 space equals space 1 half open square brackets 16 minus 0 close square brackets space equals space 8
      straight A subscript 2 space equals space integral subscript 0 superscript 4 square root of 2 end superscript straight y space dx space equals space integral subscript 0 superscript 4 square root of 2 end superscript square root of 32 minus straight x squared end root space dx space equals space integral subscript 4 superscript 4 square root of 2 end superscript square root of left parenthesis square root of 32 right parenthesis squared minus straight x squared end root space dx
             equals space open square brackets fraction numerator straight x square root of 32 minus straight x squared end root over denominator 2 end fraction plus fraction numerator left parenthesis square root of 32 right parenthesis squared over denominator 2 end fraction sin to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator square root of 32 end fraction close parentheses close square brackets subscript 4 superscript 4 square root of 2 end superscript
equals space open square brackets fraction numerator 4 square root of 2 square root of 32 minus 32 end root over denominator 2 end fraction plus 32 over 2 sin to the power of negative 1 end exponent open parentheses fraction numerator 4 square root of 2 over denominator 4 square root of 2 end fraction close parentheses close square brackets minus open square brackets fraction numerator 4 square root of 32 minus 16 end root over denominator 2 end fraction plus 32 over 2 sin to the power of negative 2 end exponent open parentheses fraction numerator 4 over denominator 4 square root of 2 end fraction close parentheses close square brackets
space equals space left square bracket 0 plus 16 space sin to the power of negative 1 end exponent left parenthesis 1 right parenthesis right square bracket space minus space open parentheses fraction numerator 4 cross times 4 over denominator 2 end fraction plus 16 space sin to the power of negative 1 end exponent fraction numerator 1 over denominator square root of 2 end fraction close parentheses
equals space 16 space cross times space straight pi over 2 minus 8 minus 16 cross times straight pi over 4 space equals space 8 straight pi minus 8 minus 4 straight pi space equals 4 straight pi minus 8
therefore space space from space left parenthesis 1 right parenthesis comma space we space get
space space space space space space required space area space equals space 8 space plus space 4 straight pi minus 8 space equals space 4 straight pi space sq. space units.

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