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Vector Algebra

Question
CBSEENMA12032774

Find the area of the region bounded by the circle x2 + y2 = 1 and x + y = 1. Also draw a rough sketch.

Solution

Consider the equations
                straight x squared plus straight y squared space equals space 1                               ...(1)
and         straight x plus straight y space equals space 1                                    ...(2)
From (2), y = 1 - x                                       ...(3)
Putting this value of y in (1), we get, 
               straight x squared plus left parenthesis 1 minus straight x right parenthesis squared space equals space 1 space space space space space space space rightwards double arrow space space space 2 space straight x squared minus 2 straight x space equals space 0
rightwards double arrow space space space space space straight x squared minus straight x space equals space 0 space space space space space space space rightwards double arrow space space space straight x left parenthesis straight x minus 1 right parenthesis space equals space 0 space space space space rightwards double arrow space space straight x space equals space 0 comma space space 1
therefore space space space from space left parenthesis 3 right parenthesis comma space space straight y space equals space 1 minus 0 comma space space space 1 minus 1 space equals space 1 comma space 0

therefore circle (1) and st . line x + y = 1 intersect in the points A(1, 0) and B(0, 1).
Required area = Area of shaded region = integral subscript 0 superscript 1 square root of 1 minus straight x squared end root dx space minus integral subscript 0 superscript 1 left parenthesis 1 minus straight x right parenthesis space dx
                                      equals space open square brackets fraction numerator straight x square root of 1 minus straight x squared end root over denominator 2 end fraction plus 1 half sin to the power of negative 1 end exponent straight x close square brackets subscript 0 superscript 1 space minus space open square brackets straight x minus straight x squared over 2 close square brackets subscript 0 superscript 1
equals space open square brackets open parentheses 0 plus 1 half sin to the power of negative 1 end exponent 1 close parentheses minus open parentheses 0 plus 1 half sin to the power of negative 1 end exponent 0 close parentheses close square brackets space minus space open square brackets open parentheses 1 minus 1 half close parentheses minus left parenthesis 0 minus 0 right parenthesis close square brackets
equals space 1 half open parentheses straight pi over 2 close parentheses minus 1 half space equals open parentheses straight pi over 4 minus 1 half close parentheses space sq. space units. space
           

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