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Vector Algebra

Question
CBSEENMA12032762

Calculate the area of the region bounded by the y = x2 and x = y2.

Solution
The equations of parabolas are
               straight y squared space equals space straight x                                ...(1)
and       straight x squared space equals space straight y                                  ...(2)
From (2), straight y equals straight x squared                               ....(3)
Putting this values of y in (1), we get,
                         straight x to the power of 4 space equals space straight x space space space or space space space straight x left parenthesis straight x cubed minus 1 right parenthesis space equals space 0
rightwards double arrow space space space straight x space equals space 0 comma space 1
therefore space space space space from space left parenthesis 3 right parenthesis comma space space straight y space equals space 0 comma space space space 1
therefore space space space space parabolas space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis space intersect space in space straight O left parenthesis 0 comma space 0 right parenthesis comma space straight P left parenthesis 1 comma space 1 right parenthesis.

From P draw PM perpendicular x-axis
Required area = Area of region OAPB = Area of region OBPM - area of region OAPM
                       equals integral subscript 0 superscript 1 square root of straight x space dx space minus space integral subscript 0 superscript 1 straight x squared dx
                     equals space open square brackets fraction numerator straight x to the power of 3 divided by 2 end exponent over denominator begin display style 3 over 2 end style end fraction close square brackets subscript 0 superscript 1 space minus space open square brackets straight x cubed over 3 close square brackets subscript 0 superscript 1 space equals space 2 over 3 open square brackets straight x to the power of 3 divided by 2 end exponent close square brackets subscript 0 superscript 1 space minus space 1 third open square brackets straight x cubed close square brackets subscript 0 superscript 1
equals space 2 over 3 left square bracket 1 minus 0 right square bracket space minus 1 third left square bracket 1 minus 0 right square bracket space equals space 2 over 3 minus 1 third space equals space 1 third space sq. space units.
 

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