Sponsor Area

Vector Algebra

Question
CBSEENMA12032705

Using definite integrals, find the area of the ellipse straight x squared over 4 plus straight y squared over 9 space equals space 1

Solution

The equation of the ellipse is 
                          straight x squared over 4 plus straight y squared over 9 space equals space 1
or                 straight y squared over 9 space equals space 1 minus straight x squared over 4
or              straight y squared space equals space 9 over 4 left parenthesis 4 minus straight x squared right parenthesis
therefore space space space space straight y space equals space 3 over 2 square root of 4 minus straight x squared end root                         ...(1)
(in the first quadrant)
The ellipse is symmetrical about both the axes,
∴ required area = 4 (area AOB)
equals space 4 integral subscript 0 superscript 2 straight y space dx space equals space 4 space integral subscript 0 superscript 2 3 over 2 square root of 4 minus straight x squared end root dx                                       open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
equals space 6 integral subscript 0 superscript 2 square root of left parenthesis 2 right parenthesis squared minus straight x squared end root space dx space equals space 6 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 0 superscript 2
equals space 6 open square brackets open parentheses fraction numerator 2 square root of 4 minus 4 end root over denominator 2 end fraction plus 2 space sin to the power of negative 1 end exponent 1 close parentheses minus left parenthesis 0 plus 2 space sin to the power of negative 1 end exponent 0 right parenthesis close square brackets space equals space 6 open square brackets open parentheses 0 plus 2 cross times straight pi over 2 close parentheses minus left parenthesis 0 plus 0 right parenthesis close square brackets
equals space 6 straight pi space sq. space units.

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