Advertisement

Find the area of the region bounded by the ellipse straight x squared over 9 plus straight y squared over 4 space equals space 1.


The equation of ellipse is
                       straight x squared over 9 plus straight y squared over 4 space equals space 1

or                   straight y squared over 4 space equals space 1 minus straight x squared over 9

or              straight y squared space equals space 4 over 9 left parenthesis 9 minus straight x squared right parenthesis     

therefore space space space space space straight y space equals space 2 over 3 square root of 9 minus straight x squared end root                                                       [In the first quadrant]

The ellipse is symmetrical about both the axes
∴ required area = 4 (area AOB)
equals space 4 integral subscript 0 superscript 3 space straight y space dx space equals space 4 cross times 2 over 3 integral subscript 0 superscript 3 square root of 9 minus straight x squared end root space dx space equals space 8 over 3 integral subscript 0 superscript 3 square root of left parenthesis 3 right parenthesis squared minus straight x squared end root dx
equals space 8 over 3 open square brackets fraction numerator straight x square root of left parenthesis 3 right parenthesis squared minus straight x squared end root over denominator 2 end fraction plus open parentheses 3 close parentheses squared over 2 sin to the power of negative 1 end exponent open parentheses straight x over 3 close parentheses close square brackets subscript 0 superscript 3
equals space 8 over 3 open square brackets open curly brackets fraction numerator 3 square root of 9 minus 9 end root over denominator 2 end fraction plus 9 over 2 sin to the power of negative 1 end exponent left parenthesis 1 right parenthesis close curly brackets space minus space open curly brackets 0 plus 9 over 2 sin to the power of negative 1 end exponent left parenthesis 0 right parenthesis close curly brackets close square brackets
equals space 8 over 3 open square brackets open curly brackets 0 plus 9 over 2 cross times straight pi over 2 close curly brackets minus open curly brackets 0 plus 9 over 2 cross times 0 close curly brackets close square brackets space equals space 8 over 3 cross times 9 over 2 cross times straight pi over 2 space equals space 6 space straight pi space sq. space units

136 Views

Advertisement

Application of Integrals

Hope you found this question and answer to be good. Find many more questions on Application of Integrals with answers for your assignments and practice.

Mathematics Part II

Browse through more topics from Mathematics Part II for questions and snapshot.