IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

vinodchills

FIND THE AREA BOUNDED BY THE ELLIPSE X2/A2 +Y2/B2=1 AND THE ORDINATES X=0 AND X=AE, WHERE B2=A2(1-E2) AND E<1???

raulrag009

Consider the diagram ,the blue line is x=ae

Now

$A=2\int_{0}^{ae}{\frac{b}{a}(\sqrt{a^{2}-x^{2}})}dx$

now u can easily integrate and solve

raulrag009

$Ans \;should \;be\\\\ A=b(be+a^{2}\sin^{-1}{e})$

forgive my calculation mistakes

vinodchills

ANSWER:AB[E^{[ ]}

1-E2 +SININV E] SOLVE IT FULLY RAUL!!PLS

raulrag009

$Here\;i\;go\\\\ As\\\\ A=2\frac{b}{a}\int_{0}^{ae}{\sqrt{a^{2}-x^{2}}}dx\\\\ Apply formulae\\\\ A=2\frac{b}{a}[\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}{\frac{x}{a}}]_{0}^{ae}\\\\ A=2\frac{b}{a}[\frac{ae}{2}\sqrt{a^{2}-a^{2}e^{2}}+\frac{a^{2}}{2}\sin^{-1}{e}\\\\ A=be\sqrt{a^{2}(1-e^{2})}+ab\sin^{-1}{e}\\\\ A=ab[e\sqrt{1-e^{2}}+\sin^{-1}{e}]$

Now u must hav understood

vinodchills

THANKS RAUL GREATJOB!!! WHICH SCHOOL WHERE R U FROM?

dheer_07

ya vinod ur right this yr ncert is coollll!!!!

sum very nice ques r there

r u getting probability if yes plz help me!!!!!!!!

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