Answer to Question #199243 in Calculus for Sarita bartwal

Question #199243

Show that the value of the integral

∫ [(x^2+3y)dx + (5x-3y^2)dy]

C

where C is the ellipse x^2/a^2+y^2/b^2=1, is twice the area enclosed by C.


1
Expert's answer
2021-06-01T06:13:37-0400

The above equation is -


c[(x2+3y)dx+(5x3y2)dy]\int_{c}[(x^{2}+3y)dx+(5x-3y^{2})dy] ,


where C is the ellipse x2a2+y2b2=1\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1


Now comparing this equation with green's theorem , we can only solve this question by green's theorem.



Now statement of green's theorem is that ,


c(Mdx+Ndy)=R(NxMy)dxdy\oint_{c}(Mdx+Ndy)=\iint_{R}(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y})dxdy ...............A)



now this is the green's theorem now we will compare the given equation with the given equation , we will get -


M=x2+3yM=x^{2}+3y


N=5x3y2N=5x-3y^{2}


My\dfrac{\partial M}{\partial y} =3=3


Nx=5\dfrac{\partial N}{\partial x}=5



Now putting the value in equation A) , we get our right hand side as -


=2Rdxdy=2\iint_{R}dxdy


wherewhere Rdxdy\iint_{R}dxdy =area of ellipse=area\ of\ ellipse


=2(=2( area of ellipse)


== area of given ellipse =πab{\pi}ab


== so , our required answer is =2πab=2{\pi}ab



which is the twice the area enclosed by C.


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