Answer to Question #199243 in Calculus for Sarita bartwal

Question

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.

Answer 1

The above equation is -

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

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

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 ,

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

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

$M=x^{2}+3y$

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

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

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

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

$=2\iint_{R}dxdy$

$where$ $\iint_{R}dxdy$ $=area\ of\ ellipse$

$=2($ area of ellipse)

$=$ area of given ellipse =${\pi}ab$

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

which is the twice the area enclosed by C.