Eat green vegetables. They are healthy

Calculus Level 4

c ( cos ( x ) sin ( y ) x y ) d x + ( sin ( x ) cos ( y ) ) d y \oint _{ c }^{ }{ \left ( \cos (x) \sin(y)-xy \right ) \ dx+\left ( \sin(x)\cos(y) \right ) } \ dy

Evaluate the integral above where c c denotes the circle x 2 + y 2 = 1 { x }^{ 2 }{ +y }^{ 2 }=1


The answer is 0.

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3 solutions

Mainak Pal
Mar 30, 2015

As I said eat green vegetables. Oh and also use Green's Theorem in plane. c M d x + N d y = R [ N / x M / y ] d x d y \oint _{ c }^{ }{ Mdx+Ndy=\iint _{ R }^{ }{ [∂N/∂x- } { ∂M/∂y]dxdy } }

Where R is the region bounded by the closed curve "c" and M, N are functions of x and y Hence, for given problem, N / x M / y = x ∂N/∂x-∂M/∂y=x So given integral becomes, R x d x d y = r = 0 1 θ = 0 2 π c o s θ r d r d θ = 0 \iint _{ R }^{ }{ xdxdy=\int _{ r=0 }^{ 1 }{ \int _{ \theta =0 }^{ 2π }{ cos\theta rdrd\theta =0 } } }

Puneet Mehra
Apr 14, 2015

When we calculate...then u will find that there is not any kind of roll of this function inside the circle .....so then apply gauss law that net flux should be zero....

Priyesh Pandey
Mar 30, 2015

Replace x with cos(theta) and y for sin(theta) now integral over whole circle's point means theta changing from 0 to 2pi now substituting these values to obtain limits of x,y which corresponds to 0. now, since both upper limit and lower limit coincides therefore integral resolves to 0.

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