Find the work done by F=(2xy^3)i+(4x^2y^2)j in moving a particle once counterclockwise around the curve C: the?

∇ x F = ∂/∂x [4x²y²] - ∂/∂y [2xy³]

= 2xy² k

∮c F·dr = ∫∫ ∇ x F · n dS ............. Stokes' or Green's in the x-y plane

... n dS = <0,0,1> dx dy for positively oriented curves the x-y plane

= ∫∫ <0,0,(2xy²)>·<0,0,1> dx dy

= ∫∫ 2xy² dx dy

{(x,y) | 0 ≤ x ≤ 1 , 0 ≤ y ≤ x³}

= 2/3 ∫ x^10 dx

{(x) | 0 ≤ x ≤ 1}

= 2/33

Answer: 2/33

Work is defined as the definite integral of F(Force) w.r.t. s(displacement), and remember that F and s are vectors. Work done is a scalar quantity, so just calculate separately for displacement along the x-axis, the line x = 1 and the curve y = x^3. e.g., for y = x^3, s = (x)i + (y)j = (x)i + (x^3)j. So, ds = { i + 3(x^2)j }dx. And F = (2(x^4))i + 4(x^4)j. Now find the definite integral of F w.r.t. s with x ranging from 1to 0( draw graph to understand ). Similarly calculate Work(s) in rest of the 2 cases( don't ignore signs ) and just add all of them up. You will get a negative answer

counterclockwise. yet as quickly as I write my calling enjoying cards @ paintings, I purely began to jot down them clockwise by way of fact it leaves me a tiny little bit of greater area to end w/the date. VERY exciting question.