Find the work done by F=(2xy^3 + (4x^y^2)j in moving a particle once counterclockwise around the curve C: the boundary of the "triangular" region in the first quadrant enclosed by the x-axis, the line x=1, and the curve y=x^3.
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∫c F · dr
= ∫c [0 dx + (2xy^3 + 4xy^2) dy]
= ∫∫ [(∂/∂x)(2xy^3 + 4xy^2) - 0] dA, by Green's Theorem
= ∫(x = 0 to 1) ∫(y = 0 to x^3) (2y^3 + 4y^2) dy dx
= ∫(x = 0 to 1) (y^4/2 + 4y^3/3) {for y = 0 to x^3} dx
= ∫(x = 0 to 1) (1/6) (3x^12 + 8x^9) dx
= (1/6) (3x^13/13 + 4x^10/5) {for x = 0 to 1}
= 67/390.
I hope this helps!
dontas
artwork is defined because of the fact the sure quintessential of F(rigidity) w.r.t. s(displacement), and remember that F and s are vectors. artwork completed is a scalar volume, so in simple terms calculate separately for displacement alongside the x-axis, the line x = a million 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 locate the sure quintessential of F w.r.t. s with x commencing from 1to 0( draw graph to comprehend ). further calculate artwork(s) in relax of the two circumstances( do not overlook approximately indications ) and in simple terms upload all of them up. you will get a destructive answer