Recent content by Rhduke

yes u can just integrate the easier way, but the point is to use greens theorem, oops its y = 2x^3 the question wants me to use this equation to solve it: ok ill use I(a,b) as the integral I(Greens) (-y/2 dx + x/2 dy) = I I (1/2 + 1/2) dA = A(s) i guess my real question is how do u...

Homework Statement Find area of indicated region: S is bounded by the curves y = 4x and y = 2x^3 actual answer: 8/3 note: }a:b{ is what i mean by the definite integral Homework Equations A(s) = 1/2 }Green's Int{ (-ydx + xdy) = 1/2 }{ }{ (dN/dx - dM/dy)dxdy The Attempt at a...

er oh... i just realized i did that wrong, strangely it worked, ok i fixed it: 1/(e^xy + xy) (e^xy) (xy' + y) + (xy' + y) = 0 (xy' + y)[(e^xy)/(e^xy + xy) + 1] = 0 xy' + y = 0 y' = -(y/x) my brain gets big picture but misses the small details :cry:

Oh i keep forgeting u can't distribute ln :rolleyes: ok i c what your saying so then i get: ln(e^xy(x)(y)) = ln2 1/[e^xy(x)(y)] (e^xy) (xy' + y) + e^xy (xy' + y) = 0 (xy' + y)(1/xy + e^xy) = 0 xy' + y = 0 y' = -(y/x) thank u all for your help!

Homework Statement e^(xy) + xy = 2, find y' the actual answer is -(y/x) The Attempt at a Solution xy + lnxy = ln2 xy' + y + 1/x + y'/y = 1/2 xy' + 1/y (y') = 1/2 - y -1/x y'(x + 1/y) = 1/2 - y - 1/x y' = (1/2 - y - 1/x)/(x+1/y) i simplified it further but could not get the...

Homework Statement Well i got the formula down, I am having trouble with the integration though: Q: {(x^3 + y) ds , x = 3t, y = t^3, 0<=t<=1 = {[ (27t^3 + t^3) sqrt(9t^2 + t^6)] dt Homework Equations f(x,y)ds = f(x(t), y(t)) sqrt[ (x'(t))^2 + (y'(t))^2 ] The Attempt at a...