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Ed25
 one year ago
Best ResponseYou've already chosen the best response.0Using Green's theorem I got 2 as result, but Wolfram says it is 5/3. Can somebody check on this?

myko
 one year ago
Best ResponseYou've already chosen the best response.0express y as function of x and substitute into the integral: y=1x

Ed25
 one year ago
Best ResponseYou've already chosen the best response.0I did it already. thats how I get 2

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1can you walk me thru your application of greens thrm

Ed25
 one year ago
Best ResponseYou've already chosen the best response.0partial derivatives are: dp/dy=3 and dq/dx=y, (i know that d should be written different)

Ed25
 one year ago
Best ResponseYou've already chosen the best response.0and then integral is: dw:1370887152616:dw

Ed25
 one year ago
Best ResponseYou've already chosen the best response.0boundaries for X goes from 0 to 1 and for Y they are 0 and 1x, am I right?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1y = 0 to y = x+1 x = 0 to x = 1

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dydx is intx[0,1] inty[0,1x]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1(1x)^2 / 2 from 0 to 1

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i forgot to yup the 3 :/

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\int_{0}^{1}\frac{(1x)^2}{2}+3(1x)~dx\]

Ed25
 one year ago
Best ResponseYou've already chosen the best response.0ok, i've found a mistake in my calculation, i'm doing it now directly by parametrization of curves, will see the result

Ed25
 one year ago
Best ResponseYou've already chosen the best response.0for C1: x=t, y=0, 0<=t<=1 and Integral is = 1 C2: x=1t, y=t, 0<=t<=1, int=1/3 c3: x=0, y=1t, 0<=t<=1, int=0

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1about surface integrals, \[ \iint dx dy \] shoudl you give areadw:1370889753653:dw choose any way you like to integrate ... if this gives you your required area then your parametrization if correct. just put the function inside and integrate it.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\int_{0}^{1}\int_{0}^{1y}y+3~dx~dy\] \[\int_{0}^{1}y(1y)+3(1y)~dy\] \[\int_{0}^{1}y^22y+3~dy\] \[\frac{1}{3}1+3\]

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1dw:1370889905032:dw
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