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anonymous
 2 years ago
Little help with line integral
anonymous
 2 years ago
Little help with line integral

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anonymous
 2 years 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?

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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I did it already. thats how I get 2

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

anonymous
 2 years 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)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0do we get dqdp = y+3

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0and then integral is: dw:1370887152616:dw

anonymous
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1y = 0 to y = x+1 x = 0 to x = 1

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

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

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

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

anonymous
 2 years 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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1370889421386:dw

anonymous
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1370889905032:dw
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