anonymous
  • anonymous
Little help with line integral
Calculus1
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
Using Green's theorem I got 2 as result, but Wolfram says it is 5/3. Can somebody check on this?
anonymous
  • anonymous

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anonymous
  • anonymous
express y as function of x and substitute into the integral: y=1-x
anonymous
  • anonymous
I did it already. thats how I get 2
amistre64
  • amistre64
can you walk me thru your application of greens thrm
anonymous
  • anonymous
partial derivatives are: dp/dy=-3 and dq/dx=y, (i know that d should be written different)
anonymous
  • anonymous
do we get dq-dp = y+3
anonymous
  • anonymous
and then integral is: |dw:1370887152616:dw|
anonymous
  • anonymous
boundaries for X goes from 0 to 1 and for Y they are 0 and 1-x, am I right?
amistre64
  • amistre64
y = 0 to y = -x+1 x = 0 to x = 1
amistre64
  • amistre64
so yes
amistre64
  • amistre64
well, dxdy or dydx?
amistre64
  • amistre64
dydx is intx[0,1] inty[0,1-x]
amistre64
  • amistre64
(1-x)^2 / 2 from 0 to 1
amistre64
  • amistre64
i forgot to yup the 3 :/
amistre64
  • amistre64
\[\int_{0}^{1}\frac{(1-x)^2}{2}+3(1-x)~dx\]
anonymous
  • anonymous
ok, i've found a mistake in my calculation, i'm doing it now directly by parametrization of curves, will see the result
anonymous
  • anonymous
|dw:1370889421386:dw|
anonymous
  • anonymous
for C1: x=t, y=0, 0<=t<=1 and Integral is = 1 C2: x=1-t, y=t, 0<=t<=1, int=-1/3 c3: x=0, y=1-t, 0<=t<=1, int=0
anonymous
  • anonymous
hm?
experimentX
  • experimentX
about surface integrals, \[ \iint dx dy \] shoudl you give area|dw: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
  • amistre64
\[\int_{0}^{1}\int_{0}^{1-y}y+3~dx~dy\] \[\int_{0}^{1}y(1-y)+3(1-y)~dy\] \[\int_{0}^{1}-y^2-2y+3~dy\] \[-\frac{1}{3}-1+3\]
experimentX
  • experimentX
|dw:1370889905032:dw|

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