Ed25
Little help with line integral
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Ed25
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Ed25
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Using Green's theorem I got 2 as result, but Wolfram says it is 5/3. Can somebody check on this?
Ed25
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@experimentX
myko
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express y as function of x and substitute into the integral:
y=1-x
Ed25
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I did it already. thats how I get 2
amistre64
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can you walk me thru your application of greens thrm
Ed25
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partial derivatives are: dp/dy=-3 and dq/dx=y, (i know that d should be written different)
Ed25
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do we get dq-dp = y+3
Ed25
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and then integral is: |dw:1370887152616:dw|
Ed25
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boundaries for X goes from 0 to 1 and for Y they are 0 and 1-x, am I right?
amistre64
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y = 0 to y = -x+1
x = 0 to x = 1
amistre64
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so yes
amistre64
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well, dxdy or dydx?
amistre64
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dydx is intx[0,1] inty[0,1-x]
amistre64
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(1-x)^2 / 2 from 0 to 1
amistre64
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i forgot to yup the 3 :/
amistre64
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\[\int_{0}^{1}\frac{(1-x)^2}{2}+3(1-x)~dx\]
Ed25
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ok, i've found a mistake in my calculation, i'm doing it now directly by parametrization of curves, will see the result
Ed25
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|dw:1370889421386:dw|
Ed25
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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
Ed25
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hm?
experimentX
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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
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\[\int_{0}^{1}\int_{0}^{1-y}y+3~dx~dy\]
\[\int_{0}^{1}y(1-y)+3(1-y)~dy\]
\[\int_{0}^{1}-y^2-2y+3~dy\]
\[-\frac{1}{3}-1+3\]
experimentX
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|dw:1370889905032:dw|