## Ed25 one year ago Little help with line integral

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1. Ed25

2. Ed25

Using Green's theorem I got 2 as result, but Wolfram says it is 5/3. Can somebody check on this?

3. Ed25

@experimentX

4. myko

express y as function of x and substitute into the integral: y=1-x

5. Ed25

I did it already. thats how I get 2

6. amistre64

can you walk me thru your application of greens thrm

7. Ed25

partial derivatives are: dp/dy=-3 and dq/dx=y, (i know that d should be written different)

8. Ed25

do we get dq-dp = y+3

9. Ed25

and then integral is: |dw:1370887152616:dw|

10. Ed25

boundaries for X goes from 0 to 1 and for Y they are 0 and 1-x, am I right?

11. amistre64

y = 0 to y = -x+1 x = 0 to x = 1

12. amistre64

so yes

13. amistre64

well, dxdy or dydx?

14. amistre64

dydx is intx[0,1] inty[0,1-x]

15. amistre64

(1-x)^2 / 2 from 0 to 1

16. amistre64

i forgot to yup the 3 :/

17. amistre64

$\int_{0}^{1}\frac{(1-x)^2}{2}+3(1-x)~dx$

18. Ed25

ok, i've found a mistake in my calculation, i'm doing it now directly by parametrization of curves, will see the result

19. Ed25

|dw:1370889421386:dw|

20. Ed25

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

21. Ed25

hm?

22. 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.

23. amistre64

$\int_{0}^{1}\int_{0}^{1-y}y+3~dx~dy$ $\int_{0}^{1}y(1-y)+3(1-y)~dy$ $\int_{0}^{1}-y^2-2y+3~dy$ $-\frac{1}{3}-1+3$

24. experimentX

|dw:1370889905032:dw|