- Luigi0210

Stoke's Theorem:

- chestercat

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- Luigi0210

Use Stoke's theorem to solve \(\large \int_{C} F~ dr \) where F(x,y,z) = and C is the boundary of part of plane 3x+y+z=3 in the first octant.

- Astrophysics

I had trouble understanding Stoke's and I've always wanted to go over it again and learn it a better way...mhm lets see if we can figure this out \[\int\limits_{C} \vec F \cdot d \vec r = \int\limits \int\limits_S curl \vec F \cdot d \vec S\] right

- Luigi0210

Not sure, I'm a bit confused because of these:
http://prntscr.com/7t4ipd

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## More answers

- Luigi0210

Same thing?

- dan815

yes all same thing

- Astrophysics

Yes, you should understand the notation and what it means then I think it would be more clear hmm. I'm thinking we can pick any surface S with the boundary C.

- Astrophysics

This would be a nice question to observe haha, I sort of have an idea, but I think @dan815 @ganeshie8 @Empty would be better in teaching this, plus I can learn from them to xD.

- dan815

|dw:1436995421444:dw|

- Astrophysics

Projection?!

- dan815

we gotta see those intersections with xz yz and xy plane

- dan815

we can see the 3 intersections with the 3 axis by setting x,y=0 y,z=0 x,z=0

- dan815

|dw:1436996249413:dw|

- dan815

thre u go not integrate over that surface area

- dan815

|dw:1436996425783:dw|

- dan815

F is a function of x,y
so we want to refine our new surface area in terms of x and y too

- dan815

ds=|F_x X Fy| dxdy

- dan815

i dont remember that exactly, u can work it out though

- dan815

|dw:1436996594088:dw|

- dan815

there fore ds= |fx X fy| dxdy

- dan815

your domain will be that triangle

- dan815

|dw:1436996772371:dw|

- dan815

|dw:1436996788233:dw|

- dan815

|dw:1436996930190:dw|

- dan815

finally note how Gx and Gy vectors are formed

- dan815

|dw:1436997056399:dw|

- dan815

|dw:1436997128852:dw|

- dan815

and finally u can look up a proof for why the determinant of the 2 vectors gives you the area composed by them

- dan815

but for starters u can take the formula at face cvalue how
u dot v = |U||V| cos theta
and
U X V = |U||V| sin theta <--- area

- dan815

as |V| sin theta is the height

- dan815

so that would the formula for the area of a parallelogram

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