I am not sure should i use Green's Theorem or else?
Thanks.
Thanks.
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No need for Green's Theorem (that is used when you are comparing an integral over a surface to the integral on its boundary). This is just a standard line integral.
The first curve is c(t) = (0,t) as t goes from 0 to 1. The second is d(t) = (1,2t) as t goes from 0 to 1.
c'(t) = (0,1)
d'(t) = (0,2)
F(c(t)) = F(0,t) = (t,0)
F(d(t)) = F(1,2t) = (2t,1)
F(c(t)) *dot* c'(t) = 0
F(d(t)) *dot* d'(t) = 2
Integrate each of the expressions directly above with respect to t, as t varies from 0 to 1.
Integral(0) from 0 to 1 = 0
Integral(2) from 0 to 1 = 2
The sum of the two integrals gives you the total integral, so it is 0 + 2 = 2.
The first curve is c(t) = (0,t) as t goes from 0 to 1. The second is d(t) = (1,2t) as t goes from 0 to 1.
c'(t) = (0,1)
d'(t) = (0,2)
F(c(t)) = F(0,t) = (t,0)
F(d(t)) = F(1,2t) = (2t,1)
F(c(t)) *dot* c'(t) = 0
F(d(t)) *dot* d'(t) = 2
Integrate each of the expressions directly above with respect to t, as t varies from 0 to 1.
Integral(0) from 0 to 1 = 0
Integral(2) from 0 to 1 = 2
The sum of the two integrals gives you the total integral, so it is 0 + 2 = 2.
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Sure. We want to describe the line segment from (0,0) to (1,0) using a parameter t. So the x value has to increase from 0 to 1, while the y value stays a constant at 0. So c(t) = ti + 0j as t goes from 0 to 1.
Now for (1,0) to (1,2), we want x to stay a constant at 1 and y to increase from 0 to 2.
Now for (1,0) to (1,2), we want x to stay a constant at 1 and y to increase from 0 to 2.
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So d(t) = 1i + 2tj. This will keep the x value at 1, and the y value will go from 0 to 2 as t goes from 0 to 1. Hope this helps!
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