Do you know Stokes' Theorem? .
You should start by drawing a sketch of your region. Can you get an equation for the surface bounded by your contour?
Hello,
What method should I employ here/how to do it ???(apparently this is stokes problem).cheers.
Given F= (2x+sinyz)i+2x+xzcosyz)j+(y^2+2z+xycosyz)k
evaluate ∮F.dr where C is the cruve C consisting of the straight line segment from (0,0,0) to (0,1,0) followed by the quarter circle y²+z²=1 from (0,1,0) to (0,0,1) followed by the straight line segment from (0,0,1) to (0,0,0).
Yes, Stoke's theorem is probably simplest but it is not all that hard to integrate directly, on the path. On the first part, (0, 0, 0) to (0, 1, 0), we can use the parameterization x= 0, y= t, z= 0, with t from 0 to1, so that dx= dz= 0, dy= dt so only the dy or "j" part is necessary. And the integrand for dy is xz cos(yz)= 0 cos(0)= 0 so the first integral is 0.
Similarly, on the third portion, the line from (0, 0, 1) to (0, 0, 0), we can take x= 0, y= 0, z= 1- t, dx=dy= 0, dz= -dt with t from 0 to 1. Now we need only look at the dz integral. The integrand is [tex]y^2+ 2z+ xy cos(yz)= 0^2+ 2(1- t)+ 0cos(0)= 2- 2t. The integral is [tex]\int_0^1 (2- 2t)(-dt)= \int_0^1(2t- 2) dt= \left[t^2- 2t\right]_0^1= 1- 2= -1.
The middle portion is the "hard one"- but not that hard. The quarter circle, from (0, 1, 0) to (0, 0, 1) can be parameterized as x= 0, y= cos(t), z= sin(t) with t from 0 to . dx= 0, dy= -sin(t)dt, and dz= cos(t)dt. 2x+ xzcos(yz)= 0 so that part is 0. so the integral becomes
Let so so that the integral becomes which is easy to integrate.
(Actually, after trying to do it with Stokes' theorem, I think doing it directly iseasiest!)