1) Find the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ wit outward orientation where σ is the portion of the elliptic cylinder r(u,v) = (2cos v) i + (sin v) j + (u) k with 0 ≤ u ≤ 5, 0 ≤ v ≤ 2pi. 2) Find the work done by the force field F(x, y, z) = (x + y) i + (xy) j - (z^2) k on a particle that moves along line segments from (0,0,0) to (1,3,1) to (2,-1,4).

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1) Find the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ wit outward orientation where σ is the portion of the elliptic cylinder r(u,v) = (2cos v) i + (sin v) j + (u) k with 0 ≤ u ≤ 5, 0 ≤ v ≤ 2pi. 2) Find the work done by the force field F(x, y, z) = (x + y) i + (xy) j - (z^2) k on a particle that moves along line segments from (0,0,0) to (1,3,1) to (2,-1,4).

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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sorry wat 2 high 4 me
\[\int\limits{\int\limits{s F dS}} = ∫∫∫v ∇·F dV\] \[∇·F = x \cos(z) - 1\] \[= ∫∫∫v x \cos(z) - 1 dx dy dz\] \[v: x² + 4y² ≤ 4 ; 0 ≤ z ≤ 5\] let u = x , v = 2y , z = z ∂(u,v)/∂(x,y) = 2 = 1/2 ∫∫∫v u cos(z) - 1 du dv dz v: u² + v² ≤ 4 ; 0 ≤ z ≤ 5 let u = r cos(theta) ; v = r sin(theta) z = z \[\frac{∂(u,v)}{∂(r,θ)} = r\] \[= 1/2 ∫∫∫ (r \cos(θ) \cos(z) - 1) r dr dθ dz\] \[{(r,θ,z) | 0 ≤ r ≤ 2 ; 0 ≤ θ ≤ 2π ; 0 ≤ z ≤ 5}\] \[= 1/3 ∫∫ (4 \cos(θ) \cos(z) - 3) dθ dz\] \[{(θ,z) | 0 ≤ θ ≤ 2π ; 0 ≤ z ≤ 5}\] \[= -2π ∫ dz\](z) | 0 ≤ z ≤ 5 \[= -10π\]
oh 2 smaalllllllll

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Other answers:

you know the second one
is the first one right? or u dont know tha answer?
im just checking what i got wrong on my hw and yeah the answer is right
oh ok im thinkin
k thanks lana :)
(0,0,0) -->> (1,3,1) x = t y = 3t z = t dx =dt........ dy = 3 dt...... dz = dt work done is integration of F.dr int{ (x + y)dx + (xy)dy - (z^2) dz
now substitute x=t y=3t and z=t and the dx dy and dz values in terms of dt
then integrate

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