Caculus Homework Help!? All related to FLUX?

Q1 (part b; the part you missed).
We need to parameterize x^2 + y^2 = 9 with z in [-4, 4].
Use R(u, v) = <3 cos u, 3 sin u, v> for u in [0, 2π], v in [-2, 2].

Note that R_u x R_v = <3 cos u, 3 sin u, 0>.

So, the flux ∫∫s <2x - 4y + 3, 2y + 4x + 3, z^3> · dS equals
∫(v = -2 to 2) ∫(u = 0 to 2π) <6 cos u - 12 sin u + 3, 6 sin u + 12 cos u + 3, v^3> · <3 cos u, 3 sin u, 0> du dv

= ∫(v = -2 to 2) ∫(u = 0 to 2π) [18 + 9 cos u + 9 sin u] du dv
= ∫(v = -2 to 2) (2π * 18 + 0) dv
= 144π.
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Q2.
1) Using cartesian coordinates:
∫∫s F · dS
= ∫∫ <6x+y, 4y, 2z> · <-z_x, -z_y, 1> dA
= ∫∫ <6x + y, 4y, 2(36 - x^2 - y^2)> · <2x, 2y, 1> dA
= ∫∫ (10x^2 + 2xy + 6y^2) dA

Convert to polar coordinates (as the region of integration is in x^2 + y^2 = 36):
∫(r = 0 to 6) ∫(θ = 0 to 2π) (10r^2 cos^2(θ) + 2r^2 cos θ sin θ + 6r^2 sin^2(θ)) * r dθ dr
= ∫(r = 0 to 6) r^3 dr * ∫(θ = 0 to 2π) (10 cos^2(θ) + 2 cos θ sin θ + 6 sin^2(θ)) dθ
= (1/4)r^4 {for r = 0 to 6} * ∫(θ = 0 to 2π) (4 cos^2(θ) + 2 cos θ sin θ + 6) dθ
= 324 * ∫(θ = 0 to 2π) (2 (1 + cos(2θ)) + 2 cos θ sin θ + 6) dθ
= 324 * ∫(θ = 0 to 2π) (8 + 2 cos(2θ) + 2 cos θ sin θ) dθ
= 324 * (2π * 8 + 0)
= 5184π.
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2) Parameterize the bottom region z = 0 with x^2 + y^2 = 36 by
R(u, v) = <u cos v, u sin v, 0> with u in [0, 6], v in [0, 2π].

Since R_u x R_v = <0, 0, u>, the flux ∫∫s F · dS equals
∫∫ <6x+y, 4y, 0> · <0, 0, u> dA = 0.
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Q3) Parameterize S by r(x, z) = <x, x^2 + z^2, 2z>.

Since r_x x r_z = <4x, -2, 2z>, we take n = <-4x, 2, -2z> so that it points in the positive y-direction.

So, the flux ∫∫s F · dS equals (over the disk x^2 + z^2 = 11)
∫∫ <x^2 + z^2, 2, -xz> · <-4x, 2, -2z> dA
= ∫∫ (4x^3 - 2xz^2 + 4) dA.

Convert to polar coordinates (x = r cos θ, z = r sin θ):
∫(r = 0 to √11) ∫(θ = 0 to 2π) (4r^3 cos^3(θ) - 2r^3 cos θ sin^2(θ) + 4) * r dθ dr
= ∫(r = 0 to √11) (0 + 2π * 4) * r dr
= 4πr^2 {for r = 0 to √11}
= 44π.

I hope this helps!

ask a friend in the class to come over and help you out