Hey guys, I need help checking my answers on a practice test.
http://www.math.ufl.edu/~zapletal/teachi…
I'm just posting the URL because there are too many symbols to write but I basically need to know #2-#5
http://www.math.ufl.edu/~zapletal/teachi…
I'm just posting the URL because there are too many symbols to write but I basically need to know #2-#5
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I'll answer the rest after breakfast.
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4) ∫c F · dr = ∫∫s curl F · dS, by Stokes' Theorem.
Check that curl F = <-2z, -2x, -2y>.
Parameterize S, which is enclosed by c(t) = by
r(u,v) = , for u in [0, 1] ("thickening factor" and v in [0, 2π].
(Note how z = x^2 is used above, along with the parameterization.)
Since r_u x r_v = <-2u^2 cos v, 0, u>,
∫∫s curl F · dS
= ∫(u = 0 to 1) ∫(v = 0 to 2π) <-2u^2 cos^2(v), -2u cos v, -2u sin v> · <-2u^2 cos v, 0, u> dv du
= ∫(u = 0 to 1) ∫(v = 0 to 2π) [4u^4 cos v cos^2(v) - 2u^2 sin v] dv du
= ∫(u = 0 to 1) ∫(v = 0 to 2π) [4u^4 (1 - sin^2(v)) cos v - 2u^2 sin v] dv du
= 0.
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4) ∫c F · dr = ∫∫s curl F · dS, by Stokes' Theorem.
Check that curl F = <-2z, -2x, -2y>.
Parameterize S, which is enclosed by c(t) =
r(u,v) = , for u in [0, 1] ("thickening factor" and v in [0, 2π].
(Note how z = x^2 is used above, along with the parameterization.)
Since r_u x r_v = <-2u^2 cos v, 0, u>,
∫∫s curl F · dS
= ∫(u = 0 to 1) ∫(v = 0 to 2π) <-2u^2 cos^2(v), -2u cos v, -2u sin v> · <-2u^2 cos v, 0, u> dv du
= ∫(u = 0 to 1) ∫(v = 0 to 2π) [4u^4 cos v cos^2(v) - 2u^2 sin v] dv du
= ∫(u = 0 to 1) ∫(v = 0 to 2π) [4u^4 (1 - sin^2(v)) cos v - 2u^2 sin v] dv du
= 0.
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2) Parameterize M by r(u,v) = for u in [0, 2π] and v in [-1, 1].
Since r_u x r_v = <-sin u, -cos u, 0>,
∫∫s x^2 dS
= ∫∫ cos^2(u) * ||r_u x r_v|| dA
= ∫(v = -1 to 1) ∫(u = 0 to 2π) (1/2)(1 + cos(2u)) * 1 du dv
= 2 * (1/2)(2π + 0)
= 2π.
Since r_u x r_v = <-sin u, -cos u, 0>,
∫∫s x^2 dS
= ∫∫ cos^2(u) * ||r_u x r_v|| dA
= ∫(v = -1 to 1) ∫(u = 0 to 2π) (1/2)(1 + cos(2u)) * 1 du dv
= 2 * (1/2)(2π + 0)
= 2π.
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3) By Green's Theorem,
A = |∫c x dy|
...= 2 |∫(t = 0 to 1) (t^2 - 1) * 4t^3 dt|, via symmtery
...= 2 |∫(t = 0 to 1) (4t^5 - 4t^3) dt|
...= 2 |(4/6 - 1)|
...= 2/3.
(To drop the absolute value bars, make sure C has counterclockwise orientation.)
A = |∫c x dy|
...= 2 |∫(t = 0 to 1) (t^2 - 1) * 4t^3 dt|, via symmtery
...= 2 |∫(t = 0 to 1) (4t^5 - 4t^3) dt|
...= 2 |(4/6 - 1)|
...= 2/3.
(To drop the absolute value bars, make sure C has counterclockwise orientation.)
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5) ∫∫m F · dS
= ∫∫∫ div F dV, by the Divergence Theorem
= ∫∫∫ (2x + 1 + 0) dV
= ∫(φ = 0 to π) ∫(ρ = 0 to 1) ∫(θ = 0 to 2π) (2ρ cos θ sin φ + 1) * (ρ^2 sin φ dθ dρ dφ), via spherical coordinates
= ∫(φ = 0 to π) ∫(ρ = 0 to 1) (0 + 2π) * ρ^2 sin φ dρ dφ
= 4π/3.
= ∫∫∫ div F dV, by the Divergence Theorem
= ∫∫∫ (2x + 1 + 0) dV
= ∫(φ = 0 to π) ∫(ρ = 0 to 1) ∫(θ = 0 to 2π) (2ρ cos θ sin φ + 1) * (ρ^2 sin φ dθ dρ dφ), via spherical coordinates
= ∫(φ = 0 to π) ∫(ρ = 0 to 1) (0 + 2π) * ρ^2 sin φ dρ dφ
= 4π/3.
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