Week 13 - Solution

# E we have to choose 0 0 r instead of 0 0 r curl f

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Unformatted text preview: 2π , 0 ≤ θ ≤ 2π . 3 x2 + y 2 + z 2 = 1 =⇒ ρ2 = 1 =⇒ ρ = 1 1 For z = − 2 , we want to ﬁnd φ, 1 2π z = ρ cos φ =⇒ − = (1) cos φ =⇒ φ = . 2 3 S1 is the circular plane with z = − 1 , so 2 √ 3 3 = 1 =⇒ x2 + y 2 = r2 = =⇒ r = and 4 2 √ 3 1 with 0 ≤ r ≤ and 0 ≤ θ ≤ 2π . r (r, θ) = r cos θ, r sin θ, − 2 2 1 x2 + y 2 + z 2 = 1 =⇒ x2 + y 2 + − 2 6 2 Using Divergence theorem, curl F = S1 S E S +S1 i j ∂ ∂ ∂x ∂y −y x2 k ∂ ∂z z3 curl F • dS. curl F • dS + div (curl F) dV = curl F • dS = = [0 − 0, 0 − 0, 2x − (−1)] = [0, 0, 2x + 1] div (curl F) = div [0, 0, 2x + 1] = 0 + 0 + 0 = 0 (0) dV = 0 div (curl F) dV = E E rr = [cos θ, sin θ, 0] and rθ = [−r sin θ, r cos θ, 0] i j k −r sin θ r cos θ 0 cos θ sin θ 0 rθ × rr = = [0, 0, −r] The Divergence theorem states that the normal on the surface is outward orientation and S1 is the surface at the bottom, so, S1 is downward orientation, i.e., we have to choose [0, 0, −r] instead of [0, 0, r]. curl F (r (r, θ)) = [0, 0, 2 (r cos θ) + 1] = [0, 0, 2r cos θ + 1] curl F • dS = S1 curl F (r (r, θ)) • (rθ × rr ) dA D √ 3 2 2π [0, 0, 2r cos θ + 1] • [0, 0, −r] dr dθ = 0 0 √ 3 2 2π −2r2 cos θ − r dr dθ = − = 0 0 3π 4 Therefore, 0= curl F • dS + − 3π 4 =⇒ S curl F • dS = 3π . 4 S ii. The Stoke’s theorem states that C is the boundary or edge of the opened surface S , so we can use the theorem to ﬁnd the surface integral by the line integral directly. √ Note that C is the boundary of S1 , a circle with radius of 23 , so r (t) = √ √ 3 3 1 cos t, sin t, − 2 2 2 √ 3 1 3 sin t, cos2 t, − F (r (t)) = − 2 4 8 7 with 0 ≤ t ≤ 2π . √ √ 3 3 and r′ (t) = − sin t, cos t, 0 2 2 curl F • dS = S...
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