FM5002-HW5-2.22.12

Let t 5 3 t t 5 3t 4 t 2 4 t

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (t) = (2 + 3 t, 3), for 0 <= t <= 1. Let a = Α (0) = ∆ (1). Let b = Β (0) = Α (1). Let c = Γ (0) = Β (1). Let d = ∆ (0) = Γ (1). Let K be the 1 − chain {(a, b), (b, c), (c, d), (d, a)}. Let Ω = P x + Q y. Ω. K Parameterizin g K reduces us to 0042 − 3, from which we conclude that the answer is 4. 0042− . 5 Let P = p (x, y) = −y. Let Q = q (x, y) = x. Let R be the rectangle [2, 5] x[3, 4]. a. Compute R. 3 t, 3 3, 0 1 4 Let P = p (x, y) = −y. Let Q = q (x, y) = x. Let Ω = P x + Q y. V2 0 0042− . 4 Compute t 1 4, 5 1 5t t 0 t 0 ∆’ t 1 V 2, 4 0 1 t , 5 0, 1 V∆t t 1 V5 0 3 Γ’ t 0 1 t t. 1 VΓt 0 1 ∆’ t 0 t 0 V 5, 3 V∆t t 1 V Βt t Γ’ t 0 1 VΑ t 1 VΓt t 3, 2 0 3t 3, 0 t t 3 4 FM5002−HW5−2.22.12.nb Ω. b. Compute R dΩ. c. Compute R a. We compute the boundary R = (((2, 3), (2, 4)), ((2, 4), (3, 4)), ((3, 4), (3, 3)), ((3, 3), (2, 3))). b. From problem 0042 Ω 4, we have that dΩ 6. R R dΩ c. By Stokes ’ theorem, we have that Ω 6. R R 0042− 6. Let P p x, y x2 y2. Let Q q x, y 2 x y. Let Α (t) = (5, 3 + t), Β (t) = (5 − 3t, 4), Γ (t) = (2, 4 − t), ∆ (t) = (2 + 3t, 3), for 0 <= t <= 1. Let a = Α (0) = ...
View Full Document

This note was uploaded on 01/19/2014 for the course MATH 5002 taught by Professor Adams during the Spring '08 term at Minnesota.

Ask a homework question - tutors are online