A i 2 change to 2 1 y 2 dx x ey dy 0 b

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Unformatted text preview: e can describe the region by 1 ≤ y ≤ x + 1, 0 ≤ x ≤ 2. Since γ is clockwise, we have y2 2 γ (1 − y ) dx + (x + e ) dy = − y2 2 (x + ey ) dy = − R R (1 + 2y ) dA = − − (y + y 2 ) 0 26 . 3 1 2 dA = 3 y x 1 x 2 2 R 1 y x+1 1 1 2 3x (1 + 2y ) dy dx = 0 1 x+1 2 − −γ (1 − y ) dx + ∂ (x + e ) ∂ (1 − y ) − ∂x ∂y 2 − 2 y 2 dx = − 0 (x2 + 3x) dx = − 13 32 x+ x 3 2 2 0 =− 8 +6 3 = MATB42H page 4 Solutions # 5 (b) We will again use Green’s Theorem. 2y ) dA 2π (ii) polars γ 2π 1 0 0 (r + 2 r 2 sin θ ) dr dθ = 0 1 dθ = 0 2π 1 2 0 (1 + R 1 (1 + 2 r sin θ ) r dr dθ = = 12 2 3 r + r sin θ 2 3 0 6. (a) (i) 2π...
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