Tutorial 9 solution

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Unformatted text preview: ves from 0 to 2, it is CLOCKWISE. • For C4 in the following graph, it is obvious to see that x = 0 is constant, and 0 ≤ y ≤ 3, so we deﬁne the parameter equations by x = 0, y = t, 0 ≤ t ≤ 3. When t moves from 0 to 2, it is CLOCKWISE. Summarily, we can get r(t) and r (t) for each line segments: Along C1 : x = t, y = 0 dy dx = 1, = 0, dt dt Along C2 : x = 2, y = t dy dx = 0, = 1, dt dt Along C3 : x = t, y = 3 dy dx = 1, = 0, dt dt Along C4 : x = 0, y = t dy dx = 0, = 1, dt dt . . . . . . . . . . . 3 . . . . . . .. ................................. . .................................. . . . .................................. .... ... ... ..................... . . . . . . . . . ... . . . . . . .. . . .. . . . . . . . . . . . . . . . . . .................................. . . . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . . . . .................................. . . . . . . . . . . . . . . . . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . .. . . 4 ....................................... 2 . .................... .............. . .. . . . . . . . . . . . . . . ... . . .................................. . . .................................. . . . . . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . . . . . .................................. . . .................................. . . .................................. . . . . . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . . . . . .................................. . . .................................. . . .................................. . . . . . . . . . . . . . . . . . . .. . .. . . . . . . ..... . . . . . . . . . . . .............................................................. .............................................................. . . . . . . . . . . . . . . . 1 y ............. for 0 ≤ t ≤ 2. (0, 3) C (2, 3) for 0 ≤ t ≤ 3. C for 0 ≤ t ≤ 2. D C for 0 ≤ t ≤ 3. 194 0 C (2, 0) x If C is given by anti-clockwise, then C = C1 + C2 − C3 − C4 . Thus xy 2 dx + x3 dy C = C1 +C2 −C3 −C4 xy 2 dx + x3 dy xy 2 dx + x3 dy + C1 C2 = = C1 xy 2 dx + x3 dy − 2 · t2 23 0 · t2 03 3 C3 xy 2 dx + x3 dy − 0 1 dt 0 1 dt xy 2 dx + x3 dy C4 t · 02 t3 t · 32 t3 3 0 · · 1 0 1 0 dt + C2 · · − 2 C3 dt − 2 C4 (Substitute r(t) and r (t)) = 0 0 dt + 8 dt − = 0 + 24 − 18 − 0 = 6. (2) Green’s theorem: We can check that: 0 9t dt − 0 dt 0 • The two integrands are both polynomials, and so both of them have continuous partial derivatives. • Since C is anticlockwise, and it gives positive orientation, and so the curve C is positive orient...
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