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: 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...
View Full Document

Ask a homework question - tutors are online