Tutorial08-sol(1)

Thus we conclude that 2 2 2 2 0 c e 3 3 0 4 4

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Unformatted text preview: 0 2 (1 + cos ) √ cos 1 − cos On setting = cos θ , we can see easily that the above integral equals to 0. Thus, we conclude that 2 + 2 − 2 2 + =0− C E π3 π3 +0=− 4 4 :) Evaluate ( 2 ) +( 2 2 − ) +( 2 − 2 ) C where C encloses the part of the sphere 2 + 2 + 2 = 1 lying in the ﬁrst octant ≥ 0. The curve C is oriented such that the piece of the sphere with ≥ 0 is on the left when one is moving along C . S The contour C consists of three parts: on the − plane, it can be parameterized by = cos = sin = 0 with 0 ≤ ≤ π/2 and the part of the integral corresponding to this is given by π /2 (sin2 − 02 ) cos + (02 − cos2 ) sin + (cos2 − sin2 ) 0 0 π /2 (1 − cos2 ) cos − (1 − sin2 ) sin = 0 = cos − cos3 3 π /2 − sin − 0 sin3 3 π /2 0 =− 4 3 On the − plane, it can be parameterized by = 0 = cos and the part of the integral corresponding to this is given by = sin with 0 ≤ ≤ π/2 π /2 (cos2 − sin2 ) 0 + (sin2 − 02 ) cos + (02 − cos2 ) sin = − 0 On the − plane, it can be parameterized by = sin =0 and the part of the integral corresponding to this is given by 4 3 = cos with 0 ≤ ≤ π/2 π /2 (02 − cos2 ) sin + (cos2 − sin2 ) 0 + (sin2 − 02 ) cos = − 0 4 3 Thus, we conclude that ( 2 − 2 ) +( 2 − 2 ) +( C 8 2 − 2 ) 4 = 3 · (− ) = − 4 3 :) 5 Answers (1) Arc length. (2) C . C (4) 0. (5) 0. (6) 256 15 3 2 (7) −2π (10) 2 2 . . . (11) 0. 3 (12) − π4 . (13) −4. 9...
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This document was uploaded on 02/10/2014.

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