Chapter_3

# Chapter_3 - III 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17...

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III 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8 1

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III.1.1 Evaluate R ° y 2 dx + x 2 dy along the following paths ° from (0 ; 0) to (2 ; 4) , (a) the arc of the parabola y = x 2 , (b) the horizontal interval from (0 ; 0) to (2 ; 0) , followed by the ver- tical interval from (2 ; 0) to (2 ; 4) , (c) the vertical interval from (0 ; 0) to (0 ; 4) , followed by the horizontal interval from (0 ; 4) to (2 ; 4) . Solution (a) Z ° y 2 dx + x 2 dy = ° x = t dx = dt; 0 ° t ° 2 y = t 2 dy = 2 t dt ± = Z 2 0 ² t 2 ³ 2 dt + Z 2 0 t 2 2 t dt = 72 5 : (b) Z ° 1 y 2 dx + x 2 dy = ° x = t dx = dt 0 ° t ° 2 y = 0 dy = 0 dt ± = Z 2 0 0 2 dt + Z 2 0 t 2 0 dt = 0 Z ° 2 y 2 dx + x 2 dy = ° x = 2 dx = 0 dt 0 ° t ° 4 y = t dy = dt ± = Z 4 0 t 2 0 dt + Z 4 0 2 2 dt = 16 Z ° y 2 dx + x 2 dy = Z ° 1 y 2 dx + x 2 dy + Z ° 2 y 2 dx + x 2 dy = 16 : (c) Z ° 1 y 2 dx + x 2 dy = ° x = 0 dx = 0 dx 0 ° t ° 4 y = t dy = dt ± = Z 4 0 y 2 0 dt + Z 4 0 0 2 dt = 0 Z ° 2 y 2 dx + x 2 dy = ° x = t dx = dt 0 ° t ° 2 y = 4 dy = 0 dt ± = Z 2 0 4 2 dt + Z 2 0 t 2 0 dt = 32 Z ° y 2 dx + x 2 dy = Z ° 1 y 2 dx + x 2 dy + Z ° 2 y 2 dx + x 2 dy = 32 : 2
III.1.2 Evaluate R ° xy dx both directly and using Green°s theorem, where ° is the boundary of the square with vertices at (0 ; 0) , (1 ; 0) , (1 ; 1) , and (0 ; 1) . Solution (A. Kumjian) 1 1 γ γ γ γ 1 2 3 4 VII.1.2 Denote the square by D and note that xy dx = P dx + Q dy where P = xy and Q = 0 . Then P and Q are continuously di∕erentiable on ° D and ° = @D , hence by Green°s Theorem we have, Z ° xy dx = Z @D P dx + Q dy = ZZ D ´ @Q @x ± @P dy µ dx dy = Z 1 0 Z 1 0 ± x dx dy = ± 1 2 : Observe that ° = ° 1 + ° 2 + ° 3 + ° 4 where ° 1 and ° 3 are the bottom and top of the square while ° 2 and ° 4 are the last two sides taken in the order indicated by the order of the vertices in the statement of the problem (so the boundary is oriented counter clockwise). Note that the path integrals on ° 2 and ° 4 are zero because the edges are vertical. Z ° xy dx = Z ° 1 xy dx + Z ° 3 xy dx = Z 1 0 x ² 0 dx ± Z 1 0 x ² 1 dx = 0 ± 1 2 = ± 1 2 : 3

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III.1.3 Evaluate R @D x 2 dy both directly and using Green°s theorem, where D is the quarter-disk in the ±rst quadrant bounded by the unit circle an the two coordinate axes. Solution Evaluate R @D x 2 dy directly, set @D = ° 1 + ° 2 + ° 3 . Z ° 1 x 2 dy = ° x = t dx = dt 0 ° t ° 1 y = 0 dy = 0 dt ± = Z 1 0 t 2 0 dt = 0 Z ° 2 x 2 dy = ° x = cos t dx = ± sin t dt 0 ° t ° ±= 2 y = sin t dy = cos t dt ± = Z ±= 2 0 cos 2 t cos t dt = 2 3 Z ° 3 x 2 dy = ° x = 0 dx = 0 dt 0 ° t ° 1 y = 1 ± t dy = ± dt ± = ± Z 1 0 0 2 dt = 0 Z @D x 2 dy = Z ° 1 x 2 dy + Z ° 2 x 2 dy + Z ° 3 x 2 dy = 2 3 Now we evaluate R @D x 2 dy , this time using Green°s theorem. In this case, P ( x; y ) = 0 and Q ( x; y ) = x 2 . Z @D x 2 dy = ZZ D 2 x dxdy = = 2 ZZ D x dxdy = ° x = r cos ² dx dy = r dr d² 0 ° r ° 1 y = r sin ² 0 ° ² ° ±= 2 ± = 2 Z 1 0 Z ±= 2 0 r cos ² rdrd² = = 2 Z 1 0 r 2 dr ² Z ±= 2 0 cos ²d² = 2 3 : 4
III.1.4 Evaluate R ° y dx both directly and using Green°s theorem, where ° is the semicircle in the upper half-plane from R to ± R .

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