CHAPTER 5 VECTOR INTEGRATION INTEGRAL THEOREMS AND EAF EBF x=X1(y) x=X2(y) Similarly let the equations of curves and be and respectively. Then f ye e f f f (ii) e e C From (i) and (ii) we have dxdy. R Example 5.22 2x 2 C Evaluate around the boundary of the ellipse ydx  y  xdy x  9 y  16 , 2 2 , , , M  2x y  N  2 y  x  1 2x 2     ydx y xdy dxdy R R But the ellipse R i.e. x  9 y  16 , 2 2  1  9 y x 2 2 16 16 A  4 a  4, b  3 3 2x 2 Hence ydx  y  xdy  dxdy 32  3 R Example 5.23 Using Green’s Theorem, find the area enclosed by the ellipse .  x  a cos , y  b sin ,0 2 M=-y, N=x If we take and apply Green’s Theorem, we obtain 2 2 2 2 where we have as equation of an ellipse.   1 a b
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS Similarly let the equations of curves EAF and EBF be x=X1(y) and x=X2(y) respectively. Then f ye e f f f e e C From (i) and (ii) we have (ii) R dxdy. Example 5.22 Evaluate 2x ydx  2 y  xdy around the boundary C of the ellipse x 2  9 y 2  16 , M  2x y ,  , N  2 y  x ,  1 2x ydx  2 y  xdy   dxdy R R But the ellipse R x 2  9 y 2  16 , i.e. x 2 16  9 y 2 16  1 a  4, b  4 3 A  3 Hence 2x ydx  2 y  xdy  dxdy  R 32 3 Example 5.23 Using Green’s Theorem, find the area enclosed by the ellipse x  a cos , y  b sin ,0  2 . If we take M=-y, N=x and apply Green’s Theorem, we obtain 2 2 2 2 where we have   1 as equation of an ellipse. a b