MA102: Multivariable Calculus Rupam Barman and Shreemayee Bora - PowerPoint PPT Presentation

Line integral and Green’s Theorem Path independence of line integral MA102: Multivariable Calculus Rupam Barman and Shreemayee Bora Department of Mathematics IIT Guwahati R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Example: ��� Evaluate V 2 xdV where V is the region bounded by the planes x = 0 , y = 0 , z = 0 and 2 x + 3 y + z = 6 . Note that V is Type-I: 0 ≤ z ≤ 6 − 2 x − 3 y and ( x , y ) ∈ D , where D is special domain given by 0 ≤ x ≤ 3 and 0 ≤ y ≤ − 2 3 x + 2 . Thus � 6 − 2 x − 3 y ��� �� 2 xdV = ( 2 x dz ) dA V D 0 � 3 � − 2 3 x + 2 = ( 6 − 2 x − 3 y ) 2 xdydx = 9 . 0 0 R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Example: Find the volume of the region bounded by the surfaces z = x 2 + 3 y 2 and z = 8 − x 2 − y 2 . ��� The volume is V = Ω dzdydx , where Ω is bounded above by the surface z = 8 − x 2 − y 2 and below by the surface z = x 2 + 3 y 2 . Therefore, the limits of z are from z = x 2 + 3 y 2 to z = 8 − x 2 − y 2 . The projection of Ω on xy -plane is the solution of 8 − x 2 − y 2 = x 2 + 3 y 2 ⇒ x 2 + 2 y 2 = 4 . = Therefore the limits of x and y are to be determined by R : x 2 + 2 y 2 = 4 . R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Example (cont.): � 8 − x 2 − y 2 �� V = z = x 2 + 3 y 2 dzdA R � √ � 2 ( 4 − x 2 ) / 2 ( 8 − 2 x 2 − 4 y 2 ) dydx = − √ − 2 ( 4 − x 2 ) / 2 � √ � 2 ( 4 − x 2 ) / 2 � ( 8 − x 2 ) y − 4 3 y 3 = y = − √ − 2 ( 4 − x 2 ) / 2 √ � 2 √ 4 2 ( 4 − x 2 ) 3 / 2 dx = 8 π = 2 . 3 − 2 R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Example: Change of order of integration sin x �� Consider the evaluation of integral x dA over the R triangle formed by y = 0 , x = 1 and y = x . If we write R = { ( x , y ) : 0 ≤ y ≤ 1 , y ≤ x ≤ 1 } , then � 1 �� 1 � �� sin x sin x dA = dx dy . x x R 0 x = y The innermost integral is difficult to evaluate. If we change the order of integration, by taking R = { ( x , y ) : 0 ≤ x ≤ 1 , 0 ≤ y ≤ x } , then � 1 � x � 1 �� sin x sin x dA = dydx = sin xdx = 1 − cos 1 . x x R 0 y = 0 0 R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Example: Change of order of integration Find the volume of the region bounded by x + z = 1, y + 2 z = 2 in the first quadrant. Figure : Region bounded by x + z = 1 and y + 2 z = 2, x , y , z ≥ 0 R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Example: Change of order of integration Draw line parallel to z -axis and note that the upper surfaces are: 2 z + y = 2 over triangle bounded by x = 0 , y = 1 y = 2 x and z = 1 − x over the triangle bounded by y = 0 , x = 1 , y = 2 x . Therefore, � 2 � y / 2 � 1 � 2 x � 1 − x 2 − y � dz dy dx = 2 2 V = dz dx dy + 3 y = 0 x = 0 z = 0 x = 0 y = 0 z = 0 On the other hand, by first drawing the line parallel to x -axis, we get � 1 � 2 − 2 z � 1 − z dx dy dz = 2 V = 3 z = 0 y = 0 x = 0 R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Example (cont.): Taking the line parallel to y -axis we get � 1 � 1 − x � 2 − 2 z dy dz dx = 2 V = 3 x = 0 z = 0 y = 0 � 4 � 1 � 2 2 cos ( x 2 ) Example: Evaluate I = √ z dx dy dz . z = 0 y = 0 x = 2 y The integral is difficult to evaluate in the given order of integration. We change the order of integration and evaluate the integral: � 4 � 2 � x / 2 2 cos ( x 2 ) √ z I = dy dx dz . z = 0 x = 0 y = 0 � 4 � 2 x cos ( x 2 ) = √ z dx dz = 2 sin 4 . z = 0 x = 0 R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Change of variable Let T : R 2 → R 2 be C 1 given by T ( u , v ) = ( x ( u , v ) , y ( u , v )) . Then the Jacobian matrix J ( u , v ) of T is given by � � x u x v J ( u , v ) := . y u y v Define the Jacobian of T by ∂ ( x , y ) ∂ ( u , v ) := x u y v − x v y u = det J ( u , v ) . Polar coordinates: Define T ( r , θ ) := ( r cos θ, r sin θ ) . Then � � ∂ ( x , y ) cos θ − r sin θ � � ∂ ( r , θ ) = � = r . � � sin θ r cos θ � R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Change of variable Let T : R 2 → R 2 be C 1 given by T ( u , v ) = ( x ( u , v ) , y ( u , v )) . Let T − 1 : R 2 → R 2 be the inverse given by T − 1 ( x , y ) = ( u ( x , y ) , v ( x , y )) . Let J ( u , v ) and J ′ ( x , y ) be the Jacobian matrices of T and T − 1 . Applying Chain rule, we have 1 = x u u x + x v v x ; 0 = x u u y + x v v y 0 = y u u x + y v v x ; 1 = y u u y + y v v y . Thus, � � 1 1 0 J ′ ( x , y ) · J ( u , v ) = ⇒ det ( J ′ ( x , y )) = det ( J ( u , v )) . 0 1 R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Change of variable for double integrals Suppose T is injective and J ( u , v ) is nonsingular. Let D ⊂ R 2 and G := T ( D ) . Suppose that f is integrable on G . Then � � ∂ ( x , y ) � � dA = dxdy = � dudv � � ∂ ( u , v ) � and � � ∂ ( x , y ) �� �� � � f ( x , y ) dxdy = f ( x ( u , v ) , y ( u , v )) � dudv . � � ∂ ( u , v ) � G D Polar coordinates: �� �� f ( x , y ) dxdy = f ( r cos θ, r sin θ ) rdrd θ. G D R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Example � 1 � 1 − x √ x + y ( y − 2 x ) 2 dA . Evaluate the integral I = 0 0 The given domain is the triangle bounded by x = 0 , y = 0 and x + y = 1. We take the transformation u = x + y and v = y − 2 x . Under this transformation, the given triangle will be transformed into triangle bounded by v = u , v = − 2 u and u = 1. The inverse of this transformation is x = u − v and 3 y = 2 u + v 3 . Hence the Jacobian � � 1 / 3 − 1 / 3 � � J = � = 1 / 3 . � � 2 / 3 1 / 3 � Hence � 1 � u √ I = 1 uv 2 dv du 3 0 v = − 2 u R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Example Using the transformation u = 2 x + 3 y and v = x − 3 y , find �� e 2 x + 3 y cos ( x − 3 y ) dx dy the value of the integral I = R where R is the region bounded by the parallelogram with vertices ( 0 , 0 ) , ( 1 , 1 / 3 ) , ( 4 / 3 , 1 / 9 ) , ( 1 / 3 , − 2 / 9 ) . Under the given transformations, R will be transformed into the rectangle with vertices ( 0 , 0 ) , ( 3 , 0 ) , ( 3 , 1 ) and ( 0 , 1 ) . Also, | J | = 1 9 . Thus, � 1 � 3 I = 1 e u cos v du dv = 1 9 ( e 3 − 1 ) sin 1 . 9 v = 0 u = 0 R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Example √ x 2 + z 2 dV where G is the region ��� Evaluate G bounded by the paraboloid y = x 2 + z 2 and y = 4 . We have �� 4 � ��� �� � x 2 + z 2 dy f ( x , y , z ) dV = dxdz , G D x 2 + z 2 where D = { ( x , z ) : x 2 + z 2 ≤ 4 } . Setting x = r cos θ and z = r sin θ for ( r , θ ) ∈ [ 0 , 2 ] × [ 0 , 2 π ] , � 2 π � 2 ��� r ( 4 − r 2 ) rdrd θ = 128 π f ( x , y , z ) dV = . 5 G 0 0 R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Change of variable for multiple integrals Let D ⊂ R n be open and bounded. Let T : D → R n be such that T is C 1 , injective and the Jacobian J ( U ) is nonsingular for U ∈ D . Let G := T ( D ) and f : G → R be integrable over G . Then � � ∂ ( x 1 , · · · , x n ) � � dx 1 · · · dx n = � du 1 · · · du n � � ∂ ( u 1 , · · · , u n ) � and � � � � ∂ ( x 1 , · · · , x n ) � � f ( X ) dx 1 · · · dx n = f ( X ( U )) � du 1 · · · du n � � ∂ ( u 1 , · · · , u n ) � G D � dX � � � � = f ( X ( U )) � dU . � � dU � D R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Cylindrical coordinates Consider T ( r , θ, z ) = ( r cos θ, r sin θ, z ) . Then � � cos θ − r sin θ 0 � � ∂ ( x , y , z ) � � ∂ ( r , θ, z ) = sin θ r cos θ 0 = r . � � � � 0 0 1 � � Thus dV = rdrd θ dz and ��� ��� f ( x , y , z ) dV = f ( r cos θ, r sin θ, z ) rdrd θ dz . G D R. Barman & S. Bora MA-102 (2017)

Line integral and Green’s Theorem Path independence of line integral Example x 2 + y 2 dV , where G is the region � ��� Evaluate G bounded by x 2 + y 2 = 1 , z = 4 and z = 1 − x 2 − y 2 . Consider cylindrical coordinates D := { ( r , θ, z ) : ( r , θ ) ∈ [ 0 , 1 ] × [ 0 , 2 π ] , 1 − r 2 ≤ z ≤ 4 } . Then � 1 � 2 π �� 4 � ��� r rdrd θ = 12 π f ( x , y , z ) dV = 1 − r 2 dz 5 . G 0 0 R. Barman & S. Bora MA-102 (2017)