ESE 318 02 Spring 2016 Homework Set 14 Due May 3 2016 1 Zill 9 14 5 2 Find work integral over the closed path as indicated 2 2 F dr F 3 y 3x z C x y 4 and z 1 C 3 Zill 9 14 8 4 Zill 9 14 9 5 Let F z 2 x 2 2 xz Let C be the curve that goes in straight line segments from 0 0 3 to 4 2 2 to 4 4 0 to 0 2 1 and back to 0 0 3 These 4 points are coplanar and the curve C traces out a parallelogram Find the line integral F dr C 6 Find work integral over the closed path as indicated 2 2 F dr F y x x z C 0 0 1 to 1 0 1 to 1 1 1 back to 0 0 1 C 7 Zill 9 16 3 8 Zill 9 16 7 9 Zill 9 16 9 10 Zill 9 16 11 11 Zill 9 16 14 Hint Cylindrical coordinates 1 Zill 9 14 5 The region of integration is the triangle in the x y plane z z F 2 z x y z x y z 1 x y 1 x y i curlF x 2z x j y y z k 1 1 i 1 2 j 0 0 k 2 1 0 P Q R z x y z z F dr curlF ndS C S R P x Q y R dxdy R 2 1 0 dA 3 dA 3 Area triangle 3 12 1 1 3 2 R 1 1 x 1 Also 3 dydx 3 1 x dx 3 12 1 x 0 0 0 2 1 0 3 12 3 2 ESE 318 02 Spring 2016 2 Find work integral over the closed path as indicated 2 2 F dr F 3 y 3x z C x y 4 and z 1 C See that the unit normal is constant in the z or k direction CurlF is also constant so the surface integral can be taken without the usual full formula n k 0 0 1 i j k 3 3 k 0 0 6 curlF x y z 3 y 3x z curlF n 0 0 6 0 0 1 6 F dr 6dS 6 Area of C circle 6 2 2 24 S 3 Zill 9 14 8 F x 2 z 3x y 2 y z z f x y 4 x 2 y z z 1 2 x y i j k curlF 2i 2 j 3k 2 2 3 x y z x 2 z 3x y 2 y z z z F dr curlF ndS 2 x 2 y 3 dxdy 2 4 3 dxdy C S R 9 dA 9 Area triangle 9 R 1 2 4 2 36 R 2 4 2 y Also 9 0 dxdy 9 4 2 y dy 9 4 y y 2 2 0 0 2 0 9 4 36 ESE 318 02 Spring 2016 4 Zill 9 14 9 Drawing the contour C in 3D was rather difficult but the region in the xy plane is easy the unit circle Convert to polar for convenience F y 3 x 3 z 3 i curlF x y3 j y x3 z f x y 1 x y z z 1 1 x y k 3x 2 3 y 2 k 0 0 3 x 2 y 2 z z3 z z 2 2 F dr curlF ndS C S R P x Q y R dxdy R 3 x y dxdy 2 1 2 1 3 r 2 rdrd 3 d r 3 dr 3 2 14 32 0 0 0 0 ESE 318 02 Spring 2016 5 Let F z 2 x 2 2 xz Let C be the curve that goes in straight line segments from 0 0 3 to 4 2 2 to 4 4 0 to 0 2 1 and back to 0 0 3 These 4 points are coplanar and the curve C traces out a parallelogram Find the line integral F dr C We could find an equation of the plane and express it as surface z f x y Plane a 4 2 2 0 0 3 4 2 1 b 0 2 1 0 0 3 0 2 2 i j k 1 4 2 i 8 0 j 8 0 k 2 8 8 a b 4 2 0 2 2 Plane 2 x 8 y 8 z 3 0 14 x y z 3 0 z 14 x y 3 However when we apply Stokes Theorem we see that we don t even need f That is we really didn t need to find the equation of the plane because P and Q are 0 i curlF x z2 j y x2 k 0 0 i 2 z 2 z j 2 x 0 k 0 0 2 x z 2 xz z z F dr curlF ndS P x Q y R dxdy 2 xdxdy C S R R See that the region of integration is the parallelogram in the xy plane 4 x 2 2 4 0 0 F dr C 2 xdydx 2 xy y x 2 dx x 2 x 2 2 4 4 2 x x 2 2 x x dx 4 xdx 2 x 2 1 2 0 1 2 0 4 0 32 ESE 318 02 Spring 2016 6 Find work integral over the closed path as indicated 2 2 F dr F y x x z C 0 0 1 to 1 0 1 to 1 1 1 back to 0 0 1 C See that the curve is a flat triangle That is z is constant over the enclosed surface z 1 The region in the x y plane is a simple right triangle where y x along the hypotenuse i j k 0 1 2 x 2 y z z 0 curlF x y z x y y2 x2 x z 1 x z z F dr S curlF ndS R P x Q y R dxdy 0 0 2 x 2 y dydx C 1 2 xy y 2 1 1 2 2 2 y 0 dx 2 x x dx x dx x 0 0 1 3 0 7 Zill 9 16 3 Use spherical coordinates F x 3 y 3 z 3 divF 3x 2 3 y 2 3z 2 3 x 2 y 2 z 2 F ndS divFdV 3 x S D 2 y 2 z 2 dV D dV 2 sin d d d 2 x2 y2 z2 2 a 2 a 4 2 4 F ndS 3 dV 3 sin d d d 3 d d sin d S D 0 0 0 0 0 0 3 2 15 a 5 cos 0 3 2 15 a 5 2 125 a 5 8 Zill 9 16 7 Use cylindrical coordinates F y 3 x 3 z 3 divF 0 0 3z 2 3z 2 z 4 x2 y2 4 r …