MVC Chapter 6 Quiz NAME: Instructions: SHOW ALL WORK !! We don’t really need a calculator, do we? 2t 3 Let the curve C be given by: x(t ), y(t ), z (t ) t , t 2 , with 1 t 2 . Set up each 3 integral below as a single integral in terms of t. #1. a. ds xyz ds x(t ) 2 y(t ) 2 z(t ) 2 dt xyz ds (t ) t 1 b. C C 2 C C 2 1 (2t ) 2 (2t 2 ) 2 dt 2t 3 2 2 2 1 (2t ) (2t ) dt 3 2 1 2 6 t 1 4t 2 4t 4 dt 3 F d s given Fx, y, z = zi + yj + xk. F d s zdx ydy xdz C 2 2t 3 dt t 2 (2tdt ) t (2t 2 dt ) 3 1 2 2t 3 4t 3 dt 3 1 #2. Find a parameterization for each of the following curves: a. The line segment from (1, 2, 4) to ( 2, 2, 0) x 3t 1 y 2 0 t 1 z 4t 4 b. The curve that starts at the point (0, 0, 0) and goes to the point (1, 0,1) along the circular paraboloid z x 2 y 2 . x t cos t y t sin t 0 t 1 z t2 MVC Let C be a curve in the plane starting at (0,0), moving to (1,1) along the curve y = x3, and then returning to the origin along the straight line y = x. #3. a. Parameterize the path (in two pieces, most likely) to express the line integral 2 2 x y dx xy dy as an integral or integrals in the single variable t. DON’T integrate. C 2x 2 y dx xy dy C 2x 2 C1 y dx xy dy 2x 2 y dx xy dy C2 1 1 0 0 2t 2t 3 dt tt 3 3t 2 dt 2t 2t dt tt dt 1 (2t 5 3t 6 2t 3 t 2 ) dt 0 b. Apply Green’s Theorem to the integral in part a to obtain a double integral, making sure to provide appropriate limits of integration. DON’T integrate. 1 x N M 2 2 x y dx xy dy C 0 3 x y x 1 x N M x y 0 x3 1 x y 2 x 2 dydx 0 x3 MVC #3. (continued) Let C be a curve in the plane starting at (0,0), moving to (1,1) along the curve y = x3, and then returning to the origin along the straight line y = x. c. Given the vector field F( x, y) 2 x 2 y i xy j , write the integral(s) in the single variable t you would need to evaluate to find the outward flux of F across the curve C. DON’T integrate. N dx M dy xy dx 2 x C 2 y dy C xy dx 2 x 2 y dy xy dx 2 x 2 y dy C2 C1 1 1 tt dt 2t t 3t dt tt dt 2t 2t dt 3 2 3 2 0 0 1 (t 4 6t 7 t 2 2t 3 )dt 0 d. Apply the divergence form of Green’s theorem to obtain a double integral that would calculate this flux. DON’T integrate. 1 x N dx Mdy div( F )dA C 0 x3 1 x M N dA x y 0 x3 1 x 4 xy x dA 0 x3 MVC #4 Use Green’s Theorem (with a thoughtful choice of F ( x, y ) M ( x, y )iˆ N ( x, y ) ˆj ) to set up a line integral that gives the area of the region inside the curve C : x(t ) t sin t , y (t ) 3cos t;0 t 2 . Write the line integral in terms of t. [Note that C is simple, smooth, closed curve]. Let F ( x, y ) M ( x, y )iˆ N ( x, y ) ˆj xjˆ; i.e. M ( x, y ) 0 and N ( x, y ) x . Then, N M Area( D) 1dA dA M ( x, y)dx N ( x, y)dy y D D x C 2 xdy (t sin t )(3sin t )dt C 0 #5. Set up a double integral that represents the area of the parametrized surface: X (2 s 3t 1, t , s t ); 1 s 1,0 t 1 MVC x2 y 2 #6. Find a parameterization for the portion of the surface z that lies above the region x2 1 1 4 in the xy-plane bounded by the curves y x, y 3x, y , and y . 3 x x MVC