# Sin x e y 10 y is actually a gradient field by

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, sin( x ) e y 10 y > is actually a gradient field by producing a function φ ( x , y ) such that ∇φ ( x , y ) = F ( x , y ) for all ( x , y ) in the plane. Evidently y (cos( x ) e y ) cos( x ) e y x (sin( x ) e y 10 y ) for each ( x , y ) 2 . Then ∂φ x ( x , y ) cos( x ) e y φ ( x , y ) cos( x ) e y dx sin( x ) e y c ( y ). Therefore, sin( x ) e y 10 y ∂φ y ( x , y ) y (sin( x ) e y ) dc dy ( y ) dc dy ( y ) 10 y c ( y ) 5 y 2 c 0 for some number c 0 . Hence, φ ( x , y ) = sin( x )e y + 5 y 2 + c 0 . (b) Using the Fundamental Theorem of Line Integrals, evaluate the path integral below, where C is any smooth path from the origin to the point ( π /2,ln(2)). [ WARNING: You must use the theorem to get any credit here.] C (cos( x ) e y ) dx (sin( x ) e y 10 y ) dy φ ( x , y ) ( π /2,ln(2)) (0,0) φ ( π /2,ln(2)) φ (0,0) sin( π /2) e ln(2) 5(ln(2)) 2 2 5(ln(2)) 2 2 5ln 2 (2).

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TEST3/MAC2313 Page 6 of 6 _________________________________________________________________ 10. (10 pts.) Use the substitution u = (1/2)( x + y ) and v = (1/2)( x - y ) to evaluate the integral below, where R is the bound region enclosed by the triangular region with vertices at (0,0), (2,0), and (1,1). R cos 1 2 ( x y ) dA R cos 1 2 ( x y ) dA x , y T 1 ( R ) cos( u ) ( x , y ) ( u , v ) dA u , v 1 0 u 0 2cos( u ) dvdu 1 0 2 u cos( u ) du 2 u sin( u ) 1 0 1 0 2sin( u ) du 2sin(1) 2cos(1) 2 T -1 : u = ( x + y )/2, v = ( x - y )/2 T : x = u + v , y = u - v Bounding Curves: y = x v = 0 y = 2 - x u = 1 y = 0 v = u ( x , y ) ( u , v ) x u x v y u y v 1 1 1 1 (1)( 1) (1)(1) 2. Note that S = T -1 (R). _________________________________________________________________ Silly 10 Point Bonus: Become a polar explorer. Reveal the details of how one can obtain the exact value of the definite integral 0 e x 2 dx even though there is no elementary anti-derivative for the function f( x ) e x 2 .
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