1873_solutions

# Function s is riemann integrable on a b if and only

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Unformatted text preview: ich the equations D 1 F x, y, z  22xz 2 x y 2yz D 2 F x, y, z  2  sin yz  yz cos yz x  y2 D 3 F x, y, z  log x 2  y 2  y 2 cos yz  2z hold whenever x 2  y 2 0. From the first of these equations we see that D 1 F x, y, z z log x 2  y 2  0 for each point x, y, z and therefore, for each point y, z , the expression F x, y, z is independent of x. We write F x, y, z z log x 2  y 2 as  y, z . In other words, F x, y, z  z log x 2  y 2   y, z for each point x, y, z . Thus D 2 F x, y, z  2yz  D 2  y, z x2  y2 and we conclude that 2yz 2yz  D 2  y, z  2  sin yz  yz cos yz 2 x  y2 y x2 which gives us D 2  y, z  sin yz  yz cos yz and so we can express  y, z in the form y sin yz   z which gives us F x, y, z  z log x 2  y 2  y sin yz   z . Therefore D 3 F x, y, z  log x 2  y 2  y 2 cos yz   z which gives us log x 2  y 2  y 2 cos yz   z  log x 2  y 2  y 2 cos yz  2z and we conclude that the function F defined by the equation 407 z log x 2  y 2 F x, y, z  z log x 2  y 2  y sin yz  z 2 whenever x 2  y 2 0 is a potential function for f. b. Given that f is the function defined in part a and that  is any piecewise smooth curve in the set x, y, z R3 x2  y2  0 running from the point 1, 0, 1 to the point 3,  6 , 2 , evaluate the integral Þ f x  dx. Using the function v that we found in part a we see that Þ f x  dx  v 3,  , 3 v 1, 0, 1  3 log 36  3 log 324   2  1   8 6 6 2. Find a potential function for the function f defined to be 3 2ux 2 yz ux 2 y 2 vux 2 x2u2 , , 2ux ln x 2  y 2 z  uv  2 2ux , , 2 z  uv x 2  y 2 z  uv 2 z  uv x 2  y 2 z  uv x 2  y 2 z  uv y x y at every point u, v, x, y, z or R 5 for which the latter expression is defined. If F is a potential function for f then, for each point u, v, x, y, z we want 2 D 1 F u, v, x, y, z  x 2 ln x 2  y 2 z  uv  2 vux 2 z  uv x y x2u2 D 2 F u, v, x, y, z  2 x  y 2 z  uv 3 D 3 F u, v, x, y, z  2ux ln x 2  y 2 z  uv  2 2ux 2 z  uv x y 2ux 2 yz D 4 F u, v, x, y, z  2 x  y 2 z  uv ux 2 y 2 D 5 F u, v, x, y, z  2 x  y 2 z  uv The last of these five equations gives us F u, v, x, y, z  ux 2 log x 2  y 2 z  uv   1 u, v, x, y . x 2 ln x 2  y 2 z  uv  x2 To cut a long story short we observe that if F u, v, x, y, z  ux 2 log x 2  y 2 z  uv for each point u, v, x, y, z then the function F is a potential for f. 3. In this exercise you will show that the necessary condition for exactness is sufficient if the domain of the given function is a rectangle. Suppose that f  f 1 , f 2 has continuous partial derivatives on the rectangle a 1 x b 1 and a 2 y b 2 E  x, y and that the condition holds for every point x, y D 1 f 2 x, y  D 2 f 1 x, y E. Prove that the function F defined at each point x, y F x, y  Þa x 1 f 1 t, y dt  Þ E by the equation y a2 f 2 a 1 , t dt is a potential of f on E. The result follows at once from the theorem on differentiation of a partial Riemann integral. 4. Repeat the preceding exerci...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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