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Unformatted text preview: MATH 52 FINAL EXAM SOLUTIONS (AUTUMN 2003) 1. Evaluate the integral by reversing the order of integration integraldisplay 1 integraldisplay 3 3 y e x 2 dxdy. Solution. integraldisplay 1 integraldisplay 3 3 y e x 2 dxdy = integraldisplay 3 integraldisplay x/ 3 e x 2 dxdy = integraldisplay 3 bracketleftBig e x 2 y bracketrightBig y = x/ 3 y =0 dx = integraldisplay 3 parenleftBig x 3 parenrightBig e x 2 dx = 1 6 e x 2 vextendsingle vextendsingle vextendsingle vextendsingle 3 = e 9 1 6 2. Rewrite the following integral in rectangular coordinates. integraldisplay π/ 4 integraldisplay 2 cos θ +sin θ r 2 cos θ dr dθ Do not evaluate the integral. Solution. The equation r = 2 / (cos θ + sin θ ) becomes x + y = 2 in rectangular co ordinates. Since r drdθ becomes dA , the integrand r cos θ becomes x in rectangular coordinates, so the integral equals integraldisplay 1 integraldisplay 2 y y xdxdy or integraldisplay 1 integraldisplay x x dydx + integraldisplay 2 1 integraldisplay 2 x xdydx 3. (a) Consider the change of variables defined by u = y x , v = xy. Find the Jacobian ∂ ( x, y ) ∂ ( u, v ) in terms of u and v . Solution. First we find the Jacobian ∂ ( u,v ) ∂ ( x,y ) . It is equal to the determinant bracketleftbigg y/x 2 y 1 /x x bracketrightbigg = 2 y/x = 2 u Hence the Jacobian ∂ ( x, y ) ∂ ( u, v ) = 1 2 u . 1 (b) Use your answer to part (a) to find the area of the region in the first quadrant bounded by the lines y = x and y = 2 x and the curves xy = 2 and xy = 3. Solution. In the uvcoordinates the region becomes 1 ≤ u ≤ 2 and 2 ≤ v ≤ 3. Hence its area is: integraldisplay 2 u =1 integraldisplay 3 v =2 1 2 u dv du = integraldisplay 2 u =1 1 2 u du = bracketleftbigg 1 2 ln u bracketrightbigg 2 u =1 = 1 2 ln2 4. Find the area of the part of hyperbolic paraboloid z = y 2 x 2 that lies between the cylinders x 2 + y 2 = 1 and x 2 + y 2 = 4. Solution. z = f ( x, y ) = y 2 x 2 with 1 ≤ x 2 + y 2 ≤ 4. Then A ( S ) = integraldisplayintegraldisplay D radicalbig 1 + 4 x 2 + 4 y 2 dA = integraldisplay 2 π integraldisplay 2 1 √ 1 + 4 r 2 r drdθ = integraldisplay 2 π dθ integraldisplay 2 1 √ 1 + 4 r 2 r dr = 2 π bracketleftbigg 1 12 (1 + 4 r 2 ) 3 / 2 bracketrightbigg 2 1 = π 6 (17 √ 17 5 √ 5) 5. (a) Let E be the solid bounded by the paraboloids...
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 Winter '08
 Demanet,L;Wieczorek,W
 Math, Calculus, Vector Calculus, Cos, dx, Stokes' theorem

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