c3-t3-a

# Y ? g x y 2 xy 2 2 yx 2 ? 2 x 2 y 0 2 xy y x y x x 0

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y ) = λ∇ g ( x , y ) < 2 xy 2 , 2 yx 2 > = λ < 2 x ,2 y > 0 = 2 xy ( y - x )( y + x ) x = 0 or y = 0 or y = x or y = - x by performing a little routine algebraic magic. Solving each of the systems consisting of (a) x 2 + y 2 = 1 and x = 0 , (b) x 2 + y 2 = 1 and y = 0 , (c) x 2 + y 2 = 1 and y = x , (d) x 2 + y 2 = 1 and y = - x , yields the eight (8) desired (constrained) critical points. Since the Extreme-value Theorem ensures we have absolute extrema, we only need to know the function values at these points. Here are the solutions to the systems: If (x,y) is an (a) or (b) pair, then f(x,y) = 0, the minimum value. Taking into account the interior analysis, we see the minimum actually occurs on the x- and y- axis points in the closed disk. If (x,y) is a (c) or (d) pair, then f(x,y) = 1/4, the maximum value.//

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TEST3/MAC2313 Page 5 of 5 ______________________________________________________________________ 9. (10 pts.) Reveal all the details in using the substitution u = x / a and v = y / b with a > 0 and b > 0 to evaluate the integral below, where R is the elliptical region defined by x 2 a 2 y 2 b 2 1. R 1 dA R 1 dA x , y T 1 ( R ) 1 ( x , y ) ( u , v ) dA u , v S ab dA u , v ( ab ) Area ( S ) ab π , since S = T -1 (R) is the closed disk of radius 1 centered at the origin in the uv-plane. T -1 : u = x / a , v = y / b T : x = au , y = bv Bounding Curve: x 2 a 2 y 2 b 2 1 u 2 v 2 1 ( x , y ) ( u , v ) x u x v y u y v a 0 0 b ( a )( b ) (0)(0) ab ______________________________________________________________________ 10. (10 pts.) Compute the surface area of the portion of the paraboloid defined by z = 16 - x 2 - y 2 that lies above the xy-plane. Obviously we have passed to polar coordinates along the way and then used the u-substitution u = 1 + 4 r 2 to accomplish our goal. The intersection of the paraboloid with the xy-plane is an easy to understand circle centered at the origin with radius 4. This provides the r and θ limits of integration. ______________________________________________________________________ Silly 10 Point Bonus: Become a polar explorer. Compute the exact value of the double integral where R is the region in the first quadrant bounded by the lines y = x /3 and y = 2 x and the circle centered at the origin with a radius of 2. [A brief indication of an answer is on the bottom of Page 1 of 5.]
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y Î g x y 2 xy 2 2 yx 2 Î 2 x 2 y 0 2 xy y x y x x 0 or y 0...

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