smt3 - Ma 233: Calculus III Solutions to Midterm...

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Ma 233: Calculus III Solutions to Midterm Examination 3 Profs. Krishtal, Ravindra, and Wickerhauser 17 questions on 17 pages Tuesday, April 12th, 2005 1. Evaluate the iterated integral Z 2 1 Z 2 1 ( x + 2 y ) - 3 dxdy (a) 1 4 (b) 1 15 (c) 1 20 (d) 3 20 (e) 1 24 (f) 1 60 (g) 1 80 (h) 1 240 Solution: Z 2 y =1 Z 2 x =1 ( x + 2 y ) - 3 dxdy = Z 2 y =1 " ( x + 2 y ) - 2 - 2 # 2 x =1 dy = Z 2 y =1 " (2 + 2 y ) - 2 - 2 - (1 + 2 y ) - 2 - 2 # dy = " (2 + 2 y ) - 1 ( - 2)( - 1)(2) - (1 + 2 y ) - 1 ( - 2)( - 1)(2) # 2 1 = 1 4 h (2 + 4) - 1 - (2 + 2) - 1 + (1 + 4) - 1 - (1 + 2) - 1 i 2 1 = 1 80 . 2 1
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2. Calculate the volume under the elliptic paraboloid z = 2 x 2 + 3 y 2 and over the rectangle R = [ - 2 , 2] × [ - 3 , 3]. (a) 180 (b) 210 (c) 240 (d) 270 (e) 280 (f) 300 (g) 320 (h) 360 Solution: The volume is the integral of the height function f ( x,y ) = z - 0 over the base rectangle R . This may be evaluated by iterated integration: Z 2 x = - 2 Z 3 y = - 3 (2 x 2 + 3 y 2 ) dydx = Z 2 x = - 2 h 2 x 2 y + y 3 i 3 - 3 dx = Z 2 x = - 2 h 12 x 2 + 54 i dx = h 4 x 3 + 54 x i 2 x = - 2 = 280 . 2 2
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3. Evaluate the double integral I = Z Z D xy dA where D is the triangular region with vertices (0 , 0) , (2 , 0) , (2 , 1). (a) 1 (b) 1 / 2 (c) 1 / 3 (d) 1 / 4 (e) 1 / 5 (f) 1 / 6 (g) 1 / 8 (h) 1 / 16 Solution: Write D = { ( x,y ) : 0 x 2 , 0 y x/ 2 } and compute the double integral by Fubini’s theorem as an iterated integral in the y-variable first: I = Z 2 x =0 Z x/ 2 y =0 xy dydx = Z 2 x =0 " xy 2 2 # x/ 2 0 dx = Z 2 x =0 " x 3 8 # dx = " x 4 32 # 2 0 = 1 2 . Also by Fubini’s theorem, the iterated integral in the other order gives the same result. 2 3
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4. Evaluate the integral by reversing the order of integration. Z 2 y =0 Z 2 x = y e - x 2 dxdy (a) (1 - e - 4 ) / 2 (b) (1 - e - 2 ) / 2 (c) (1 - e - 2 ) / 4 (d) ( e - 2 - e - 4 ) / 2 (e) ( e - 4 - e - 2 ) / 2 (f) ( e - 4 - e - 2 ) / 4 (g) ( e - 4 - e - 2 ) / 3 (h) 1 - e - 4 Solution: The domain of integration has the two descriptions { 0 y 2 ,y x 2 } and { 0 x 2 , 0 y x } . Use the second to compute the iterated integral in the other order as suggested: Z 2 y =0 Z 2 x = y e - x 2 dxdy = Z 2 x =0 Z x y =0 e - x 2 dydx = Z 2 x =0 xe - x 2 dx = - 1 2 e - x 2 ± ± ± ± 2 0 = (1 - e - 4 ) / 2 . 2 4
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Z Z R cos( x 2 + y 2 ) dA where R is the region 4 x 2 + y 2 9. (a) 2 π sin 9 (b) 2 π sin 5 (c) 2 π sin 4 (d) 2 π sin 1 (e) π sin 9 (f) 2 π (sin 9 - sin 4) (g) π (sin 9 - sin 4) (h) 2 π (sin 3 - sin 2) (i) π (sin 3 - sin 2) Solution: Let x = r cos θ , y = r sin θ , so x 2 + y 2 = r 2 . In these polar coordinates ( r,θ ), dA = r drdθ and R = { ( r,θ ) : 2 r 3 } , giving Z Z R cos( x 2 + y 2 ) dA = Z 2 π θ =0 Z 3 r =2 cos( r 2 ) r drdθ = 2 π " sin( r 2 ) 2 # 3 2 = π [sin 9 - sin 4] . 2
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smt3 - Ma 233: Calculus III Solutions to Midterm...

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