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HW_13_Key - Version 084 – Homework 13 – Gilbert...

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Unformatted text preview: Version 084 – Homework 13 – Gilbert – (59825) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 10.0 points Evaluate the iterated integral I = integraldisplay 3 π/ 2 integraldisplay cos θ 3 e sin θ drdθ . 1. I = 3 parenleftBig 1 e- 1 parenrightBig correct 2. I = 0 3. I = 3( e- 1) 4. I = e- 3 5. I = 3 e 6. I = 1 e- 3 Explanation: After simple integration integraldisplay cos θ 3 e sin θ dr = bracketleftBig 3 r e sin θ bracketrightBig cos θ = 3 cos θ e sin θ . In this case, I = integraldisplay 3 π/ 2 3 cos θ e sin θ dθ = bracketleftBig 3 e sin θ bracketrightBig 3 π/ 2 . Consequently, I = 3 parenleftBig 1 e- 1 parenrightBig . 002 10.0 points Find the value of the integral I = integraldisplay integraldisplay A (6 x 2- 2 y 2 ) dxdy when A = braceleftBig ( x, y ) : 0 ≤ y ≤ 2 x, ≤ x ≤ 2 bracerightBig . 1. I = 163 6 2. I = 80 3 correct 3. I = 27 4. I = 161 6 5. I = 82 3 Explanation: As an iterated integral, I = integraldisplay 2 bracketleftbiggintegraldisplay 2 x (6 x 2- 2 y 2 ) dy bracketrightbigg dx = integraldisplay 2 bracketleftbigg 6 x 2 y- 2 3 y 3 bracketrightbigg 2 x dx = integraldisplay 2 20 3 x 3 dx . Consequently, I = 80 3 . 003 10.0 points The graph of f ( x, y ) = 4 xy over the bounded region A in the first quad- rant enclosed by y = radicalbig 9- x 2 and the x, y-axes is the surface Version 084 – Homework 13 – Gilbert – (59825) 2 Find the volume of the solid under this graph over the region A . 1. Volume = 81 8 cu. units 2. Volume = 81 2 cu. units correct 3. Volume = 27 cu. units 4. Volume = 81 4 cu. units 5. Volume = 81 cu. units Explanation: The volume of the solid under the graph of f is given by the double integral V = integraldisplay integraldisplay A f ( x, y ) dxdy, which in turn can be written as the repeated integral integraldisplay 3 parenleftBig integraldisplay √ 9- x 2 4 xy dy parenrightBig dx. Now the inner integral is equal to bracketleftBig 2 xy 2 bracketrightBig √ 9- x 2 = 2 x (9- x 2 ) . Thus V = 2 integraldisplay 3 x (9- x 2 ) dx = bracketleftBig- 1 2 (9- x 2 ) 2 bracketrightBig 3 . Consequently, Volume = 81 2 cu. units. 004 10.0 points Evaluate the double integral I = integraldisplay integraldisplay A (2 x + 3 y ) dxdy when A is the region enclosed by the graphs of x = 1 , x- y = 1 , y = 1 . 1. I = 4 3 2. I = 5 3 3. I = 2 4. I = 1 5. I = 7 3 correct Explanation: The graphs of x = 1 , x- y = 1 , y = 1 are straight lines intersecting at the points (1 , 0) , (1 , 1) , (2 , 1) . Thus the region of integration is the shaded triangular region (1 , 0) (2 , 1) (1 , 1) ( x, x- 1) ( x, 1) ( x, 0) Version 084 – Homework 13 – Gilbert – (59825) 3 so I can be written as the repeated integral I = integraldisplay 2 1 parenleftbiggintegraldisplay 1 x- 1 (2 x + 3 y ) dy parenrightbigg dx , integrating first with respect to y from y = x- 1 to y = 1. Now the inner integral is equal= 1....
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