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# h13 - moore(jwm2685 – HW13 – gilbert –(55485 1 This...

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Unformatted text preview: moore (jwm2685) – HW13 – gilbert – (55485) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This assignment (for credit) as well as the (non-credit) Integration Review will you lots of practice with integration. 001 10.0 points The graph of the function z = f ( x, y ) = 7- x is the plane shown in z 7 x y Determine the value of the double integral I = integraldisplay integraldisplay A f ( x, y ) dxdy over the region A = braceleftBig ( x, y ) : 0 ≤ x ≤ 3 , ≤ y ≤ 4 bracerightBig in the xy-plane by first identifying it as the volume of a solid below the graph of f . 1. I = 69 cu. units 2. I = 66 cu. units correct 3. I = 68 cu. units 4. I = 67 cu. units 5. I = 70 cu. units Explanation: The double integral I = integraldisplay integraldisplay A f ( x, y ) dxxdy is the volume of the solid below the graph of f having the rectangle A = braceleftBig ( x, y ) : 0 ≤ x ≤ 3 , ≤ y ≤ 4 bracerightBig for its base. Thus the solid is the wedge z 3 7 x y (3 , 4) and so its volume is the area of trapezoidal face multiplied by the thickness of the wedge. Consequently, I = 66 cu. units . keywords: 002 10.0 points Determine the value of the iterated integral I = integraldisplay 3 braceleftBig integraldisplay 3 1 (1 + 2 xy ) dx bracerightBig dy . 1. I = 44 2. I = 38 3. I = 36 4. I = 42 correct 5. I = 40 moore (jwm2685) – HW13 – gilbert – (55485) 2 Explanation: Integrating with respect to x and holding y fixed, we see that integraldisplay 3 1 (1 + 2 xy ) dx = bracketleftBig x + x 2 y bracketrightBig x =3 x =1 . Thus I = integraldisplay 3 braceleftBig 2 + 8 y bracerightBig dy = bracketleftBig 2 y + 4 y 2 bracketrightBig 3 . Consequently, I = braceleftBig 6 + 36 bracerightBig = 42 . keywords: 003 10.0 points The volume of a solid can often be deter- mined by integration even when the cross- section of the solid is not necessarily a circle. In the following figure x y z the base is the circle x 2 + y 2 = 4 and the cross-section perpendicular to the y- axis is a square. Set up a definite integral expressing the volume, V , of the solid. 1. volume = integraldisplay 2- 2 (4- y 2 ) dy cu.units 2. volume = integraldisplay 2- 2 4(4- y 2 ) dy cu.units correct 3. volume = integraldisplay 2- 2 (4 + y 2 ) dy cu.units 4. volume = integraldisplay 2- 2 2(4- y 2 ) dy cu.units 5. volume = integraldisplay 2- 2 4(4 + y 2 ) dy cu.units Explanation: At the point P = ( x, y ) b P x y z on the base of the solid the cross-section is a square having side-length 2 x . Thus the area of cross-section is a square having area 4 x 2 , so the volume of the solid is given by the definite integral V = integraldisplay 2- 2 4 x 2 dy....
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h13 - moore(jwm2685 – HW13 – gilbert –(55485 1 This...

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