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Unformatted text preview: moore (jwm2685) – HW13 – gilbert – (55485) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This assignment (for credit) as well as the (noncredit) 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 xyplane 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 crosssection 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 crosssection is a square having sidelength 2 x . Thus the area of crosssection 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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 Spring '07
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