# 009 100 points evaluate the double integral i

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Unformatted text preview: 009 10.0 points Evaluate the double integral I = integraldisplay integraldisplay D (3 x + 4) dA when D is the bounded region enclosed by y = x and y = x 2 . 1. I = 5 12 2. I = 3 4 3. I = 7 12 4. I = 1 4 5. I = 11 12 correct Explanation: The area of integration D is the shaded region in the figure To determine the limits of integration, there- fore, we have first to find the points of inter- section of the line y = x and the parabola y = x 2 . These occur when x 2 = x , i.e. , when x = 0 and x = 1. Thus the double integral can be written as a repeated integral I = integraldisplay 1 bracketleftbiggintegraldisplay x x 2 (3 x + 4) dy bracketrightbigg dx, integrating first with respect to y . After inte- gration this inner integral becomes bracketleftBig (3 x + 4) y bracketrightBig x x 2 = (3 x + 4)( x − x 2 ) = 4 x − x 2 − 3 x 3 . Thus I = integraldisplay 1 (4 x − x 2 − 3 x 3 ) dx = bracketleftbigg 2 x 2 − 1 3 x 3 − 3 4 x 4 bracketrightbigg 1 . Consequently, I = 11 12 . 010 10.0 points meier (rkm597) – HW 13 – Odell – (57000) 7 Find the volume, V , of the solid under the graph of the function f ( x, y ) = 1 x + y + 4 and over the triangular region A enclosed by the graphs of x = 1 , x = 6 , x + y = 2 , y + 4 = 0 . 1. V = 6 − ln 6 2. V = 7 − ln 6 3. V = 5 + ln 6 4. V = 6 + ln 6 5. V = 5 − ln 6 correct Explanation: The volume of the solid is the value of the double integral I = integraldisplay integraldisplay A f ( x, y ) dydx . As the region of integration, A , is given by (1 , − 4) (6 , − 4) (1 , 1) ( x, − 4) ( x, 2 − x ) (not drawn to scale) the double integral can be written as the repeated integral I = integraldisplay 6 1 parenleftbiggintegraldisplay 2- x- 4 1 x + y + 4 dy parenrightbigg dx , integrating first with respect to y from y = − 4 to y = 2 − x . Now the inner integral is equal to bracketleftBig ln( x + y + 4) bracketrightBig 2- x- 4 = ln6 − ln x. Thus I = integraldisplay 6 1 braceleftBig ln 6 − ln x bracerightBig dx = 5 ln6 − bracketleftBig x ln x − x bracketrightBig 6 1 . Consequently, V = 5 − ln 6 . 011 10.0 points The graph of f ( x, y ) = 3 xy over the bounded region A in the first quad- rant enclosed by y = radicalbig 16 − x 2 and the x, y-axes is the surface Find the volume of the solid under this graph over the region A . 1. Volume = 192 cu. units 2. Volume = 48 cu. units 3. Volume = 64 cu. units meier (rkm597) – HW 13 – Odell – (57000) 8 4. Volume = 96 cu. units correct 5. Volume = 24 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 4 parenleftBig integraldisplay √ 16- x 2 3 xy dy parenrightBig dx....
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