But after integration the inner integral be-comesbracketleftBig6yex2bracketrightBigx0= 6xex2,soI=integraldisplay206xex2dx=bracketleftbigg126ex2bracketrightbigg20.Consequently,I= 3(e4-1).02310.0pointsEvaluate the double integralI=integraldisplay integraldisplayA8x36 +y2dydxwhenAis the shaded region inxy

perez (cp32376) – HW10 – hong – (52960)14enclosed by the graphs ofy=x2,x= 0,y= 4.1.I= 2 ln1792.I= 2 ln1493.I= ln1494.I= 2 ln139correct5.I= ln1796.I= ln139Explanation:Since a trigonometric substitution is neededto integrate the functionf(x, y) =8x36 +y2with respect toy, we representIas a repeatedintegral, integrating first with respect toxandthen with respect toy.Now integrating first with respect toxforfixedymeans integrating along the dashedline segment from (0, y) to (y1/2, y) lying in-side the shaded regionxyand then fromy= 0 toy= 4.Thus as arepeated integralI=integraldisplay40parenleftBigintegraldisplayy1/208x36 +y2dxparenrightBigdyNow after integration the inner integral be-comesbracketleftBig4x236 +y2bracketrightBigy1/20=4y36 +y2,and soI=integraldisplay404y36 +y2dy=bracketleftBig2 ln(36 +y2)bracketrightBig40.Consequently,I= 2 ln139.

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