Integraldisplay 6 1 parenleftbig integraldisplay ln y

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integraldisplay61parenleftBigintegraldisplaylny0f(x, y)dxparenrightBigdy6.I=integraldisplayln60parenleftBigintegraldisplay6eyf(x, y)dxparenrightBigdyExplanation:The region of integration is the set of allpointsbraceleftBig(x, y) : 1yex,0xln 6bracerightBigin the plane bounded by the graph ofy=exand the the linesy= 1,y= 6,x= ln 6.This is the shaded region inxy16ln 6Integration is taken first with respect toyforfixedxalong the solid vertical line.
byrne (hcb539) – Homework 11 – spice – (54070)5To change the order of integration, now fixyand letxvary along the solid horizontal lineinxy16ln 6(not drawn to scale).Since the inverse ofy=exisx= lny, integration inxis alongthe line from (lny, y) to (ln 6, y) for fixedy,and then fromy= 1 toy= 6.Consequently, after changing the order ofintegration,I=integraldisplay61parenleftBigintegraldisplayln6lnyf(x, y)dxparenrightBigdy.keywords: double integral, reverse order inte-gration, exponential function, log function,00910.0pointsEvaluate the double integralI=integraldisplay integraldisplayD(1-cos(x+y))dAwhenDis the bounded region enclosed by thelinesx+y=π,x= 0 andy= 0.1.I=π+ 12.I=12π2+ 2correct3.I=12π2-24.I=π2-15.I=π+ 26.I=12π-1Explanation:The area of integrationDis the shadedregion in the figurexyππsince the linex+y=πintersects the axesatx=πandy=π.To expressIas arepeated integral, therefore, we can first fixxand then integrate with respect toyalong thedashed line fromy= 0 toy=π-x; finally,we integrate with respect toxfromx= 0 tox=π. For thenI=integraldisplayπ0parenleftBigintegraldisplayπ-x0(1-cos(x+y))dyparenrightBigdx.After integration the inner integral becomesbracketleftBigy-sin(x+y)bracketrightBigπ-x0= (π-x)-(sinπ-sinx)=π-x+ sinx .ThusI=integraldisplayπ0(π-x+ sinx)dx=bracketleftBigπx-12x2-cosxbracketrightBigπ0.Consequently,I=12π2+ 2.
byrne (hcb539) – Homework 11 – spice – (54070)601010.0pointsEvaluate the integralI=integraldisplay integraldisplayDbraceleftBig(π+ 2 tan-1parenleftBigyxparenrightBigbracerightBigdxdywhenDis the region in the first quadrantinside the circlex2+y2= 4.1.I=34π2.I=32π3.I= 3π24.I= 3π5.I=34π26.I=32π2correctExplanation:In Cartesian coordinates the region of inte-gration isbraceleftBig(x, y) : 0yradicalbig4-x2,0x2bracerightBig,which is the shaded region inxy2θrOn the other hand, in polar coordinates theregion of integration isbraceleftBig(r, θ) : 0r2,0θπ/2bracerightBig,whiletan-1parenleftBigyxparenrightBig=θ .I=integraldisplay20integraldisplayπ/20(π+ 2θ)rdθdr=integraldisplay20bracketleftBigπθ+θ2bracketrightBigπ/20rdr=34π2integraldisplay20r dr .Thus in polar coordinates,

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