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(a) Each value ofxbetween 0 andπgives a vertical strip, of thicknessdx, thatgenerates a disk when rotated about thex-axis. Thus we use the formula for volume ofrotation via the disk method,V=Zπ0πR2dxIn our case, the radius of the disk is given by the distance from (x,sinx) to (x,0) foreachx, and is therefore given byR=sinx. HenceV=Zπ0π(sinx)2dxWe can compute this integral by applying the identity sin2(x)=12(1-cos(2x)) (seeProblem 1, part (2)):V=πZπ0sin2(x)dx=πx2-sin(2x)4!x=πx=0=ππ2-sin(2π)4-02+sin(0)4!=ππ2-0-0+0=π2220
(b) Since we are now rotating about the (vertical) linex=-2, our vertical stripsno longer generate disks, but instead generate cylindrical shells. The formula for thevolume of rotation via the cylindrical shell method isV=Zπ02π·R·hdxHereRis the radius of rotation, given by the distance fromxto the line of rotation at-2; this distance isR=x+2. The heighthis the height of the vertical strip, equal tosinxas before in part (a). ThusV=Zπ02π(x+2) sin(x)dx=2πZπ0xsin(x)dx+2πZπ02 sin(x)dx=2πZπ0xsin(x)dx-4πcos(x)|x=πx=0=2πZπ0xsin(x)dx-(4π·(-1)-4π·(1))=2πZπ0xsin(x)dx+8πThe integralRπ0xsin(x)dxcan be evaluated by an integration by parts, settingu=xanddv=sin(x)dx, wherebydu=dxandv=-cos(x). This givesZπ0xsin(x)dx=-xcos(x)|x=πx=0-Zπ0(-cos(x))dx=-πcos(π)-0+Zπ0cos(x)dx=-π·(-1)+sin(x)|x=πx=0=π+(0-0)=πPutting all this together, we haveV=2πZπ0xsin(x)dx+8π=2π(π)+8π=2π2+8π2.As in part (1), we will use the disk/washer method for part (a) and the cylindricalshell method for part (b); although the functiony=x-5is more easily inverted than thefunction in part (a), and we could exchange disk/washer and cylindrical shell methodsby consideringx=y-1/5, it seems easier to integrate overxsince we are given theinformation about the region in terms ofyas a function ofx.21
(a) We have as above vertical strips stretching from (x,x-5) to (x,0) for eachxin[0,∞). However, unlike part (a), the axis of rotation is nowy=-2, not thex-axis;therefore the region under consideration does NOT border the axis of rotation, and wehave to use the washer methodV=Z∞1(πR2-πr2)dxHereRis the outer radius of rotation, given by the distance of the graph of the functiony=x-5from the liney=-2; soR=x-5+2. The inner radiusris the distance fromthe inner curvey=0 to the liney=-2, which isr=2 for allx. ThereforeV=πZ∞1((x-5+2)2-22)dx=πZ∞1(x-10+2x-5+4-4)dx=πZ∞1(x-10+2x-5)dx=πx-9-9+2x-4-4!x=∞x=1=π(0+0-(-1/9-1/4))=π(1/9+1/4)=13π36(b) Once again we use the cylindrical shells formulaV=Z∞12π·R·hdxwhereRis the radius of rotation, given by the distance fromxto they-axis; thereforeR=x. The height is given by the height of the vertical strip which, as in part (a), isgiven byh=x-5.