APPLICATIONS OF THE DEFINITE INTEGRAL


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1 APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its crosssection in plne pssing through (x,, ) nd prllel to the yzplne hs re A(x). To first order pproximtion, the volume of the slice of the object on the right to the plne of thickness x is then A(x) x so tht the volume of the solid is the limit of the elementl sum A(x) x. This being Riemnn sum, the volume is given by the formul V = A(x) dx. Exmple.. () The Pyrmid Aren in Memphis hs squre bse of side pproximtely 6 feet nd height of pproximtely 3 feet. Find the volume of the pyrmid with these mesurements. () Find the volume of sphere of rdius R... The Method of Disks. Suppose tht f(x) nd f is continuous on the intervl [, b]. Tke the region bounded by the curve y = f(x) nd the xxis, for x b, nd revolve it bout the xxis, generting solid. We cn find the volume of this solid by slicing it perpendiculr to the xxis nd recognizing tht ech cross section is circulr disk of rdius r = f(x), we then hve tht the volume of the solid is V = π[f(x)] dx.
2 APPLICATIONS OF THE DEFINITE INTEGRAL Since the cross sections of such solid of revolution re ll disks, we refer to this method of finding volume s the method of disks. Exmple.. () Revolve the region under the curve y = x on the intervl [, 4] bout the xxis nd find the volume of the resulting solid of revolution. () Find the volume of sphere of rdius R. (3) Find the volume of the solid generted by revolving the region bounded by y = x +, y =, x =, x = bout the xxis. In similr wy, suppose tht g(y) nd g is continuous on the intervl [c, d]. Then, revolving the region bounded by the curve x = g(y) nd the yxis, for c y d, bout the yxis genertes solid. Notice tht the cross sections of the resulting solid of revolution re circulr disks of rdius r = g(y). All tht hs chnged here is tht we hve interchnged the roles of the vribles x nd y. The volume of the solid is then given by V = ˆ d c π[g(y)] dy. Exmple.3. () Find the volume of the solid resulting from revolving the region bounded by the curves y = 4 x nd y = from x = to x = 3 bout the yxis. () Find the volume of the solid obtined by revolving the region bounded by the curves y = x 3 nd yxis nd y = 8 bout the yxis. (3) Find the volume of the solid generted by revolving the region bounded by y = 4 x, the positive xxis, nd the positive yxis, bout the yxis..3. The Method of Wshers. One compliction tht occurs in computing volumes is tht the solid my hve cvity or hole in it. Another occurs when region is revolved bout line other thn the xxis or the yxis. For exmple, consider the solid obtined by revolving the region bounded by the grphs of y = 4 x, x = nd y = bout the xxis, nd the line y =, respectively:
3 APPLICATIONS OF THE DEFINITE INTEGRAL 3 Exmple.4. () Let R be the region bounded by the grphs of y = 4 x, x = nd y =. Compute the volume of the solid formed by revolving R bout () the yxis, (b) the xxis, nd (c) the line y =. () Let R be the region bounded by y = 4 x nd y =. Find the volume of the solids obtined by revolving R bout ech of the following: () the yxis, (b) the line y = 3, (c) the line y = 7, nd (d) the line x = 3. (3) Find the volume of the solid obtined by revolving the region R bounded by the curves () y = x nd y = x bout the xxis, (b) y = x nd y = 8x bout the xxis, (c) y = 6 x, y = bout the y =, (d) x = 4 y nd yxis bout the x =, The technique used to solve the problems bove is slight generliztion of the method of disks nd is referred to s the method of wshers, since the cross sections of the solids look like wshers.. VOLUMES BY CYLINDRICAL SHELLS Let R denote the region bounded by the grph of y = f(x) nd the xxis on the intervl [, b], where < < b nd f(x) on [, b]. If we revolve this region bout the yxis, we get the solid shown below: If insted of tking cross section perpendiculr to the yxis, we tke cross section perpendiculr to the xxis, nd revolve it bout the yxis, we get cylinder. Recll tht the re of cylinder is given by: A(x) = πrh, where r is the rdius of the cylinder nd h is the height of the cylinder. We cn see tht the rdius is the x coordinte of the point on the curve, nd the height is the y coordinte of the curve. Hence A(x) = πxy = πxf(x), Therefore the volume is given by πxf(x) dx.
4 4 APPLICATIONS OF THE DEFINITE INTEGRAL Remrk.. Note tht for given solid, the vrible of integrtion in the method of shells is exctly opposite tht of the method of wshers. So, your choice of integrtion vrible will determine which method you use. Exmple.. () Revolve the region bounded by the grphs of y = x nd y = x in the first qudrnt bout the yxis. () Find the volume of the solid formed by revolving the region bounded by the grph of y = 4 x nd the xxis bout the line x = 3. (3) Let R be the region bounded by the grphs of y = x, y = x nd y =. Compute the volume of the solid formed by revolving R bout the lines () y =, (b) y =, (c) x = 3. (4) Let R be the region bounded by the grphs of y = x(x ) nd the xxis. Compute the volume of the solid formed by revolving R bout the yxis. (5) Let R be the region bounded by the grphs of y = 4, x =, x = 4, y =. Compute x the volume of the solid formed by revolving R bout the yxis. (6) Let R be the region bounded by the grphs of x = y, y =, x =. Compute the volume of the solid formed by revolving R bout the xxis. (7) Let R be the region bounded by the grphs of y = x nd y = x in the first qudrnt. Compute the volume of the solid formed by revolving R bout the yxis. (8) Set up n integrl for the volume of the solid tht results when the region bounded by the curve y = 3 + x x, the xxis, nd the yxis, is revolved bout () the xxis, (b) the yxis, (c) the line y =... Summry. We close this section with summry of strtegies for computing volumes of solids of revolution. Sketch the region to be revolved. Determine the vrible of integrtion (x if the region hs welldefined top nd bottom, y if the region hs welldefined left nd right boundries). Bsed on the xis of revolution nd the vrible of integrtion, determine the method (disks or wshers for xintegrtion bout horizontl xis or yintegrtion bout verticl xis, shells for xintegrtion bout verticl xis or yintegrtion bout horizontl xis). Lbel your picture with the inner nd outer rdii for disks or wshers; lbel the rdius nd height for cylindricl shells. Set up the integrl(s) nd evlute. 3. Arc length nd surfce re 3.. Arc Length. Let f(x) be continuous on [, b] nd differentible on (, b). Our im is to find the length of the curve y = f(x), x b. We begin by prtitioning the intervl [, b] into n equl pieces: = x < x < < x n = b, where x i x i = x = b for n ech i =,,, n. Between ech pir of djcent points on the curve, (x i, f(x i )) nd
5 APPLICATIONS OF THE DEFINITE INTEGRAL 5 (x i, f(x i )), we pproximte the rc length s i by the strightline distnce between the two points. From the usul distnce formul, we hve s i d((x i, f(x i ) ), (x i, f(x i )) = (x i x i ) + [f(x i ) f(x i )]. Since f is continuous on ll of [, b] nd differentible on (, b), f is lso continuous on the subintervl [x i, x i ] nd is differentible on (x i, x i ). By the Men Vlue Theorem, we then hve f(x i ) f(x i ) = f (c i )(x i x i ), for some number c i (x i, x i ). This gives us the pproximtion s i (x i x i ) + [f(x i ) f(x i )] = (x i x i ) + [f (c i )(x i x i )] = + [f (c i )] (x i x i ) = + [f(c i )] x. Adding together the lengths of these n line segments, we get n pproximtion of the totl rc length, n s + [f (c i )] x. i= Notice tht s n gets lrger, this pproximtion should pproch the exct rc length, tht is, n s = lim + [f (c i )] x. n So, the rc length is given exctly by the definite integrl: i= whenever the limit exists. s = + [f (x)] dx, Exmple 3.. () A cble is to be hung between two poles of equl height tht re feet prt. It cn be shown tht such hnging cble ssumes the shpe of ctenry, the generl form of which is y = cosh x = (ex/ + e x/ ). In this cse, suppose tht the cble tkes the shpe of y = 5(e x/ + e x/ ), for x. How long is the cble? () Find the rc length of the curve y = 3 (x + ) 3 between x 3. (3) Find the length of the curve x = 3 y 4 3 3y 3 between y Surfce Are. One cn esily show tht the curved surfce re of the right circulr cone of bse rdius r nd slnt height l is A = πrl,
6 6 APPLICATIONS OF THE DEFINITE INTEGRAL nd so the curved surfce re of the frustum of the cone shown below is A = π(r + r )L. Now, suppose tht f is nonnegtive nd continuous on [, b] nd differentible on (, b). If we revolve the grph of y = f(x) bout the xxis on the intervl [, b], we get the surfce of revolution seen below: We prtition [, b] into n mny pieces of equl size s we hve done so mny times. On ech subintervl, we cn pproximte the curve by the stright line segment joining the points (x i, f(x i )) nd (x i, f(x i )). Notice tht revolving this line segment round the xxis genertes the frustum of cone. The surfce re of this frustum will give us n pproximtion to the ctul surfce re on the intervl [x i, x i ]. First, observe tht the slnt height of this frustum is L i = d((x i, f(x i )), (x i, f(x i ))) = (x i x i ) + [f(x i ) f(x i )],
7 APPLICATIONS OF THE DEFINITE INTEGRAL 7 from the usul distnce formul. Becuse of our ssumptions on f, we cn pply the Men Vlue Theorem, to obtin f(x i ) f(x i ) = f (c i )(x i x i ), for some number c i (x i, x i ). This gives us L i = (x i x i ) + [f(x i ) f(x i )] = + [f (c i )] (x i x i ). The surfce re S i of tht portion of the surfce on the intervl [x i, x i ] is pproximtely the surfce re of the frustum of the cone, S i π[f(x i ) + f(x i )] + [f (c i )] x πf(c i ) + [f (c i )] x. since if x is smll, f(x i ) + f(x i ) f(c i ). Repeting this rgument for ech subintervl [x i, x i ], i =,,, n, gives us n pproximtion to the totl surfce re S, n S πf(c i ) + [f (c i )] x. i= As n gets lrger, this pproximtion pproches the ctul surfce re, n S = lim πf(c i ) + [f (c i )] x. n Recognizing this s the limit of Riemnn sum gives us the integrl whenever the integrl exists. S = i= πf(x) + [f (x)] dx. Exmple 3.. () Find the re of the surfce obtined by revolving the curve y = 5 x, x 3 bout the xxis. () Find the re of the surfce obtined by revolving the curve y = x, x bout the yxis. (3) Find the re of the surfce obtined by revolving the curve x = y 3, y bout the yxis. (4) Find the re of the surfce obtined by revolving the curve y = 6x, x bout the xxis. (5) Find the re of the surfce obtined by revolving the curve y = x, x bout the yxis.
8 8 APPLICATIONS OF THE DEFINITE INTEGRAL Answers.. () () ˆ R R Answers.. () () (3) ˆ R ˆ R ˆ 3 π( R x ) dx = 4πR3 3. ˆ 4 Answers ( 6 5 ) dx = π( x) dx = 8π. π( R x ) dx = 4πR3 3. π(x + ) dx = 8π Answers.3. () () (3) ˆ 8 ˆ 4 Answers.4. (b) (c) () () (b) (c) (d) (3) () (b) (c) (d) ˆ 4 π(y 3 ) dy = 96π π( 4 y) dy = 8π. ˆ 4 ˆ 4 ˆ π π () () ( π( 4 y) dy = 9π. ˆ ( 4 x ( ( 4 x π( 4y) dy = π. ) ) dx = 8π ) ) dx = 56π π( 4 y) dy = 8π. (( ( ) ) π 4 x ( 3)) 3 dx = 47π ( ( ) ) π 7 7 (4 x ) dx = 576π (( π 3 ( ( 4 y)) 3 ) ) 4 y dy = 64π. ( ) ) π x (x dx = π ) 8x ) ) π(( (x dx = 48π (( ( ) ) π 6 x ) dx = 83π 4 ( ) ) π(( y ( )) dy = 3π 3 + 4π.
9 Answers.. () () (3) () (4) (5) (6) (7) (b) (c) ˆ ˆ 4 ˆ ˆ ˆ APPLICATIONS OF THE DEFINITE INTEGRAL 9 πx(x x ) dx = π 6. π(3 x)(4 x ) dx = π 6 = 64π. ˆ ˆ ˆ π( y)(( y) y) dy = π 3. π(y ( ))(( y) y) dy = 8π 3. π(3 x)x dx + πx(x(x ) ) dx = π ( ) 4 πx dx = 4π. x πy(y ) dy = 8π. πx(( x ) x ) dx = π. (8) () Disk Method: (b) Shell Method: ˆ 3 (c) Wsher Method: ˆ ˆ 3 π(3 x)( x) dx = 4π. π(3 + x x ) dx = 53π πx(3 + x x ) dx = 45π. (( π 3 + x x ( )) ˆ 3 Answers 3.. () ˆ 3 () ( + x ) dx =. ˆ 8 ( 3 y (3) (e x/ 4 ) + dy = 9. 3 y ˆ 3 ) e x/ + dx = e e. Answers 3.. () π x 5 x + dx = 5π. 5 x ˆ () π y + 4y dy = π π. 6 (3) (4) (5) ˆ ˆ ˆ 3 πy 3 + 9y 4 dy = π 7 ( ). π6x 37 dx = 6π 37. π (y + ) + 5π dy = (y + ) 3. ( ) ) ( ) dx = 43π
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