l 2 p2 n 4n 2, the total surface area of the

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1 Week 6 Lectures Sections 7.5, 7.6 Section 7.5: Surfce re of Revolution Surfce re of Cone: Let C be circle of rdius r. Let P n be n n-sided regulr polygon of perimeter p n with vertices on C. Form cone C n of slnt length l by glueing together n iscoseles tringles with sides of length l, l, pn. n Given tht ech tringle hs re 1 p n n l p n n, the totl surfce re of the cone is 1p n l p n. The cone C n n is n pproximtion of the right circulr cone corresponding to circle of rdius r nd slnt length l. Given tht 1 lim n p n l p n n = πrl, the surfce re of the cone with circulr bse is πrl. Surfce re of Conic Frustrum: To obtin the surfce re of conic frustrum with slnt length l nd rdii r nd R, imgine subtrcting two cones, the smll one with slnt length l 1 nd rdius r 1, nd the lrge one with slnt length l nd rdius r. Then the net surfce re is πr l πr 1 l 1. Given the reltionships l 1 = r 1 l r nd l l 1 = l, the net surfce re (fter simplifiction) is πl(r 1 + r ). Surfce re of Revolution: Let y = f(x) be curve from x = to x = b. Rotting bout the x-xis yields surfce of revolution. pproximting the curve by line segments with slope f (x i ), we obtin conic frustrums with slnt length 1 + f (x i ) x nd rdii f(x i 1 ) nd f(x i ). Summing, we obtin the pproximtion π 1 + f (x i ) (f(x i 1 ) + f(x i )) x. 1

2 Using the further pproximtion f(x i 1 ) + f(x i )) f(x i ), we obtin the Riemnn sum πf(x i ) 1 + f (x i ) x. This yields surfce re = πf(x) 1 + f (x) dx. For exmple, using f(x) = x, x [, 1], we obtin 1 π x 1 + x dx = π (sec 6 θ sec θ) cos θ dθ = I 3 (u) I (u) π = 1 16 tn θ sec 3 θ dθ = ( 6 log π 1 (1 u ) 1 3 sec 5 θ sec 3 θ dθ = ( + 3 (1 u ) du = )). Homework: Section 7.5, problems 7, 9, 11, 13, 5. Section 7.6: pplictions to Physics nd Engineering Things tht cn be mesured: distnce (d), time (t), mss (m). Things tht cn be clculted: re (), velocity (v), ccelertion (), force (F ), pressure (P ), work (W ). English Units: d ft, t sec, v ft/sec, ft/sec, F lbs, P = F/ lbs/ft, F d ft-lbs. Metric Units: d m, t sec, m kg, v m/sec, m/sec, F = m N, P = F/ N/m, F d J. Note tht N is short for netwons (kg m/sec ) nd J is short for Joules (newton-meters or kg m /sec ). Constnts:

3 Grvity ner erth cuses constnt ccelertion of 3 ft/sec nd 9.8 m/sec. Wter hs weight density 6.5 lbs/in 3 nd mss density 1 kg/m 3. Work clcultions: If you lift n object of weight 3 ounces (force) through 75 inches (distnce), then the work done is F d = = 8.98 ft lbs. If you lift n object of mss 7 g through 5 cm, the work done is F d = md = = J. Vrible Work Clcultion: Exmple 3, pge. pound rope tht is 1 feet long is suspended from the top of building. Find the work done in pulling up the rope to the top of the building, discounting other forces. Solution: Think of the rope s being prtitioned into N pieces of length x feet. We will clculte the work done to lift ech segment, then dd. Ech segment weighs x pounds, nd the i th segment from the top is lifted through x i feet, so the x i x foot pounds of work is done, for totl of x i x foot pounds. Since x i is vrying in [, 1], the exct mount the work is foot-pounds. 1 x dx The Leky Bucket Problem: Imgine tht the rope bove is supporting bucket of 5 gllons of wter, tht n empty bucket weighs 1 pounds, tht we pull up the bucket t rte of 1 feet per second, nd tht t the moment we strt pulling the bucket leks wter t rte of gllons per 3

4 second. Clculte the work done in pulling up the rope nd the bucket nd the wter in the bucket. Solution: We will just clculte the mount of work to pull up the bucket nd the wter in it, then dd to the previous nswer. It will tke 1 seconds to pull up the bucket. Think of time s being prtitioned into N subintervls of t seconds. We will clculte the work done in ech subintervl of time. From time t i to time t i+1 we hve lifted the bucket 1 t feet. Choosing ny t i in this intervl, the bucket nd wter weighs pproximtely t i = 6 t i pounds, so in this time intervl we hve done pproximtely (6 t i ) t foot-pounds of work. Totl pproximte work done is (6 t i ) t. Since t i is vrying in [, 1], the exct mount of work done is 1 6 t dt. Pumping Wter out of Tnk Formed by Volume of Revolution: Imgine tht tnk of wter is formed by revolving the region bounded by y = x nd y = 3 bout the y-xis. The tnk is filled with wter, nd the wter is to be pumped out. How much work does this tke, ssuming tht the units long the x nd y xes re given in feet? Solution: We will prtition the tnk into slices of width y nd clculte the work done to pump ech slice out. The slice tht extends from y i 1 to y i cn be pproximted by wsher tht hs thickness of y feet, cross-sectionl rdius of pproximtely x i feet corresponding to vlue of yi [y i 1, y i ], cross-sectionl re of pproximtely π(x i ) = π yi squre feet, volume of pproximtely π yi y cubic feet, nd weight of 6.5π yi y pounds. Since this slice is to be lifted 3 y i feet, the work done is 6.5π yi (3 y i ) y foot-pounds. Totl pproximte work done is 6.5π yi (3 y i ) y

5 foot-pounds Exct work done is foot-pounds 3 6.5π y(3 y) dy Exmple, pge : The tnk is the volume of revolution formed by the line y =.5x. The slice tht extends from y i 1 to y i cn be pproximted by wsher tht hs thickness of y meters, cross-sectionl rdius of pproximtely x i meters corresponding to vlue of yi [y i 1, y i ], cross-sectionl re of pproximtely π(x i ) = π (y i ) squre meters, volume of pproximtely 6.5 π (y i ) y cubic meters, mss of 1π (y i ) y kilogrms, nd represents force of 98π (y i ) y newtons. Since this slice is to be lifted y i feet, the work done is 98π (y i ) 6.5 (1 y i ) y foot-pounds. Totl pproximte work done is 98π (y i ) 6.5 (1 y i ) y. Since y i vries in [, 8], exct work done is 8 98π y (1 y) dy 6.5 joules. We get the sme nswer s in the book. Remrk: We cn use the sme ides for other shpes, so long s we cn pproximte the typicl slice of volume. Hydrosttic Pressure: Consider rectngulr continer tht hs bse re squre feet nd depth d feet. When the contininer is filled with wter, the pressure on the bse of the continer from the weight of the wter bove it is P = F/. The weight of the wter 6.5 pounds per cubic foot times d cubic feet, which yields F = 6.5d. The re of the bse is squre feet. Hence the pressure is 6.5d = 6.5d pounds per squre foot. 5

6 If the continer hs bse re squre meters nd depth d meters, then the mss of the wter is d kg, the ccelertion is 9.8 meters per squre second, hence force of the wter on the bse is F = 9.8d newtons, hence the pressure on the bse is 9.8d = 9.8d newtons per squre meter. Hydrosttic pressure is regrded to be the sme in ll directions t ny given depth. Hydrosttic Force Problem: metl plte in the shpe bounded by the curves y = x nd y = 18 x (dimensions re feet) is submerged in wter so tht the top is 1 feet below wter level. Clculte the totl mout of hydrosttic force exerted by the wter ginst the plte. Solution: Since hydrosttic pressure (pounds of force per squre foot of re) cross slice of the plte is the sme, we will pproximte the totl force on given slice of the plte by multiplying the re of the slice by the force per squre foot, then dd the result. Segment ech hlf of the plte into slices of width y. ssuming the pproximte depth of the i th slice is yi, the hydrosttic pressure on the plte t depth of yi is 6.5( yi ) pounds per squre foot. For yi [, 9], the slice cn be pproximted by rectngle with length extending between the corresponding x-coordintes on the curve y = x, nmely yi, for n re of yi y squre feet nd force of 15( y i ) y i y pounds. For yi [9, 18], the slice cn be pproximted by rectngle with length extending between the corresponding x-coordintes on the curve y = 18 x, nmely 18 yi, for n re of 18 yi y squre feet nd force of 15( yi ) 18 yi y pounds. Hence totl force is F = 9 15( y) y dy ( y) 18 y dy = 9, 5 pounds. In the second integrl, mke the substitution u = 18 y to simplify the clcultion. 6

7 Center of mss: Sy objects with msses m 1 through m k nd totl mss M re locted t positions (x 1, y 1, z 1 ) through (x k, y k, z k ). The center of mss of the collection of objects is defined to be (x, y, z) where x = m 1 M x m k M x k, nd y = m 1 M y m k M y k, z = m 1 M z m k M z k. Given, for simplicity, two-dimensionl region with uniform mss density nd bounded by the curves y = f(x) nd y = g(x) over [, b], we pproximte the region by rectngulr slices nd tret ech rectngle s hving mss equl to the re of the rectngle concentrted in the center of the rectngle. This yields x f(x 1) g(x 1 ) x x (f(x n) g(x n )) x x n, y (f(x 1) g(x 1 )) x f(x 1 ) + g(x 1 ) Letting n, we obtin + + (f(x n) g(x n )) x f(x n ) + g(x n ). x = x(f(x) g(x)) dx, f(x) g(x) dx y = f(x) g(x) dx f(x) g(x) dx. Note: If there is mss density function ρ tht vries with x-coordinte, we cn modify the formuls bove suitbly. Theorem of Pppus: Rotte plne figure bout xis. Volume of revolution is re times distnce centroid (center of mss ssuming uniform density) trvels. Homework: Section 7.6, Problems 9, 11, 13,, 7, 3, 53 7

7.6 The Use of Definite Integrals in Physics and Engineering

Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 7.6 The Use of Definite Integrls in Physics nd Engineering It hs been shown how clculus cn be pplied to find solutions to geometric problems