# Math 0230 Calculus 2 Lectures

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1 Mth Clculus Lectures Chpter 7 Applictions of Integrtion Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition. Section 7. Ares Between Curves Two pproches:. We consider region S tht lies between two curves y = f(x) nd y = g(x) nd between the verticl lines x = nd x = b, where f nd g re continuous nd f g on [, b]. Then we divide the region into smll verticl rectngles with bse x nd heights f(x i ) g(x i ). The corresponding Riemnn sum is [f(x i ) g(x i )] x nd the re A of S is A = lim n [f(x i ) g(x i )] x = b [f(x) g(x)] dx To solve problems on finding res between two curves lwys drw picture. Exmple. Find the re tht lies between two curves y = x + nd y = x + nd between the verticl lines x = nd x =. Solution: First, drw the picture. f(x) = x +, g(x) = x + re continuous nd f g on [, b]. Then nd (importnt!) f nd g A = [ ] [ ] x + dx = x + (x + )/ ln(x + ) = 9/ ln 4 ( ) 5/ ln = ln. We consider region S tht lies between two curves x = f(y) nd x = g(y) nd between the horizontl lines y = c nd y = d, where f nd g re continuous nd f(y) g(y) on [c, d]. Then

2 we divide the region into smll horizontl rectngles with height y nd bses f(y i ) g(y i ). The corresponding Riemnn sum is nd the re A of S is A = lim n [f(yi ) g(yi )] y [f(yi ) g(yi )] y = d c [f(y) g(y)] To solve problems on finding res between two curves lwys drw picture. Exmple. Find the re enclosed by the prbol 4x + y = nd the line y = x. Solution: The prbol hs the vertex t (, ). Points of intersection: 4x = y = 4y, y + 4y =, y = 6 nd y =. Points re ( 6, 6) nd (, ). Consider functions y x = f(y) = = y nd x = g(y) = y on [ 6, ], f(y) g(y). Then the re is 4 4 ] ] A = [ y 4 y = [y y y 6 6 = 6 8 ( ) 6 8 = = Explin why horizontl rectngles work better here compring to verticl rectngles (with x). Exmple. Sketch the region enclosed by the curves y x = nd x = e y nd lines y = nd y =. Solution: The prbol x = y hs the vertex t (, ). Two curves do not intersect inside y. Consider functions x = f(y) = e y nd x = g(y) = y on [, ], f(y) g(y). Then the re is [ A = e y y + ] ] = [e y y + y = e ( + e + ) = + e e

3 Section 7. Volumes Let S be d solid plced in the xy-plne. Let P x be plne perpendiculr to the x-xis nd P y be plne perpendiculr to the y-xis. First consider P x type of plnes. They slice S. The intersection of S nd P x is clled cross-section nd let A(x) be its re, x b. If slice is x = (b )/n thick then its volume is A(x i ) x, where x i lies in the intervl [x i, x i ]. Then the volume of S is pproximtely V A(x i ) x or exctly V = lim n A(x i ) x = Exmple. Find the volume of sphere of rdius r. b A(x) dx Solution: r x r. A cross-section t x is circle of rdius y = r x. Then A(x) = πy = π(r x ) nd V = r r π(r x ) dx = π r (r x ) dx = π ] r [r x x = π(r r /) = 4π r Now ssume we hve solid obtined by rotting bout the x-xis the region bounded by curves y = f(x) nd y =, when x b. Then ech cross-section is circle with rdius f(x) nd the volume of the solid is b V = π f (x) dx If solid is obtined by rotting bout the y-xis the region bounded by the curve x = f(y), when c y d, then ech cross-section is circle with rdius f(y) nd the volume of the solid is d V = π f (y) c Exmple. Find the volume of the solid obtined by rotting bout the y-xis the region bounded by the curves x = y nd x =, when y.

4 Solution: V = π (y ) = π y 4 = π y5 5 = π 5 Now ssume region is bounded by the curves y = f(x) nd y = g(x), x b, where f(x) g(x) [note, the second curve is not y = nymore]. In this cse cross-section is ctully wsher with the outer (lrger) rdius f(x) nd the inner (smller) rdius g(x). The re of the wsher is π[f (x) g (x)] nd the volume of the solid obtined by rottion is V = π b [f (x) g (x)] dx In the cse of rottion bout the y-xis the coressponding volume is V = π d c [f (y) g (y)] Exmple. Find the volume of the solid obtined by rotting bout the y-xis the region bounded by the curves x = y nd x = y. Solution: First, we hve to find c nd d s y coordintes of points of intersection of the curves: the eqution y = y gives two solutions c = nd d =. Also, y y when y. Hence [ ] 4y V = π [(y) (y ) ] = π [4y y 4 ] = π y5 = 64π 5 5 Solids tht we hve considered so fr re clled solids of revolution. Now consider n exmple of not solid of revolution Exmple 4. Find the volume of pyrmid whose bse is squre with side nd whose height is h. Solution: We plce the origin O t the vertex of the pyrmid nd the x-xis long its centrl xis. Every cross-section t x ( x h) is squre with the side b = x (it cn be found h from corresponding proportion). The cross-sectionl re is A(x) = b = h x. Then V = h A(x) dx = h h x dx = x h h = h h = h 4

5 Section 7. Volumes by Cylindricl Shells Some volumes cn be clculted by the method of cylindricl shells. Picture. Let there be cylindricl shell with inner rdius r nd outer rdius r. Then its volume V is the difference of the volume V of the outer cylinder nd the volume V of the inner cylinder: V = V V = πrh πrh = π(r r)h = π(r + r )(r r )h = π r + r (r r )h We put r = r + r (the verge rdius of the shell) nd r = r r, then V = πrh r Another wy: V = π(r + r) h πr h = πrh r where r is the inner rdius nd r + r is the outer rdius. Let s pply tht to clcultion of the volume V of solid S obtined by rotting bout y-xis the region bounded by y = f(x) (f(x) must be continuous nd nonnegtive f(x) ), y =, x =, nd x = b, where < < b. As usul, we divide the intervl [, b] into subintervls [x i, x i ] ech of the length x. Let x i be midpoint of ith subintervl. If the rectngle with bse [x i, x i ] nd height f( x i ) is rotted bout the y-xis, then the result is cylindricl shell with verge rdius x i, height f( x i ), nd thickness x. Its volume is nd the volume of the solid is pproximtely or exctly V V = lim n V i = π x i f( x i ) x V i = π x i f( x i ) x b π x i f( x i ) x = π xf(x) dx If solid S is obtined by rotting bout x-xis the region bounded by x = f(y) (f(y) is continuous nd nonnegtive f(y) ), x =, y = c, nd y = d, where < c < d, then its volume is d V = π yf(y) c If solid S is obtined by rotting bout the line x = l (which is prllel to the y-xis) the region bounded by y = f(x) (f(x) is continuous nd nonnegtive f(x) ), y =, x =, nd x = b, where < < b < l, then its volume is V = π b (l x)f(x) dx 5

6 Exmple. Find the volume of sphere of rdius R. Solution: The sphere is obtined by rotting bout the y-xis the region bounded by the curves y = R x nd y = R x when x R. Then r = x, h = R x ( R x ) = R x nd V = πx R x x, V = 4π R x R x dx Substitution: u = R x, du = xdx, u() = R, u(r) = V = π R u / du = π R [ ] R u / du = π u/ = 4 πr Exmple. Find the volume of the solid obtined by rotting bout the y-xis the region bounded by the curves x = y nd x = y. [It is the exmple in the section 7.] Solution: The region is bounded by the curves y = x nd y = x/ with x/ x. Before we found tht x 4. Here we pply the method of cylindricl shells: r = x, h = x x/, V = πrh r = πx( x x/) x = π(x / x /) x. Then 4 V = π ) [ ] 4 [ ] [ (x / x dx = π 5 x5/ x = π = 64π ] = 64π 5 Exmple. Find the volume of the solid obtined by rotting bout the y-xis the region bounded by the curves y = x nd y =, nd x = 4. Solution: Drw picture. =, b = 4, f(x) = x. Then 4 V = π x 4 x dx = π x / dx = π 5 x5/ 4 = 8π 5 Exmple 4. Find the volume of the solid obtined by rotting bout the x-xis the region bounded by the curves y = x, y =, nd x = 4. Solution: x = y, y. V = π y y = π 6 y = π y4 4 = 8π

7 Exmple 5. Find the volume of the solid obtined by rotting bout the line x = the region bounded by the curves y = 4x x, y = 8x x. Solution: Drw picture(!). Vertices of both prbols re t x = : (, 4) nd (, 8). Intesections of prbols: 4x x = 8x x, x 4x =, x =, x = 4. We divide [, 4] into subintervls ov width x. Let x be in one of the subinervls. Averge rdius of cylindricl shell is the distnce between x nd which is x ( ) = x +. The height is the difference between prbols: h(x) = 8x x (4x x ) = 4x x. Then V = π 4 (x + )(4x x ) dx = π 4 (8x + x x ) dx = π = π [ ] = 56 π [4x + ] 4 x x4 4 7

8 Section 7.4 Arc Length Picture. We wnt to find the length L of curve given by y = f(x), where f (x) is continuous nd x runs through the intervl [, b]. A curve is mde of smll rcs. Consider the ith rc. Its length by the Pythgoren theorem is Then or exctly L = lim n s i L = ( x i ) + ( y i ) = s i = ( ) yi + x i = x i ( ) yi + x i x i ( ) yi + x i x i b + becuse x i dx nd y i s n nd dx = f (x). ( ) b dx = + (f dx (x)) dx Exmple. Find the length L of the curve given by y = (x ) /, when x. ( ) Solution: dx = (x )/, + = + 9(x ) = 9x 8. Hence, dx L = 9x 8 dx [u = 9x 8, du = 9xdx, u() =, u() = ] = 9 u / du = 9 u/ = 7 ( ) Exmple. Find the length L of the curve given by y = x, when x. Solution: dx = x, + x ( ) = + x dx x =. Hence, x L = dx x Trig substitution: x = sin θ, θ π/, dx = cos θ dθ, x = cos θ = cos θ. Then L = π/ cos θ dθ cos θ = π/ dθ = π/ 8

9 Notice, tht the curve is qurter of the unit circle whose length is π. Exmple. Solution: Find the length L of the curve given by y = ln(cos x), when x π/. ( ) = tn x, + = + tn x = sec dx dx x = sec x. Hence, L = π/ sec x dx = ln sec x + tn x π/ = ln( + ) = ln( + ) 9

10 Section 7.6 Applictions to Physics nd Engineering Work W = F d F = m d s dt work = force x distnce (here the force is constnt) force = mss x ccelertion If the force is vrible F = f(x), then the work done in moving n object from to b is W = b f(x) dx Exmple. When prticle is locted x meters from the origin, force of x newtons cts on it. How much work is done in moving the prticle from x = to x =? Solution: W = x dx = x / dx = x x x = x = 4 Hook s Lw The force is proportionl to x: f(x) = kx, k > is the spring constnt. Exmple. A spring hs nturl length of 5 cm. If 6-N force is required to keep it stretched to length of 8 cm, how much work is required to stretch it from 5 cm to 6 cm? Solution: 8 5 = cm =. m. Then f(.) = 6, k. = 6, k =. Hence f(x) = x nd.. W = x dx = x = J Hydrosttic Pressure nd Force Suppose tht thin horizontl plte with re A sq. meters is submerged in fluid of density ρ kg per m t depth d meters below the surfce. The fluid directly bove the plte hs volume V = Ad nd the mss m = ρv = ρad. The force exerted by the fluid on the plte is F = mg = ρgad. The pressure on the plte is P = F/A = ρgd. Exmple. A plte in n irrigtion cnl is in the form of trpezoid 4 feet wide t the bottom, 8 feet wide t the top, with the height equl to feet. It is plced verticlly in the cnl nd is submerged in wter feet deep. Find the hydrosttic force in pounds on the plte. The weight density of wter is 6.5 = 5/ lb/ft. Solution : Let s plce the coordinte xes s it is shown on the picture. We use horizontl strips of equl width y nd consider strip on the depth d = y with 5 y ccording

11 to the picture. By similr tringles: = 5 d = 5 + y, = (5 + y). The length of the strip is l = 4+ = 4+ 4 (5+y) = 4 (8+y) nd its re is A = 4 (8+y) y. The pressure on the strip is P = 6.5d = 6.5y (note tht d = y since y is negtive) nd the force is F = P A = 5 y 4 (8 + y) y = 5 (8y + y ). Then the totl force is F = 5 5 (8y + y ) = 5 = 5 ] [4y + y = 5 [ 6 8 ] [ 5 ] = 5 5 = 75 lb Solution : Now we plce the coordinte xes differently. See the picture below. As before we use horizontl strips of equl width δy nd consider strip on the depth d = y with y ccording to the picture. By similr tringles: = 5 d = 5 + y, = ( + y). The length of the strip is l = 4+ = 4+ 4 (+y) = 4 (6+y) nd its re is A = 4 (6+y) y. The pressure on the strip is P = 6.5d = 6.5( y) nd the force is F = P A = 5 Then the totl force is 4 5 ( y) (6 + y) y = ( 4y y ). F = 5 ( 4y y ) = 5 ] [y y y = 5 [ ( )] = 5 45 = 5 5 = 75 lb

12 Section 7.7 Differentil Equtions Exmples. Order of DE = order of the highest derivtive. A solution. Seprble Equtions A seprble eqution is st order DE tht cn be written in the form dx = g(x) f(y) Assuming tht f(y) neq we denote h(y) =. Then the eqution becomes f(y) h(y) = g(x) dx nd we integrte both sides to get h(y) = g(x) dx nd solve the lst for y. Initil-Vlue Problem IVP = DE together with the initil condition y(x ) = y Exmple. Solve the IVP: y = x, y() =. y Solution: y = x dx, dx = x y. It is seprble eqution. y = x dx, y where C = (C C ). Then y = ± x + C. + C = x + C, y = x + C, Initil condition: y() = >. We use the solution with + sign. y() = + C = C =, C = 4. Answer: y = x + 4. Checking: y = x x + 4 = x x + 4 = x y, y() = 4 =. Exmple. Solve the IVP: y = e y (x 5), y() =. Solution: dx = x 5. It is seprble eqution. e y e y = (x 5) dx, e y = (x 5) dx, e y = x 5x + C, y = ln(x 5x + C), Initil condition: y() = ln( 6 + C) =, 6 + C =, C = 7.

13 Answer: y = ln(x 5x + 7). Logistic Growth Exmple. Solve the IVP: dt = ky (M y) (logistic DE), < y < M, y() = y. The logistic eqution models world popultion growth. Solution: y (M y) = M [ M y + It is seprble eqution. [ y + ]. So, M y ] = kt + C, M y [recll tht < y < M], Initil condition gives Notice tht lim t y(t) = M. Direction Field ln y M y y M y = A. Hence, y (M y) = k dt. [ ln y ln (M y) ] = kt + C M = M(kt + C), y M y = em(kt+c) = Ae kmt, A = e MC. y M y = y M y e kmt or y(t) = y M y + (M y )e kmt. y = F (x, y) Exmple 4. Sketch the direction field for the DE: y = xy. Solution: Notice tht on the lines x = nd y = we hve y =. Mixing Problems y = F (x, y) Exmple 4. Consider tnk with volume liters contining slt solution. Suppose solution with. kg/liter of slt flows into the tnk t rte of 5 liters/min. The solution in the tnk is well-mixed. Solution flows out of the tnk t rte of 5 liters/min. If initilly there is kg of slt in the tnk, how much slt will be in the tnk s function of time?

14 Solution: Let y(t) denote the mount of slt in kg in the tnk fter t minutes. We use fundmentl property of rtes: Totl Rte = Rte in - Rte out To find the rte in we use 5 liters kg kg (.) =.5 min liter min The rte t which slt leves the tnk is equl to the rte of flow of solution out of the tnk times the concentrtion of slt in the solution. Thus, the rte out is 5 liters min y The differentil eqution for the mount of slt is y =.5 y, y() =. kg liter = y kg min Using the method of seprtion of vrible we find dt = y, y = dt, y = dt, y = C e.5t, (C = e C ), y = Ce.5t. ln y = t + C The initil condition gives y() = C =, C =. Hence, the mount of slt in the tnk fter t minutes is given by the formul y(t) = + e.5t. Notice tht lim t y(t) = + = 4

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