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MIT 18 01 - Geometric applications: volumes, area, arclength

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18.01A Topic 6: Geometric applications: volumes, area, arclength.Read: TB: 7.1-7.6 (7.4 to top of page 233, 7.5 lightly).Idea:1. Divide into little (infinitesimal) pieces, each of which is easy to compute.2. Sum them up.Area between curvesExamples: (won’t do all in class)1. Find the area between y = x + 2 and y = x2.answer: Find points of intersection:x + 2 = x2⇒ x = −1, 2.On top: x + 2 above x2.Area =R2−1(x + 2) − x2dx = 9/2.xywwwwwwwwwwwwwwwwwwwwwwww2. Find the area between curves y = x3and y =4x.Intersect at x = −2, 0, 2.Two pieces:R0−2x3− 4x dx +R204x − x3dx = 8.xy3. Find the area between y = 1−x2and y = x+1.Intersection: 1 − x2= x + 1 ⇒ x = 0, −1.On top: 1 − x2is above x + 1.Area =R0−1(1 − x2) − (x + 1) dx = 1/6.xy4. Find the area bounded by y = cos x, y = sin 2x, between 0 ≤ x ≤ π/2.Intersection: cos x = sin 2x = 2 sin x cos x⇒ cos x = 0 or 2 sin x = 1 ⇒ x = π/2, π/6.In [0, π/6] cos x is on top.In [π/6, π/2] sin 2x is on top.xyπ/6 π/2Area =Rπ/60cos x − sin 2x +Rπ/2π/6sin 2x − cos x= sin x +12cos 2xπ/60+ −12cos 2x − sin xπ/2π/6= (12− 0 +14−12) + (12− 1 +14+12) =12.(continued)118.01A topic 6 2Volume by Slices:Example 1: Find the volume of the sphere ofradius R.Volume of slice: dV = πr2dx = π(R2− x2) dx.Volume = ’sum’ of slices:V =RR−Rπ(R2− x2) dx= π(R2x − x3/3)|R−R=43πR3.Volume of revolution:Revolve graph of y = f(x) around x-axis.Compute volume by vertical slices.Volume of slice: dV = πy2dx = πf(x)2dxVolume = ’sum’ of slices:V =Rbaπf(x)2dxExample 2: . Find the volume of revolution ofthe curve y = 4 − x2between 0 and 2 revolvedaround the x-axis.Volume of slice: dV = πy2dx = π(4 − x2)2dx.Volume = ’sum’ of slices:V =R20π16 − 8x2+ x4dx= π(16x −83x3+15x5)20= π(32 − 64/3 + 32/5).Example 3: . Same curve around y-axis.Volume of slice: dV = πx2dy = π(4 − y) dy.Volume = ’sum’ of slices:V =R40π(4 − y) dy = π(4y −y2/2)|40= 8π.Volume by ShellsExample 4: Same curve (y = 4 − x2) aroundy-axis.Vertical rectangle in first graph sweeps out cylin-drical shell.thickness = dx, height = y, radius = x.⇒ Volume of shell = dV = 2πxy dx.⇒ Volume = ’sum’ of shells:V =R202πx(4 − x2) dx= 2π(2x2− x4/4)|20= 8π.(Same as by slices!)WashersRevolve area between curves around the x-axis.Just like slices with inside and outside radii.Volume of slice = dV = π(y21− y22) dx.Volume = ’sum’ of slices =RbadV.(continued)18.01A topic 6 3The same ideas work for volumes of revolution around other lines.Example 5: Find the volume of revolution of the curvey = 4 − x2between x = 0 and x = 2 around the liney = −1.answer: This is the same as example 2 ab ove, exceptinstead of rotating around the line y = 0 (the x-axis) werotate around y = −1.The volume of a thin disk of rotation isdV = πr2dx = π(y + 1)2dx = π(5 − x2)2dx.Volume = ’sum’ of slices:V =R20π(25 − 10x2+ x4dx= π(25x −103x3+15x5|20= π(50 −803+325).xy−14y1dx{{vvvvvvvvArclength∆s is the length along the curve.It’s approximated by the secant line.I.e. ∆s ≈p(∆x)2+ (∆y)2In the limit: ds =p(dx)2+ (dy)2This is the basic formula. It can be manipulated.E.g.dsdx=s1 +dydx2or ds =s1 +dydx2dx.Arclength = L =Zds =Zbas1 +dydx2dx.xy∆s******•AA •∆x∆yExample: Find the arclength of the curve y2= x3between (0, 0) and (4, 8).answer: y = x3/2⇒dydx=32x1/2⇒dsdx=q1 +94x.⇒ Arclength = L =R40q1 +94x dx =827(1 +94x)3/240=827(103/2− 1).Example: Find the arclength of y = sin x for x in [0, π].answer:dydx= cos x ⇒dsdx=√1 + cos2x.⇒ Arclength = L =Rπ0√1 + cos2x dx.(Not possible to evaluate in terms of elementary functions –called an elliptic integral.)(continued)18.01A topic 6 4Surface area of revolution:Trickier than volume.Main idea: rotate line to get frustrum of cone: read the book §7.6.For a curve y = f (x) rotated around the x-axis the differential of area isdA = 2πy ds = 2πyp1 + (y0)2dx.In general, for a element ds rotated in a circle of radius r we get dA = 2πr ds .Example: Find the area of the surface of revolution of y = 1/x between 1 and brevolved around the x-axis.answer: Find dA:y0= −1/x2.ds =p1 + (y0)2dx =p1 + (−1/x2)2dx.dA = 2πy ds = 2π1xp1 + (−1/x2)2dx.⇒ Surface area A =Rb12π1xp1 + 1/x4dx.Hard to compute but we can analyze:A >Rb12π1xdx = 2π ln b.⇒ A → ∞ as b → ∞.Find the volume of revolution of the same curve.Volume of slice = dV = πy2dx = π1x2dx.⇒ V =Rb1πx2dx = −πxb1= π(1 −1b).⇒ V → π as b → ∞.Finite volume and infinite surface area!What happens if you fill the volume of revolution with


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MIT 18 01 - Geometric applications: volumes, area, arclength

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