-------------------------------------------------------- Previous assignment: Page 385 Problem 4 2 x + 3x INT -------------- dx Sqrt[x+4] u = Sqrt[x+4] 2 u = x+4 2 x = u -4 dx = 2 u du 2 2 2 (u -4) + 3(u -4) INT ---------------------- 2 u du u 2 2 2 2 INT (u -4) + 3(u -4) du 4 2 2 2 INT u -8u + 16 + 3u - 12 du 4 2 2 INT u -5 u + 4 du | 5 3 | 2 | u /5 - 5u /3 + 4 u | + C | | | 4 2 | 2 u | u /5 - 5 u /3 + 4 | + C | | 2 u 4 2 ---- ( 3 u - 25 u + 60 ) 15 2 2/15 Sqrt[x+4] ( 3(x+4) - 25(x+4) + 60 ) 2 2/15 Sqrt[x+4] ( 3(x + 8 x + 16) - 25(x+4) + 60 ) 2 2/15 Sqrt[x+4] ( 3x - x + 8) <=========Answer / \/ -1/2 2 1/2 2/15 (1/2) (x+4) (3x - x + 8) + 2/15 (x+4) (6x-1) 2 3x - x + 8 2 (x+4)(6x-1) (1/15) --------------- + (1/15) ---------------- Sqrt[x+4] Sqrt[x+4] 2 2 3x - x + 8 + 12 x + 46 x - 8 ---------------------------------- 15 Sqrt[x+4] 2 2 15 x + 45 x x + 3 x ---------------- = ----------- 15 Sqrt[x+4] Sqrt[x+4] ---------------------------------------------------------- -------------------------------------- Integrate[ (x^2 + 3x)/Sqrt[x+4],x] -------------------------------------- Page 385 Problem 18 dx INT --------------------- 2 Sqrt[x + 4 x + 5 ] dx INT --------------------- 2 Sqrt[x + 4 x + 4 + 1] dx INT --------------------- 2 Sqrt[(x+2) + 1] u = x+2 du = dx du INT --------------------- 2 Sqrt[ u + 1] | /| 2 / | Sqrt[u +1] / | | / | u | / | | / | |/ t | -------------+----------------------- | 1 2 Sqrt[u +1] = Sec[t] u = Tan[t] 2 du = Sec[t] dt 2 Sec[t] dt INT --------------- = INT Sec[t] dt Sec[t] ln( Sec[t] + Tan[t] ) + C | /| 2 / | Sqrt[u +1] / | | / | u | / | | / | |/ t | -------------+----------------------- | 1 2 ln( Sqrt[u +1] + u ) + C 2 ln( Sqrt[x + 4 x + 5] + x+2 ) + C <=============== / \/ 2 -1/2 1/2 (x + 4 x + 5) (2x + 4) + 1 -------------------------------------------------------------- 2 Sqrt[x + 4 x + 5 ] + x+2 2 -1/2 (x + 4 x + 5) (2x + 4) + 2 -------------------------------------------------------------- 2 2(Sqrt[x + 4 x + 5 ] + x+2) 2 (2x + 4) + 2 Sqrt[x +4x+5] -------------------------------------------------------------- _ _ | 2 | 2 2|Sqrt[x + 4 x + 5 ] + x+2 | Sqrt[x +4x+5] |_ _| 1 ---------------- 2 Sqrt[x +4x+5] --------------------------------------------------------- Page 385 Problem 32 Two circles of radius b intersect as shown with their centers 2 a units apart (0 <= a < b). Find the area of the region of their overlap. ..... .... ' / ` ' ` . b/ .. . . / a . . . . ''''''.' . . . .. . ........ ..... The problem is the same as finding the area of a circle inside a chord. We find this area and then double it. ..... 2 2 2 ' . /| x + y = b . .b/ | . . ./ a| . . . | . ` . . .|' ''' x=b 2 2 A = INT 2 Sqrt[ b - x ] dx x=a | | /| | b/ | 2 2 | / |Sqrt[ b - x ] |/t | -----------+------------- | x | | x = b Cos[t] 2 2 Sqrt[b -x ] = b Sin[t] dx = -b Sin[t] dt x=b A = 2 INT b Sin[t] (-b Sint[t]) dt x=a 2 x=b 2 A = -2 b INT Sin[t] dt x=a 2 x=b 1 - Cos[2t] A = -2 b INT ------------ dt x=a 2 | 2 Sin[2t] | t = 0 A = | - b ( t - ------- ) | | 2 | t = ArcCos[a/b] | 2 | t=0 A = | -b ( t - Sin[t] Cos[t] ) | | | t = ArcCos[a/b] | | t=0 | 2 | A = |- b t + b Sin[t] b Cos[t] | | | | | t = ArcCos[a/b] | /| | b/ | 2 2 | / |Sqrt[b -a ] |/t | -----------+------------- | a | | 2 2 2 A = 2 b ArcCos[a/b] - 2 a Sqrt[b -a ] The area from the chord to the circle is A, the area we want is 2A 2 2 2 2A = 4 b ArcCos[a/b] - 4 a Sqrt[b - a ] <======== -------------------------------------- Integrate[ 2 Sqrt[ b^2 - x^2 ],{x,a,b}] -b^2 ArcSin[x/b] -x Sqrt[b^2-x^2] a = 1; b = Sqrt[2]; p1 = ParametricPlot[ b{Cos[t],Sin[t]},{t,0,2 Pi}]; p2 = ParametricPlot[ 2a {1,0} + b{Cos[t],Sin[t]},{t,0,2 Pi}]; p3 = ParametricPlot[ {a,y},{y,-Sqrt[b^2-a^2],Sqrt[b^2-a^2]},PlotStyle->{RGBColor[1,0,0]}]; p4 = Show[p1,p2,p3,PlotLabel->"P385 p32; a=1 b=Sqrt[2]",AspectRatio->Automatic]; Display["p32.ps",p4]; ============================================= Page 385 Problem 33 Hippocrates of Chios (ca.430 B.C. showed that the two shaded regions in Figure 7 have the same area. (He squared the lune). Not that C is the center of the lower arc of the lune. Show Hippocrates' result (a) using calculus and (b) without calculus ........ ' ` . o o o . . o o . o o o o o o . o o . . o o . ` o o ' .. .. ---------------------------------------------------------------- a = ParametricPlot[ {Cos[t],Sin[t]},{t,0,2 Pi}]; b = ParametricPlot[ {0,-1} + Sqrt[2] {Cos[t],Sin[t]},{t,Pi/4,3 Pi/4}, PlotStyle->{RGBColor[1,0,0]}]; c = ParametricPlot[ {0,-1} + t{1/Sqrt[2],1/Sqrt[2]},{t,0,Sqrt[2]}, PlotStyle->{RGBColor[1,0,0]}]; d = ParametricPlot[ {0,-1} + t{-1/Sqrt[2],1/Sqrt[2]},{t,0,Sqrt[2]}, PlotStyle->{RGBColor[1,0,0]}]; e = Show[a,b,c,d,PlotLabel->"P385 p33; Hippocrates of Chios",AspectRatio->Automatic]; Display["p33.ps",e]; ---------------------------------------------------------------- With Calculus: In the previous problem we found the area we need. We take half of the answer to have the area of the area between a chord and the circle. b = Sqrt[2] a 2 2 2 A = b ArcCos[a/b] - a Sqrt[b -a ] 2 a 2 2 A = 2 a ArcCos[----------] - a Sqrt[2 a -a ] Sqrt[2] a 2 A = 2 a ArcCos[1/Sqrt[2]] - a^2 2 2 A = 2 a Pi/4 - a 2 2 A = (1/2) a Pi - a The lune is 1/2 the circle minus the chord area or 2 2 2 2 1/2 Pi a - ( 1/2 a Pi - a ) = a . Without calculus. The area between the chord and the circle is 2 2 2 2 1/4 Pi( 2a ) - 1/2 (2a ) = Pi a /2 - a just as before. ----------------------------------------------------------