Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://orion.math.iastate.edu/hentzel/class.166.05 Textbook: Calculus by Varberg, Purcell, Rigdon, eight edition. January 19 6.4 Length p299:1,2,5-7,(8),13,(14),17,(21),23,(24),32 Main Idea: Approximate a curve with little tiny straight lines. Key Words: ds, Surface area = INT 2 Pi f(x) ds ds = Sqrt[1+(dy/dx)^2] = Sqrt[(dx/dy)^2+1]=Sqrt[dx^2+dy^2] Goal: Find arc length and surface area. -------------------------------------------------------- Previous assignment: January 14 6.3 Shells p292:1,(2),5,6,9,(10),(14),17,18,(20),22 Page 292 Problem 2 Find the volume of the solid generated when the region R bounded by the given curves is revolved about the indicated axis. Do this by performing the following steps. (a) Sketch the region R (b) Show a typical rectangular slice properly labeled (c) Write a formula for the approximate volume of the shell generated by this slice. (d) Set up the corresponding integral. (e) Evaluate this integral. y = x^2, x=1 y = 0 about the x-axis. | | | . | .| | . | | .__| | .|___|dy | . 1-x| _ | . | / |\ ---------------+-----------------------| ---- | \__/ y=1 y=1 y=1 INT 2 Pi y (1-x) dy = 2 Pi INT y (1-Sqrt[y]) dy = 2 Pi INT y-y^(3/2) dy y=0 y=0 y=0 | | y=1 = 2 Pi | y^2/2 - y^(5/2)/(5/2) | = 2 Pi (1/2 -2/5) = Pi/5 | | y=0 x=1 x=1 | | x=1 INT Pi y^2 dx = Pi INT x^4 dx = Pi | x^5/5 | = Pi/5 x=0 x=0 | | x=0 ---------------------------------------------------------------------------- Page 292 Problem 10 x = Sqrt[y] + 1, y = 4, x = 0, y = 0 about the x-axis. | 4|-------------* | 3|_1+Sqrt[y]___* |_____________| dy 2| * | * 1| * _ | * / |\ ---------------+----*----'----'----'---| ---- | \__/ y=4 y=4 INT 2 Pi y (1+Sqrt[y]) dy = 2 Pi INT y + y^(3/2) dy y=0 y=0 | | y=4 = 2 Pi | y^2/2 + y^(5/2)/(5/2) | = 2 Pi (8+2/5 32) = 208 Pi/5 | | y=0 x=3 x=3 3 Pi 4^2 - INT Pi y^2 dx = 48 Pi - Pi INT (x-1)^4 dx x=1 x=1 | | x=3 = 48 Pi - Pi | (x-1)^5/5 | = 48 Pi - Pi(32/5) = 208 Pi/5 | | x=1 -------------------------------------------------------- f[y_] := 1 + Sqrt[y]; a = ParametricPlot3D[ {f[y],y Cos[t],y Sin[t]},{y,0,4},{t,-Pi/2, Pi/2}]; b = ParametricPlot3D[ {x,4 Cos[t], 4 Sin[t]}, {x,0,3},{t,-Pi/2, Pi/2}]; c = Show[a,b,PlotLabel->"P292p10 x = 1 + Sqrt[y]"]; Display["10.ps",c]; ============================================================== Page 292 Problem 14 A region R is shown in Figure 9. Set up an integral for the volume of the solid obtained when R is revolved about each line. Use the indicated method. 3 ------------------------------ d----------------------- / \ x = g(y) \ / x = f(y) / \ \ / c------------------------------- y=d 2 2 (a) The y axis (washers) INT Pi f(y) - Pi g(y) dy y=c y=d (b) The x axis (shells) INT 2 Pi y (f(y)-g(y)) dy y=c y=d (c) The line y=3 (shells) INT 2 Pi (3-y) (f(y)-g(y)) dy y=c -------------------------------------------------------------- Page 292 Problem 20 Set up the integral (using shells) | | . a . dx | ' | `_ | ' | | | | ' | | |` | . | | | . | -------------- -a-------+---|-|--a---------------------------+------------ . | | | . b| | | | | ` | |_| | ` .|. ' | -a | | | | | x =a INT 2 Pi (b-x) (2y) dx x=-a x =a 4 Pi INT (b-x) Sqrt[a^2 - x^2] dx x=-a x =a x= a 4 Pi INT b Sqrt[a^2-x^2] - 4 Pi INT x Sqrt[a^2 - x^2] dx x=-a x=-a | (a^2-x^2)^(3/2) | x= a 4 Pi b (1/2) Pi a^2 - |------------------| | (3/2)(-2) | x=-a 2 2 Pi a^2 b ------------------------------------------------- a = 2; b = 5; p = ParametricPlot3D[ { (b+a Cos[t]) Cos[u], (b+ a Cos[t] ) Sin[u], a Sin[t] }, {t,0,2 Pi},{u,0,7 Pi/4}]; q = Show[p,PlotLabel->"P292p20, "a=2,b=5"]; Display["20.ps",q]; ======================================================================= ------------------------------------------------------------------------------ New Material ------------------------------------------------------------------------------ The length of a curve is defined to be the total length of approximating straight line segments. This idea fits nicely into the style of calculus. It also makes practical sense because when we follow a curve, we actually approximate as a path of line segments. The distance we will experience will be the limit of these types of paths. . . ' '. .' .ds' | dy ._ '________| dx From the Pythagorian theorem 2 2 ds = Sqrt[ dx + dy ] The basis formula for arc length is given in three ways and the one you use depends on how the curve is specified. (a) parametric R = (f[t], g[t]) (b) y = f(x) (c) x = g(y) t=b 2 2 (a) INT Sqrt[ (dx/dt) + (dy/dt) ] dt t=a t=b 2 (b) INT Sqrt[1+ (dy/dx) ] dx t=a t=b 2 (c) INT Sqrt[ (dx/dy) +1] dy t=a Each of these equations expresses the same idea. We approximate the curve with line segments and add the lengths of these segments. The introduction of the derivative is based on the mean value theorem that says f(b)-f(a) ------------ = f'(c) for some c where a < c < b. b-a /\y 2 Thus ds = Sqrt[ /\x^2 + /\y^2] = Sqrt[ 1 + (-----) ] /\x /\x ds = Sqrt[1+ (dy/dx)^2 ] /\x Page 295 Example 2. Find the circumference of the circle x^2 + y^2 = a^2 The path is given by (a Cos[t], a Sin[t]) t=2Pi The length = INT Sqrt[ (-a Sin[t])^2 + (a Cos[t])^2 ] dt t=0 t=2Pi = INT Sqrt[ a^2 (Sin[t]^2 + Cos[t]^2) ] dt t=0 t=2Pi = INT a dt = 2 Pi a t=0 Page 296 Example 3. Find the length of the line segment from (0,1) to (5,13). Answer: Length = Sqrt[ 12^2 + 5^2 ] = 13. Calculus should give us the same answer. y-1 = 12/5(x-0) y = 12/5 x + 1 check: when x = 0 y = 1. when x = 5. y = 13. x=5 x=5 INT Sqrt[1+(dy/dx)^2] = INT Sqrt[1+(12/5)^2] dx x=0 x=0 = Sqrt[1+144/25] 5 = Sqrt[169] = 13 Thank goodness, the answers are the same. Try working the problem using the formula x = 5/12(y-1). Page 196 Example 4. Find the length of the curve (2 Cos[t], 4 Sin[t]) from t=0 to t=Pi. t=Pi INT Sqrt[ 4 Sin[t]^2 + 16 Cos[t]^2 ] dt t=0 t=Pi INT Sqrt[ 4 + 12 Cos[t]^2 ] dt t=0 t=Pi INT 2 Sqrt[ 1 + 3 Cos[t]^2 ] dt Now we have to approximate this t=0 integral. Why not just approximate the length to begin with. NIntegrate[2 Sqrt[1+3 Cos[t]^2 ],{t,0,Pi}] = 9.68845 f[t_] := {2 Cos[t],4 Sin[t]}; a = ParametricPlot[f[t],{t,0,Pi}]; b = Show[a,PlotLabel->"Pag3 196 Ex 4 {2 Cos[t],4 Sin[t]} ",AspectRatio->Automatic]; Display["ex4.ps",b]; d[n_] := Sum[ Sqrt[(f[ Pi i/n] - f[Pi(i-1)/n] ). (f[ Pi i/n] - f[Pi(i-1)/n] )],{i,1,n}]; Do[ Print[n," ",N[d[n]]],{n,1,20}]; In[15]:= Do[ Print[n," ",N[d[n]]],{n,1,20}]; 1 4. 2 8.94427 3 9.2111 4 9.44982 5 9.52786 6 9.57865 7 9.60721 8 9.62635 9 9.63933 10 9.64866 11 9.65555 12 9.6608 13 9.66489 14 9.66813 15 9.67075 16 9.67289 17 9.67467 18 9.67616 19 9.67742 20 9.67849 Page 297 Example 5. Find the length of the curve y = x^(3/2) from (1,1) to (4,8). x=4 3/2 | x=4 3/2 3/2 L = INT Sqrt[ 1+ 9/4 x] dx = (1+9/4x) | = 8/27 (10 - (13/4) ) x=1 ---------- | (3/2)(9/4) | x=1 Integrate[Sqrt[1+9/4 x],{x,1,4}] = 7.63371 Page 298 Example 6. Find the area of the surface generated by revolving y = Sqrt[x] from x=0 to x=4 about the x axis. x=4 x=4 INT 2 Pi y ds = 2 Pi INT Sqrt[x] Sqrt[1+1/4 x^(-1)] dx x=0 x=0 3/2 | x=4 (x+1/4) | 3/2 3/2 2 Pi INT Sqrt[ x + 1/4 ] dx = 2 Pi ---------- | = 4 Pi/3 ( (17/4) - (1/4) ) 3/2 | x=0 3/2 Pi ( 17 -1) --------------- 6 --------------------------------------------------------- Write the parametric equation of a cycloid. ----------------------------------------- Find the surface area of a sphere x^2 + y^2 + z^2 = r^2 from the plane z = 0 to z = h.