MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Lecture 31 18.01 Fall 2006 Lecture 31: Parametric Equations, Arclength,Surface Area Arclength, continued Example 1. Consider this parametric equation: x = t2 y = t3 for 0 ≤ t ≤ 1 x 3 = (t2)3 = t6; y 2 = (t3)2 = t6 = ⇒ x 3 = y 2 = ⇒ y = x 2/3 0 ≤ x ≤ 1 dsdydxdsdydxFigure 1: Infinitesimal Arclength. (ds)2 = (dx)2 + (dy)2 (ds)2 = (2t dt)2 + (3t2 dt)2 = (4t2 + 9t4)(dt)2 � �� � � �� � (dx)2 (dy)2 � t=1 � 1 �� 1 � Length = ds = 4t2 + 9t4dt = t 4 + 9t2dt t=0 0 0 1 = (4 + 9t2)3/2 ��� = 1 (133/2 − 43/2)27 0 27 Even if you can’t evaluate the integral analytically, you can always use numerical methods. 1Lecture 31 18.01 Fall 2006 Surface Area (surfaces of revolution) ydsa byxFigure 2: Calculating surface area ds (the infinitesimal curve length in Figure 2) is revolved a distance 2πy. The surface area of the thin strip of width ds is 2πy ds. Example 2. Revolve Example 1 (x = t2, y = t3 , 0 ≤ t ≤ 1) around the x-axis. Refer to Figure 3. yxFigure 3: Curved surface of a trumpet. 2� Lecture 31 18.01 Fall 2006 �� 13 �� 1 �2π t t 4 + 9t2 dt 4Area = 2πy ds = 0 ����� �� � = 2π t4 + 9t2 dt y ds 0 Now, we discuss the method used to evaluate t4(4 + 9t2)1/2dt We’re going to ignore the factor of 2π. You can reinsert it once you’re done evaluating the integral. We use the trigonometric substitution 2 2 t = tan u; dt = sec2 u du; tan2 u + 1 = sec2 u3 3 Putting all of this together gives us: � �� �4 � � ��1/2 � �2 4 2 t4(4 + 9t2)1/2 dt = tan u 4 + 9 tan2 u sec2 u du 3 9 3 � �5 �2 = tan4 u(2 sec u)(sec2 u du)3 This is a tan − sec integral. It’s doable, but it will take a long time for you to work the whole thing out. We’re going to stop evaluating it here. Example 3 Let’s use what we’ve learned to find the surface area of the unit sphere (see Figure 4). abrotate the curve by 2π radians xy..Figure 4: Slice of spherical surface (orange peel, only, not the insides). 3Lecture 31 18.01 Fall 2006 For the top half of the sphere, � y = 1 − x2 We want to find the area of the spherical slice between x = a and x = b. A spherical slice has area � x=b A = 2πy ds x=a From last time, dx ds = √1 − x2 Plugging that in yields a remarkably simple formula for A: � b � dx � b A = a 2π 1 − x2 √1 − x2 = a 2π dx = 2π(b − a) Special Cases For a whole sphere, a = −1, and b = 1. 2π(1 − (−1)) = 4π is the surface area of a unit sphere. For a half sphere, a = 0 and b = 1. 2π(1 − 0) = 2π
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