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Unformatted text preview: Math 2B Dr. C. Famiglietti ************************************************************************ Section 5.4: Arc Length and Surface Area Arc Length (length of a curve) df If f = is continuous on [ a,b] , then the length of the curve (i.e., arc length) dx y = f ( x ) on a x b is:
b s = 1+ ( f ( x )) dx = a 2 b a dg If g = is continuous on [c,d ] , then the length of the curve (i.e., arc length) dy x = g( y ) on c y d is: d df 2 1+ dx . dx s = 1+ ( g( y )) dy = c 2 d c dg 2 1+ dy dy Surface Area (of a solid of revolution) df continuous dx on [ a,b] ), then the surface area of the resulting solid of revolutions is:
If curve y = f ( x ) , a x b , is rotated about the xaxis (with f = b b S= 2 f ( x )
a 1+ ( f ( x )) 2 dy 2 dx = 2 f ( x ) 1+ dx dx a Note: The above formulas lead to difficult integrals that often have to be integrated numerically. Example. Use integration to find the length of the line segment y = 2x + 3 between x = 1 and x = 3 . Check by the distance formula. Solution. f ( x ) = 2x + 3 f ( x ) = 2 Therefore,
b s= a 1+ ( f ( x )) dx
3 2 = 1 1+ 2 2 dx
3 = 5 dx
1 = 5x 3 1 = 5 ( 3 1) =2 5 2 2 Check using the distance formula, d = ( x 2  x1 ) + ( y 2  y1 ) where the points are: ( x1, y1) = (1,2(1) + 3) = (1,5) ( x 2 , y 2 ) = (3) + 3) = ( 3,9) (3,2 Therefore, d= (3 1) 2 + (9  5) 2 = 22 + 4 2 = 20 = 2 5 (the same as that obtained above) 3 4 3 2 Example. Find the length of the curve x = g( y ) = y 3  y 3 + 5 for 1 y 8 . 4 8 Solution. 3 43 3 2 g( y ) = y  y 3 + 5 4 8 1 3 4 3 2  13 g( y ) = y 3  y 4 3 8 3 1 1 1 3 =y 3 y 4 13 1  13 2 2 (g( y )) = y  4 y 1 1 1 1  1 1 1 1 1 1 1 1 = y 3y 3  y 3y 3  y 3y 3 + y 3 y 3 4 4 4 4 2 1 1 23 =y 3 + y 2 16 2 2 1 1 2 1+ ( g( y )) = 1+ y 3  + y 3 2 16 2 1 1 23 =y 3+ + y 2 16 13 1  13 2 = y + y 4 Therefore, d s= c 1+ ( g( y )) dy
8 2 = 1 13 1  13 2 y + y dy 4 1  13 y dy 4 1 8 4 2 y 3 1 y 3 = + 4 4 2 3 3 1 8 3 4 3 3 23 = y + y 1 4 8 3 4 3 2 3 4 3 2 = (8) 3 + (8) 3  (1) 3 + (1) 3 4 4 8 8 3 3 3 3 = (16) + ( 4 )   4 8 4 8 3 3 3 = 12 +   2 4 8 96 12 6 3 = +   8 8 8 8 99 = 8 = 8 y 1 3 + Example. Find the length of the curve y = sin x  x cos x for 0 x by evaluating the definite integral using Simpson's Rule with n = 4 . Solution. f ( x ) = sin x  x cos x f ( x ) = cos x  ( x (sin x ) + cos x ) = cos x + x sin x  cos x = x sin x 2 ( f ( x )) = x 2 sin2 x 1+ ( f ( x )) = 1+ x 2 sin 2 x Therefore,
2 b s= a 1+ ( f ( x )) dx 2 = and where 0 1+ x 2 sin 2 xdx x 3 0 1+ x 2 sin 2 xdx S4 = ( f ( x 0 ) + 4 f ( x1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) + f ( x 4 )) x = ba 0 = = n 4 4 x0 = a = 0 x1 = x 0 + 1 x = 0 + = 4 4 x 2 = x 0 + 2 x = 0 + 2 = 4 2 3 x 3 = x 0 + 3 x = 0 + 3 = 4 4 x4 = b = f ( x ) = 1+ x 2 sin 2 x f ( x 0 ) = f (0) = 1+ 0 2 sin 2 0 = 1
2 2 2 2 2 1 f ( x1 ) = f = 1+ sin = 1+ = 1+ 1.143864 4 4 4 16 2 32 2 2 2 2 2 f ( x 2 ) = f = 1+ sin = 1+ 1 = 1+ 1.862096 2 2 2 4 4 2 3 3 2 3 9 2 1 f ( x 3 ) = f = 1+ sin 2 = 1+ 4 4 4 16 2 9 2 1.943149 32 f ( x 4 ) = f ( ) = 1+ 2 sin 2 = 1+ 2 0 = 1 = 1+
so S4 = (1+ 4 (1.143864 ) + 2(1.862096) + 4 (1.943149) + 1) 4 3 = (18.072244 ) 12 = 4.731302
Therefore, s= 0 1+ x 2 sin 2 xdx 4.731302 x3 Example. Find the area of the surface generated by revolving y = , 1 x 7 , about 3 the xaxis. Solution. 1 f ( x) = x 3 3 f ( x ) = x 2 Therefore,
b S= 2 f ( x )
a
7 1+ ( f ( x )) dx 2 2 x3 = 2 1+ ( x 2 ) dx 3 1 2 = 3 7 x
1 3 1+ x 4 dx
u = 1+ x 4 du = 4 x 3 dx du x 3 dx = 4 when x = 1, u = 2 when x = 7, u = 1+ ( 7) 4 = 1+ 49 = 50 50 2 du = u 3 2 4 50 3 2 2 1 u = 3 4 3 2 2 3 3 2 = 50 2  2 2 6 3 3 3 = 50 2  2 2 9 Example. Find the area of the surface of revolution generated by revolving y = x , 0 x 4 , about the xaxis. Solution. f ( x) = x = x 2 1 1 1 f ( x ) = x 2 = 2 2 x Therefore,
b 1 S= 2 f ( x )
a
4 1+ ( f ( x )) dx
2 2 1 = 2 x 1+ dx 2 x 0 4 1 = 2 x 1+ dx 4x 0 4 4x +1 = 2 x dx 4x 0 4 4 x + 1 = 2 x dx 4x 0 4 4x +1 = 2 dx 4 0 4 4x +1 = 2 dx 2 0 4 = 0 4 x + 1dx 17 u = 4x +1 du = 4dx du dx = 4 when x = 0, u = 1 when x = 4, u = 17 du 4 1 17 3 u 2 = 4 3 2 1 3 2 3 = 17 2 1 2 4 3 3 = 17 2 1 6 = u ...
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This note was uploaded on 06/07/2008 for the course MATH Math2b taught by Professor Famigleitti during the Spring '08 term at UC Irvine.
 Spring '08
 Famigleitti
 Calculus, Arc Length

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