Solucao Calculo 5ed SEC.8.1e8.2

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(1)Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. 1. 1. 1. y=23x L= 1+({dy /dx) italic{dx}= 2. 2. 2. 1+(3) dx= 10 1(2) =3 10 .. 2. The arc length can be calculated using the distance formula, since the curve is a line segment, so 2. 2. L=[ distance from (2,8) to (1,1) ]= [1(2)] +[ ( 1 ) 8] = 90 =3 10 2. 2. Using the arc length formula with y= 4x 2. L=. 2. . dy dx. 1+. 0. . 2. 1. t. = 2lim sin (x/2) 0=2lim t 2. t 2. x. , we get 2. 4x. 2. 2. 1+. dx=. 0. dy = dx. x. 2. . dx=. 4x. t. 2dx. =2lim 2. . t 2 4x 1 1 sin (t/2)sin 0 =2 0 = 2 0. . dx 2. The curve is a quarter of a circle with radius 2 , so the length of the arc is. 2. 2 x. 0. 1 (2 2)= , as above. 4. 3.. From the figure, the length of the curve is slightly larger than the hypotenuse of the triangle formed 3/2 2 2 by the points ( 1,0 ) , ( 3,0 ) , and ( 3,f(3) ) ( 3,15) , where y= f (x)= . This length is about x 1 3 3/2 1/2 2 2 2 2 / 2 15 +2 15 , so we might estimate the length to be 15.5 . y= x 1 y = x 1 (2x) 3. (. (. / 2. 2. 1+(y ) =1+4x 3. L=. 1. = (183). ). (. ). ( x21) =4x44x2+1= ( 2x21) 2 , so, using the fact that 2x21>0 for 1 x 3 , 3. 3. (2x 1) dx= 2x 1 dx= (2x 1)dx= 2. ). 2. 2. 1. 1. 2. 2 3 x x 3. 3 1. 46 2 1 = =15.3 3 3 1.

(2) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. 4.. From the figure, the length of the curve is slightly larger than the hypotenuse of the triangle formed 2 3 by the points ( 0.5,f(0.5) 1 ) , ( 1,f(0.5) 1 ) and 1, , where y= f (x)=x /6+1/(2x) . This length is 3 1 2. about. 2. 2. 1 3. +. 0.6 , so we might estimate the length to be 0.65 .. 3. 2. 2. x 1 x / x y= + y = 6 2x 2 2. ( ). 1+ y. 4. 4. 4. 4. 2. x 1 x x 1 x =1+ + = + + = 4 2 4 4 2 4. / 2. 2. x x + 2 2. 2. so, using the fact that the parenthetical expression is positive, 1. . L=. 1/2. 1 1 6 2. =. 2. 2. 5. y=1+6x. 3/2. dx= . 1/2. 0. 1. L = 1+81x dx= u 1 2 3/2 u 81 3 3. 2. 1/2. dy/dx=9x 82. 2. 1. 2. x x + 2 2. 3. dx=. x 1 6 2x. 1 1/2. 1 31 1 = =0.64583 48 48. . 1. =. 2. x x + 2 2. 1/2. 82 1. =. 2. 1+(dy/dx) =1+81x . So. 1 81. du [ where u=1+81x and du=81dx ]. 2 ( 82 82 1 ) 243 3/2. 6. y =4(x+4) , y>0 y=2(x+4). 1/2. dy/dx=3(x+4) . 2. 1+(dy/dx) =1+9(x+4)=9x+37 . So. 2.

(3) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. 2. L = 9x+37 dx 0. 1 2 3/2 = u 9 3. 55. = u. u=9x+37, du=9dx. 1 9. 1/2. 37. du. 2 ( 55 55 37 37 ) 27. 55. =. 37. 5. x dy 5 4 3 4 1 = x x 7. y= + 3 6 dx 6 10 10x 25 8 1 9 8 25 8 1 9 8 2 1+ ( dy/dx ) =1+ x + x = x+ + x = 36 2 100 36 2 100 2. 5 4 3 4 x+ x 6 10. L = 1. 32 1 6 80. =. 2. 5 4 3 4 x+ x dx= 6 10. dx= 1. 1 1 6 10. . 2. =. 2. L= 2. x ln x + 2 4. dx=. dy dx. 4. 2. 1 x+ 4x. 1 5 1 3 x x 6 10. = 8+. 2. 2. 2. =x +. 1 3. 9. L = 1. =. 1 2. 2 1. 1 1 + . So 2 2 16x. 2ln 2 4. 2+. ln 2 4. =6+. 1 3/2 1/2 1 1/2 1 1/2 y y dx/dy= y y 3 2 2 1 1 1 1 1 1 1 1 1 1/2 1 1/2 2 y + y 1+ ( dx/dy ) =1+ y + y = y+ + y = 4 2 4 4 2 4 2 2 9. x=. . So. 31 7 1261 + = 6 80 240. x ln x dy 1 =x 1+ 8. y= 2 4 dx 4x 4. 2. 5 4 3 4 x+ x 6 10. ln 2 . 4. y ( y3) =. 1 1/2 1 1/2 1 y + y dy= 2 2 2 24. 8 3. =. 1 2. 64 3. 9. 2 3/2 1/2 y +2y 3 =. =. 1. 1 2. 2. . So. 2 27+2 3 3. 2 1+2 1 3. 32 3 2. 2. 2. 10. y=ln (cos x) dy/dx=tan x 1+(dy/dx) =1+tan x=sec x . So /3. L= . /3. sec x dx= sec xdx= ln sec x+tan x 2. 0. 0. /3 0. =ln ( 2+ 3 ) ln (1+0)=ln ( 2+ 3 ) .. 11. y=ln (sec x) 3.

(4) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. dy sec xtan x = =tan x 1+ dx sec x /4. L =. 2. 2. =1+tan x=sec x , so. /4. /4. 0. 0. /4. sec x dx= sec x dx= sec xdx= ln (sec x+tan x) 2. 0. =ln. 2. dy dx. (. 2 +1 ) ln (1+0)=ln. dy 1 12. y=ln x = dx x 2. 2. 1+. (. 0. 2 +1 ) 2. dy dx. =. 1+. 1 x. 2. 2. =. 1+x x. 3. . So L= 1. 2. 1+x dx . Now let x. 2. v= 1+x , so v =1+x and vdv=xdx. Thus 2. L=. 2. =. 2. v 2. v 1. v. vdv= 2. 1 ln 2. 1/2 1/2 1+ v1 v+1 2. v+1 v1. 2. =2. 1 1 dv= v+ ln v1 ln v+1 2 2 2 +1. 1 1 ln 3 2 + ln 2 2. 2 1. 2. =2 2 +ln. 2. (. 2 +1 ) . 1 ln 3 2. Or: Use Formula 23 in the table of integrals.. ( / ) 2=1+sinh 2x=cosh 2x .. /. 13. y=cosh x y =sinh x 1+ y 1. So L= cosh xdx= sinh x =sinh 1= 1. 0. 0 2. 14. y =4x , x=. 1 2 dx 1 y = y 1+ 4 dy 2 L= = u. /. x. 1 (e1/e) . 2. 2 0. dx dy 1+ 2. 2. =1+. 1 2 y . So 4. 1 1 2 2 y dy= 1+u 2du 0 4. 1+u +ln u+ 1+u. 2. 1. = 2 +ln ( 1+ 2 ). 0. ( / ) 2=1+e2x . So. x. 15. y=e y =e 1+ y 1. L=. 0. 2x. 1+e. e. dx= 1. 1+u. 2. du x u=e ,sox=ln u,dx=du/u u. 4.

(5) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. e. =. . 1+u u. 1. 2. 1+e. 2. 2. v. . udu=. 2. v 1. 2. 2. 2. vdv v= 1+u ,sov =1+u ,vdv=udu. 2. 2. 1+e. 1/2 1/2 1+ v1 v+1. . =. 2. 2. 1+e 1. 1 1+e + ln 2 2. =. 2. (. 1+e 2 +ln. x. 16. y=ln. (. (. 2. 2 1. (. 2 1 ). ). 1+u /u du , or substitute u=tan .. ) (. x. x. ). e. /. =ln e +1 ln e 1 y =. x. e 1. 2. ( e +1) = 1+ ( y ) =1+ ( e2x1) 2 ( e2x1) 2 2x. / 2. 2. 2 +1. ). 2. 1+e. 1 ln 2. 1+e 1 1ln. Or: Use Formula 23 for e +1. 2. 1+e +1. 2. =. 2. 1 v1 dv= v+ ln 2 v+1. 2x. 4e. / 2. ( ). 1+ y. x. e. . x. x. x. e +1. e 1. 2x. x. =. e +1 2x. e 1. b. cosh x b So L= dx= ln sinh x =ln sinh bln sinh a=ln a a sinh x. =. =. x. 2. 2. e 1 x. x. =. e e. sinh b sinh a. x. 2x. e +e. 17. y=cos x dy/dx=sin x 1+(dy/dx) =1+sin x . So L= 2. 2e. . cosh x . sinh x b. b. a. a. e e. =ln. .. e e 2. 1+sin x dx .. 0. ( ). .. 3. 18. y=2 dy/dx= 2 ln 2 L= x. x. 2 2x. 1+ ( ln 2 ) 2 dx. 0 4. 19. x=y+y dx/dy=1+3y 1+(dx/dy) =1+(1+3y ) =9y +6y +2 . So L= 3. 2. 20.. x. 2. a. 2. +. y. 2. b. 2. 2. 22. 4. 2. 4. 2. 9y +6y +2 dy .. 1 2. =1 , y= b 1 x. / a = ba 2. 2. 2. a x .. 5.

(6) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. b y= a. dy a x = dx 2. bx. 2. 2. 2. . 1/2. 2 2. 1+. a. b x 2. a. x. . a a x. a. So L=2. dy dx. ( a x ) 2. x. 2. x. b x. =. 2. a. a. 4 dx= a. 2 2. 2. . 2. dx .. 2. a x. 0. x. 2x. 1/2. ( b2a2) x2+a4 2x. 2. 21. y=xe dy/dx=e xe =e (1x) 1+(dy/dx) =1+e 2. .. ( a2x2). 2. (1x) . Let. 5. (1x) . Then L= f (x)dx . Since n=10 , x=. f (x)= 1+(dy/dx) = 1+e. 2. 0. LS. 10. 1/2 1 3 [ f (0)+4 f +2 f (1)+4 f +2 f (2)+4 f 3 2 2 7 9 +4 f +2 f (4)+4 f + f (5)] 5.115840 2 2. 5 2. =. 5 0 1 = . Now 10 2. +2 f (3)]. The value of the integral produced by a calculator is 5.113568 (to six decimal places). 22. x=y+ y dx/dy=1+. 1 2 y. 2. . Then L= g(y)dy . Since n=10 , y= 1. LS. 10. 1. 2. 1+ ( dx/dy ) =1+ 1+. 2 y. 2. =2+. 1 1 2 + . Let g(y)= 1+(dx/dy) y 4y. 2 1 1 = . Now 10 10. 1/10 [g(1)+4g(1.1)+2g(1.2)+4g(1.3)+2g(1.4)+4g(1.5)] 3 +2g(1.6)+4g(1.7)+2g(1.8)+4g(1.9)+g(2)] 1.732215 , =. which is the same value of the integral produced by a calculator to six decimal places. /3. 23. y=sec x dy/dx=sec xtan x L= f (x)dx , where f (x)= 1+sec xtan x . 2. 2. 0. /30 Since n=10 , x= . Now = 10 30 LS = /30 f (0)+4 f +2 f 10 3 30 6 7 +2 f +4 f +2 f 30 30. 2 +4 f 30 8 +4 f 30. 3 +2 f 30 9 +f 30. 4 30 3. +4 f. 5 30. 1.569619 . 6.

(7) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. The value of the integral produced by a calculator is 1.569259 (to six decimal places). 2. 2. 24. y=xln x dy/dx=1+ln x . Let f (x)= 1+(dy/dx) = 1+(1+ln x) . 3. Then L= f (x)dx . Since n=10 , x= 1. 3 1 1 = . Now 10 5. 1/5 [ f (1)+4 f (1.2)+2 f (1.4)+4 f (1.6)+2 f (1.8)+4 f (2)] 3 +2 f (2.2)+4 f (2.4)+2 f (2.6)+4 f (2.8)+ f (3)] 3.869618 . The value of the integral produced by a calculator is 3.869617 (to six decimal places). LS. 10. =. 25. (a). (b) 3. Let f (x)=y=x 4x . The polygon with one side is just the line segment joining the points ( 0,f(0) ) = ( 0,0 ) and ( 4,f(4) ) = ( 4,0 ) , and its length is 4 . The polygon with two sides joins the points. ( 0,0 ) , ( 2,f(2) ) = ( 2,2 2 ) and ( 4,0 ) . 3. Its length is 2. (. (20) + 2. 3. 2 0. )2 +. 2. (. (42) + 02. 3. 2. ) 2 =2. 8/3. 4+2. (. 6.43. 3. ) (. 3. ). Similarly, the inscribed polygon with four sides joins the points ( 0,0 ) , 1, 3 , 2,2 2 , ( 3,3) , and ( 4,0 ) , so its length is 1+. ( 3 3 )2 +. (. 1+ 2. 3. 3. 2 3. )2 +. (. 1+ 32. 3. 2. )2 +. 1+9 7.50 7.

(8) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. (c) Using the arc length formula with. 1 2/3 124x 3 (4x) (1) + 4x = , the length of the 2/3 3 3(4x). dy =x dx. 4. 4. curve is L= 0. 1+. 2. dy dx. . dx=. 2. 124x. 1+. dx .. 2/3. 3(4x) (d) According to a CAS, the length of the curve is L 7.7988 . The actual value is larger than any of the approximations in part (b). This is always true, since any approximating straight line between two points on the curve is shorter than the length of the curve between the two points. 0. 26. (a) Let f (x)=y=x+sin x with 0 x 2 .. (b) The polygon with one side is just the line segment joining the points ( 0,f(0) ) = ( 0,0 ) and 2. 2. (2 0) +(2 0) =2 2 8.9 . The polygon with two sides joins the points ( 0,0 ) , ( ,f( ) ) = ( , ) , and ( 2 ,2 ) . Its length is. ( 2 ,f(2 ) ) = ( 2 ,2 ) , and its length is 2. 2. 2. 2. ( 0) +( 0) + (2 ) +(2 ) = 2 + 2 =2 2 8.9 Note from the diagram that the two approximations are the same because the sides of the 2 sided 1 polygon are in fact on the same line, since f ( )= = f (2 ) . 2. The foursided polygon joins the points ( 0,0 ) ,. , +1 2 2. , ( , ) ,. 3 3 , 1 2 2. , and. ( 2 ,2 ) , so its length is 8.

(9) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. 2. +1 2. 2. +. 2. 2. +. 2. +. 1 2. 2. 2. +. 2. +. 1 2. 2. +. 2. 2. +. +1 2. 2. 9.4. (c) Using the arc length formula with dy/dx=1+cos x , the length of the curve is 2. 2. L= . 1+(1+cos x) dx= . 0. 0. 2. 2. 2+2cos x+cos x dx. (d) The CAS approximates the integral as 9.5076 . The actual length is larger than the approximations in part (b). dx 2y = 1+ 27. x=ln 1y 2 dy 1y. (. 2. ). 2 2 ( 1+y ) =1+ = 2 2 ( 1y ) ( 1y2) 2 2. 2. dx dy. 4y. . So. 1/2. 0. 4/3. dy/dx=. . ( 1+y ) ( 1y2) 2. L=. 28. y=x. 1/2. 2 2. dy=. 0. 2. 1+y. 2. 1y. dy=ln 3. 1 0.599 2. 4 1/3 16 2/3 2 x 1+(dy/dx) =1+ x 3 9. 4/3 16 2/3 4 1/3 4 2/3 9 2/3 9 9 2 81 2 2 81 2 x dx= 1+u u du u= x ,du= x dx,dx= x du= u du= u du 0 0 64 9 3 9 4 4 16 64 4/3 1 81 2 2 1 2 = u 1+2u 1+u ln u+ 1+u 0 8 64 8 1 25 1 1 41 5 1 81 32 4 25 81 = 1+ ln + = ln 3 6 9 6 9 3 8 64 9 8 3 9 64 205 81 = ln 3 1.4277586 128 512. L=. 1. 1+. (. 2/3. (. ). 2/3. (. 2/3 3/2. 29. y =1x y= 1x. ). . ). dy 3 2/3 = 1x dx 2. (. ) 1/2. . 2 1/3 1/3 2/3 x =x 1x 3. (. ) 1/2 9.

(10) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. 2. dy dx. =x. 1. (. L=4. 2/3. 1+ x. ( 1x2/3) =x2/31 . Thus. 2/3. 1. ). 1 dx=4 x. 0. 1/3. +. 0. 1. 3 2/3 x 2. dx=4lim t 0. t. =6 .. 30. (a) 2/3. (b) y=x. 1+. integral]. x=y. 3/2. 2. dy dx. 2 1/3 x 3. =1+ 2. dx dy. 1+. 1. 2. 4 2/3 =1+ x . So L= 9 0. =1+. 9 =1+ y . So L= 4 0. 4 2 9 The second integral equals 1+ y 9 3 4 integral can be evaluated as follows: 1. 0. 4 2/3 x dx [ an improper 9. 1. 2. 3 1/2 y 2. 1+. 1. 3/2. 8 27. =. 0. 1+. 9 y dy . 4. 13 13 13 13 8 1 = . The first 27 8. 1. 4 2/3 1+ x dx = lim 9 +. . t 0 t 9. =. 0. 2/3. 9x +4 1/3. 3x. u+4 1 du= 18 18. 9. dx=lim +. . t 0 9t. 2/3. 2 3/2 (u+4) 3. u+4 2/3 1/3 du u=9x ,du=6x dx 18 9. =. 0. 1 3/2 3/2 13 13 8 (13 4 )= 27 27. (c) L = length of the arc of this curve from ( 1,1 ) to ( 8,4 ) 1. = 0. 4. 9 1+ y dy+ 4 0. 13 13 8 8 9 1+ y dy= + 27 4 27. 9 1+ y 4. 3/2. 4 0. 10.

(11) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. =. 13 13 8 8 13 13 +80 10 16 + 10 10 1 ) = ( 27 27 27. 31. y=2x. 3/2. /. 1/2. y =3x. ( / ) 2=1+9x . The arc length function with starting point P0 ( 1,2 ) is. 1+ y x. 2 3/2 (1+9t) 27. s ( x ) = 1+9t dt= 1. x. =. 1. 2 3/2 (1+9x) 10 10 27. 32. (a) dy dx. (b) 1+ x. s ( x) =. . 2. 2. 4. =x +. 1 1 , + 4 2 16x. ( 2). t +1/ 4t. dt. 1. 1 3 t 1/(4t) 3. =. x 1. 1 1 1 3 x 1/(4x) 3 4 3 1 1 3 = x 1/(4x) for x 1 12 3 =. (c) 33. The prey hits the ground when y=0

(12) 180 /. positive. y =. 2 x 1+ y 45. ( / ) 2=1+. 4. 1 2 2 x =0

(13) x =45 180 x= 8100 =90 , since x must be 45. 2. 2. x , so the distance traveled by the prey is. 45. 11.

(14) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. L=. 90 0. 45 = 2 45 = 2 34. y=150. 4. 1+. x dx= 2. 2. 45. 4. 1+u. 0. 45 du 2. 2. 4 1 2 1+u + ln u+ 1+u 0 2 1 45 2 17+ ln ( 4+ 17 ) =45 17+ ln ( 4+ 17 ) 209.1m 2 4. 1 u 2. (. 2. ). 1 1 2 / (x50) y = (x50) 1+ y 40 20. ( / ) 2=1+. 1. 2. 2. (x50) , so the distance traveled by the. 20. kite is 80. . L=. 1+. 1. 3/2. 2. 2. (x50) dx=. 2. 1+u (20du) [u=. 5/2. 20. 0. . 1 1 (x50),du= dx] 20 20. 3/2 1 2 1+u + ln u+ 1+u =20 5/2 2 13 29 3 13 5 =10 +ln + + ln 4 4 2 4 2 3+ 13 15 25 = 13 + 29 +10ln 122.8ft 2 2 5+ 29. 1 u 2 3 2. (. 2. ). . 5 + 2. 29 4. 35. The sine wave has amplitude 1 and period 14 , since it goes through two periods in a distance of 2 x =sin x . The width w of the flat metal sheet needed 28 in., so its equation is y=1sin 7 14 to make the panel is the arc length of the sine curve from x=0 to x=28 . We set up the integral to dy evaluate w using the arc length formula with = cos x : dx 7 7 28. 14. 2 2 L= 1+ cos x dx=2 1+ cos x dx . This integral would be very 7 7 7 7 0 0 difficult to evaluate exactly, so we use a CAS, and find that L 29.36 inches.. 36. (a) y=c+acosh. x a. /. y =sinh. x a. ( / ) 2=1+sinh 2. 1+ y. x a. =cosh. 2. x a. . So. 12.

(15) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. b. L= . cosh. b. 2. x a. b. dx=2 cosh 0. x a. dx=2 asinh. x a. b. =2asinh. 0. b a. (b) At x=0 , y=c+a , so c+a=20 . The poles are 50 ft apart, so b=25 , and L=51 51=2asinh (b/a) . From the figure, we see that y=51 intersects y=2xsinh (25/x) at x 72.3843 for x>0 . So a 72.3843 and the wire should be attached at a distance of y=c+acosh (25/a)=20a+acosh (25/a) 24.36 ft above the ground.. x. 37. y= 1. dy 3 t 1 dt = x 1 [ by FTC1] 1+ dx 3. 4. L=. 4. x dx= x 3. 1. 1. 3/2. dx=. dy dx. 2. =1+. (. 3. x 1. 2. ) =x 3. 2 5/2 4 2 62 x = (321)= =12.4 1 5 5 5. 38. By symmetry, the length of the curve in each quadrant is the same, so we’ll find the length in the 2k 2k 2k 2k 2k 1/ ( 2k ) first quadrant and multiply by 4 . x +y =1 y =1x y= 1x (in the first quadrant), so we use the arc length formula with dy 1 2k 1/ ( 2k ) 1 2k1 = 1x 2kx dx 2k = x2k1 1x2k 1/ ( 2k ) 1 The total length is therefore. (. (. ). (. (. ). ). ). 13.

(16) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth. 1. L =4 2k. 2k1. 1+ x. 0. 2k 1/ ( 2k ) 1 2. ( 1x ). 1. dx=4. 2 ( 2k1 ). 1+x. ( 1x2k ) 1/k2 dx Now from the graph, we. 0. see that as k increases, the ‘‘corners’’ of these fat circles get closer to the points ( 1, 1 ) and ( 1, 1 ) , and the ‘‘edges’’ of the fat circles approach the lines joining these four points. It seems plausible that as k , the total length of the fat circle with n=2k will approach the length of the perimeter of the square with sides of length 2 . This is supported by taking the limit as k of the 2k 1/ ( 2k ) equation of the fat circle in the first quadrant: lim 1x =1 for 0 x<1 . So we guess that k . (. ). lim L =4 2=8 .. k . 2k. 14.

(17) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution. 3. 1. y=ln x ds= 1+(dy/dx) dx= 1+(1/x) dx S= 2 (ln x) 1+(1/x) dx 2. 2. 2. 1 /2. 2. y=sin x ds= 1+(dy/dx) dx= 1+(2sin xcos x) dx S= 2 sin x 1+(2sin xcos x) dx [ by 2. 2. 2. 2. 2. 0. (7)] /4. 3. y=sec x ds= 1+(dy/dx) dx= 1+(sec xtan x) dx S= 2 x 2. 2. 2. 1+(sec xtan x) dx [ by (8)]. 0. x. 2. 2x. 4. y=e ds= 1+(dy/dx) dx= 1+e. ln 2. dx S= 2 x. 2x. 1+e. 2. dx [ by (8)] or 2 (ln y). 0. 2. 1+ ( 1/y ) dy. 1. [ by (6)] 3. /. 2. 5. y=x y =3x . So 2. S = 2 y. / 2. ( ). 1+ y. 2. dx=2 x. 0. 2 = 36. 3. 4. 4. 3. 1+9x dx[u=1+9x ,du=36x dx]. 0 145. . 18. u du=. 1. 2 3/2 u 3. 145 1. =. ( 145 145 1 ) 27. 2. 6. The curve 9x=y +18 is symmetric about the x axis, so we only use its top half, given by y=3 x2 3 9 2 . dy/dx= , so 1+(dy/dx) =1+ . Thus, 4(x2) 2 x2 6 9 1+ dx=6 2 4(x2). S = 2 3 x2 6. 2. 2 =6 3. 3/2. 1 x+ 4. 6 2. 2. 7. y= x 1+(dy/dx) =1+ 1/ ( 2 x 9. S = 2 y 4. 1+. 25 4. =4. dy dx. 2. 9. ). 3/2. . 9 4. x+. 3/2. =4. 1/2. 1 4. dx 125 27 8 8. =4 . 98 =49 8. 2. =1+1/(4x) . So. dx= 2 x 4. 6. 9 x2+ dx=6 4 2. 9. 1 1+ dx=2 4x 4. x+. 1 dx 4. 1.

(18) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution. 2 3. = 2. 9. 3/2. 1 x+ 4. =. 4. 9. 1 3/2 (4x+1) 8. 4 3. 2. =. 4. ( 37 37 17 17 ) 6. 2. 8. y=cos 2x ds= 1+(dy/dx) dx= 1+(2sin 2x) dx /6. S= 2 cos 2x. 3. 2. 1+4sin 2x dx=2. 0. =. . 1+u. 1 4. 2. 0. 2. 1 u 2. 2. 1+u +. (. 1 2 ln u+ 1+u 2. 2. 2. ). 3. =. 0. du[u=2sin 2x,du=4cos 2xdx]. 2. 3 1 2+ ln 2 2. (. 3 +2 ). =. 3 + ln ( 2+ 3 ) 2 4. 2. 9. y=cosh x 1+(dy/dx) =1+sinh x=cosh x . So 1. 1. 0. 0. S = 2 cosh xcosh xdx=2 =. 1+. 1 sinh 2 2. or. 3. 1 (1+cosh 2x)dx= 2. 1+. 1 2 2 e e 4. (. 1. 1 x+ sinh 2x 2. 0. ). 2. x 1 dy x 1 10. y= + = 2 6 2x dx 2 2x 4. 2. 1+(dy/dx) =. x 1 + 2 2 2x. 1. . 2. x 1 + 2 2 2x. x 1 + 6 2x. 1/2 1. 3. 5. . 1 1 1 + 72 6 8. =2. 1 2 y +2 3. (. dx=2. dx=2 . . 5. x x x 1 + + + 3 12 12 4 4x. 1/2. 6. x x x + + 12 3 4. 1/2. 11. x=. 2. x 1 = + 2 2 2x. 1. 3. S = 2. =2. 2. 2. x 1 1 + + = 4 4 2 4x. 2. 2. 1. x x x + 72 6 8. 1 1 1 + 64 72 24 2. ) 3/2 dx/dy= 12 ( y2+2) 1/2(2y)=y. dx. 1/2. =2 2. 263 512. =. 263 256 2. 2. y +2 1+(dx/dy) =1+y. ( y2+2) = ( y2+1) 2 . So 2.

(19) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution. 2 1. 2. 2. 2. 1 4 1 2 y+ y 4 2. 2 S=2 y ( y +1 ) dy=2. 2. =2. 1 1 4 2. 4+2. 1. =. 21 2. 2. 12. x=1+2y 1+(dx/dy) =1+(4y) =1+16y . So 2. S = 2 y 1. 2. 1+16y dy= 16. ( 16y +1 ). 2. 2. 1/2. 32ydy=. 1. 16. 2 2 16y +1 3. (. ) 3/2. 2 1. ( 65 65 17 17 ) 24. =. 3. 3. 2. 4. 13. y= x x=y 1+(dx/dy) =1+9y . So 2. S = 2 x. 2. 1+(dx/dy) dy=2 y 2. 1. 4. 1. 18. =. 2 1+9y dy= 36. 3. 2 4 1+9y 3. (. 2. =. 1. 2. . 4. 3. 1+9y 36y dy. 1. ( 145 145 10 10 27. 2. ) 3/2. 2. ). 2. 14. y=1x 1+(dy/dx) =1+4x 1. S=2 x 0 2. 2. 15. x= a y dx/dy= 2. y. 2. 1+(dx/dy) =1+. 2. 2. a y. 1. 1+4x dx= 8x 4 0 2. =. 2. 4x +1 dx=. 4. 2 2 4x +1 3. (. ) 3/2. 1. =. 0. ( 5 5 1 ) 6. 1 2 2 1/2 2 2 (a y ) (2y)=y/ a y 2 2. 2. 2. 2. a y. 2. +. a y. 2. y 2. 2. a y. a. =. 2. 2. a y. . a/2. . S=. 0. 2. 2. 2. a y. a/2. a 2. dy=2 2. ady=2 a 0. a y. y. a/2 0. =2 a. a 2 0 = a . Note that this is 2. 1 1 the surface area of a sphere of radius a , and the length of the interval y=0 to y=a/2 is the 4 4 length of the interval y=a to y=a . 2. 2. 2. 16. x=acosh (y/a) 1+(dx/dy) =1+sinh (y/a)=cosh (y/a) . So 3.

(20) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution. a. y a. S = 2 acosh a. cosh. a y+ sinh 2. = 2 a. a. y a. dy=4 a cosh 0. a. 2y a. 2. =2 a. 0. a. y a. dy=2 a 0. a 2 a+ sinh 2 =2 a 2. 17. y=ln x dy/dx=1/x 1+(dy/dx) =1+1/x S= 2 ln x 2. 3. 2. 1. 2. 1+1/x . Since n=10 , x=. Let f (x)=ln x. 2y a. 1+cosh. 1 1+ sinh 2 2. dy 2. or. a. ( e2+4e2) 2. 2. 1+1/x dx .. 3 1 1 = . Then 10 5. 1/5 f (1)+4 f (1.2)+2 f (1.4)+ +2 f (2.6)+4 f (2.8)+ f (3) 9.023754 . 10 3 The value of the integral produced by a calculator is 9.024262 (to six decimal places).. SS =2 . 18. y=x+ x dy/dx=1+. 1 1/2 2 1/2 1 1 x 1+(dy/dx) =2+x + x 2 4. 2. S= 2 ( x+ x. 1 1 1 1 . + dx . Let f (x)=(x+ x ) 2+ + x 4x x 4x 1 2 1 1 = Since n=10 , x= . Then 10 10 1/10 SS =2 f (1)+4 f (1.1)+2 f (1.2)+ +2 f (1.8)+4 f (1.9)+ f (2) 29.506566 . 10 3 The value of the integral produced by a calculator is 29.506568 (to six decimal places).. ). 2+. 2. 2. 2. 19. y=sec x dy/dx=sec xtan x 1+(dy/dx) =1+sec xtan x /3. S= 2 sec x. 2. 2. 1+sec xtan x dx . Let f (x)=sec x. 2. 2. 1+sec xtan x .. 0. /30 . Then = 10 30 /30 2 8 9 SS =2 f (0)+4 f +2 f + +2 f +4 f +f 10 3 30 30 30 30 . The value of the integral produced by a calculator is 13.516987 (to six decimal places). Since n=10 , x=. dy 1 x 1/2 x 20. y=(1+e ) = (1+e ) e = dx 2 x 1/2. e. 3. 13.527296. x x 1/2. . 2(1+e ). 4.

(21) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution. 1+. x. 2x. 2. dy dx. e. =1+. =. x. =. x. 4(1+e ). x. 2x. 4+4e +e. 4(1+e ). 2. (e +2). . x. 4(1+e ). 1. . S=. 2. 1+e. x. 2. 0. 1. e +2. x. 1+e. x. 1. x. dx= (e +2)dx= e +2x = [(e+2)(1+0)]= (e+1) .. x. 0. 0. 1 x 1 0 1 . Then (e +2) . Since n=10 , x= = 2 10 10 1/10 SS =2 f (0)+4 f (0.1)+2 f (0.2)+ +2 f (0.8)+4 f (0.9)+ f (1) 11.681330 . 10 3 The value of the integral produced by a calculator is 11.681327 (to six decimal places). Let f (x)=. 2 2. (. 2. ). 21. y=1/x ds= 1+(dy/dx) dx= 1+ 1/x. 4. dx= 1+1/x dx. 2. S=. . 2. 1 2 x. 1. 1+. x. 1. dx=2. 4. . 1+u u. 1. =. . x. dy 22. y= x +1 = dx. x. (. ds=. 2. u. 1. 2 17 +ln ( 4+ 17 ) + ln ( 1+ 2 4 1. 2. . 1+u 2 +ln u+ 1+u u. . du=. 3. dx=2. 2. u +1. 1 2. 2. 2. du[u=x ,du=2xdx]. 4. 2. 2. 2. . x +1. 1. 4. =. 4. 4. ). dy dx. 1+. x +1. ). 1. 2. =. 17 +ln 4. 4+ 17 1+ 2. 2. 2. dx=. 1+. x 2. dx. x +1. 3. S. . =. 2. 2. 2. x +1. 1+. 2. x +1. 0. =2 2. x. 1 x 2. dx=2 . 1 1 x + + ln 2 4 2. 3. 0. x+. 3. 2x +1 dx=2 2 2. 0. 1 x+ 2 2. 2. x+. 1 2. 2. dx. 3 0. 5.

(22) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution. =2 2. 3 2. =2 2. 3 2. 9+ 19 2. 1 1 + ln 2 4. 3+. 9+. 1 ln ( 3 2 + 19 ) 4. +. 3. /. 1 2. 1 1 ln 4 2. . 19 1 + ln 2 4. 3 2. =2 2. 19 2. 3+. ln ( 3 2 + 19 ) 2. =3 19 +. 2. 23. y=x and 0 y 1 y =3x and 0 x 1 . 1. 3. 2 2. ( ). S = 2 x. dx=2 . 1+ 3x. 0. = 3. 3. . 1+u du= [or use CAS ]. 3. 3 1 10 + ln ( 3+ 10 2 2. =. 2. 0. 24. y=ln (x+1) , 0 x 1 . ds= 1. . =. 1 2 du[u=3x ,du=6xdx] 6. 2. 0. 3. =. 1+u. ). 1 u 2. (. 1 2 ln u+ 1+u 2. 3 10 +ln ( 3+ 10 6 2. dy dx. 1+. 2. 1+u +. dx=. 3 0. ). 1 x+1. 1+. ). 2. dx , so S. 2. 2 x. ( x+1 ). 0. 2. 1. 2 (u1). u. 2. . 2. du[u=x+1,du=dx] 2. 2. 1+u du=2 u 1. 1. 1. 1+. 1. 2. 2. . dx=. 2. 1+u du2 u. =2 u. . 1. 1+. 2. 1+u du2. 2. . 1+u du u. 1. 2 1 2 1 2 u 1+u + ln u+ 1+u 2 = [or use CAS ] 2 1 2 2 1 1 1 =2 5+ ln ( 2+ 5 ) 2 ln ( 1+ 2 ) 2 5 ln 2 2 2 1+ 5 2 3 1 =2 ln ( 2+ 5 ) +ln + ln ( 1+ 2 ) 2 2 2 2. (. ). 2. 1+u ln 1+ 5 2. 1+ 1+u u. 2. 2 1. 2 +ln ( 1+ 2 ). 25. S=2 y 1. 1+. dy dx. 2. dx=2. 1. 1 x. 1+. 1 x. 4. dx=2. 1. 4. x +1 x. 3. dx . Rather than trying to. 6. +. 1 4.

(23) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution. 4. 4. 2. x +1 > x =x for x>0 . Thus, if the area is finite,. evaluate this integral, note that. S=2. . 4. x +1 x. 1. 3. dx>2. 1. 2. x x. dx=2 . 3. 1. 1 dx x. But we know that this integral diverges, so the area S is infinite.. 26. S= 2 y. x. x 2. 1+(dy/dx) dx=2 e 2. 0. 1+(e ) dx .. 0 x. Evaluate I= e. x 2. x. x. 1+(e ) dx by using the substitution u=e , du=e dx .. (. ). 1 2 1 2 2 I= 1+u du = 2 u 1+u + 2 ln u+ 1+u +C 1 x 2x 1 x 2x = (e ) 1+e + ln e + 1+e +C 2 2. (. ). Returning to the surface area integral, we have t. S =2 lim. x. e. t 0. 1 x (e ) 2. =2 lim. t . =2 lim. t . =2 =2. { {. x 2. 1+(e ) dx. {. 2x. 1+e. 1 t (e ) 2. +. 2t. 1+e. (. 1 x 2x ln e + 1+e 2 +. 1 1 (0) 1 + ln ( 0+ 1 2 2 1 [0]+ 2 ln ( 2 1 ) 2 2. (. 1 t 2t ln e + 1+e 2. ). }. = 2. t. ). 0. ). . 1 1 2 + ln ( 1+ 2 2 2 2 ln. (. 1 1 (1) 1+1 + ln ( 1+ 1+1 2 2. ). ). }. }. 2 1 ) 2. 2. 27. Since a>0 , the curve 3ay =x(ax) only has points with x 0 . (3ay 0 x(ax) 0 x 0 . ) The curve is symmetric about the x axis (since the equation is unchanged when y is replaced by y ). y=0 when x=0 or a , so the curve’s loop extends from x=0 to x=a .. 7.

(24) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution. d d dy dy (ax)[2x+ax] 2 2 (3ay )= 6ay =x 2(ax)(1)+(ax) = dx dx dx dx 6ay 2. 2. dy dx. =. 2 2. 2. =. 2. (ax) (a3x) 2. 2. 2. (a3x) = 12ax. 2. 2. is 1/y. x(ax). 36a. 2. 2. the last fraction. 3a. . 2. 36a y. 2. 2. 2. a 6ax+9x 12ax a 6ax+9x a +6ax+9x (a+3x) =1+ = + = = for x

(25) 0 . 12ax 12ax 12ax 12ax 12ax. 2. dy dx. 1+. 2. (ax) (a3x). (a) a. a. x=0. 0. x (ax). S = 2 yds=2 . 3a. a. (ax)(a+3x) a+3x dx=2 dx 6a 12ax 0. a. 2. 2 2 2 2 3 a 3 3 3 3 a = (a +2ax3x )dx= a x+ax x = (a +a a )= a = . 0 3a 3a 0 3 3a 3a Note that we have rotated the top half of the loop about the x axis. This generates the full surface. (b) We must rotate the full loop about the y axis, so we get double the area obtained by rotating the top half of the loop: a. a. a+3x 4 S =2 2 xds=4 x dx= 12ax 2 3a x=0 0 2 = 3a. a. (ax. 1/2. 3/2. +3x )dx=. 0. 2 3 = 3. 2 6 + 3 5. 2 3a. a. x. 1/2. (a+3x)dx. 0. 2 3/2 6 5/2 ax + x 3 5. 2 3 a= 3. 28 15. 2. a. =. 2 3 3 a. 0. 2 5/2 6 5/2 a + a 3 5. 2. 56 3 a a= 45 2. 2. 28. In general, if the parabola y=ax , c x c , is rotated about the y axis, the surface area it generates is c. 2 x 0. 2. 1+(2ax) dx = 2. 2ac. 0. u 2a. 1+u. 2. 1 du u=2ax,du=2adx 2a. 8.

(26) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution. 2ac. . =. 2. 4a . =. 2. . . ( 1+u2) 1/22udu=. 2 2 1+u 3. (. 2. 0. 4a. ) 3/2. 2ac 0. ( 1+4a2c2) 3/21. 6a. 2 . Thus, the surface area is 25. 2. Here 2c=10 ft and ac =2 ft, so c=5 and a=. 3/2 4 625 625 1+4 25 1 = 625 24 6 4 5 2 = 41 41 125) 90.01ft ( 24. S=. 2. 29.. x. 2. 2. +. a 1+. y. 2. y(dy/dx). =1. 2. b. dy dx. b. 1+. 3/2. 16 25. 1 =. 625 24. 41 41 1 125. 2. dy b x = = 2 2 dx a y a x. 2 2 4 2 4 2 2 2 2 ( 1x /a ) a b +b x a b x = = = 1+ 4 2 4 2 4 2 2 2 4 2 2 2 2 ( ) a y a y a b 1x /a a b a b x 4 2 2 2 2 4 2 2 2 a ( a b ) x a +b x a x = 4 2 2 2 2 2 a a x a ( a x ) 4 2. b x. 2. =. =. 4 2. 4 2. b x +a y. 4 2. 4 2. b x +a b. The ellipsoid’s surface area is twice the area generated by rotating the first quadrant portion of the ellipse about the x axis. Thus, a. S=. a. 2 2 y. 1+. 0. 2. dy dx. dx=4. . b a. 4. a x. a. 4 b 2. a. a. . 4. 2. a. 2. 2. ) x2 dx= 42b a. 0. 4 b. =. (. a a b. 2. 2. a b. u 2. 4. 2. 2. ) x2. 2. 2. 2. 2. dx. a x. 2. a b. . 4. a u. du. 2. 2. 2. 2. 2. [u= a b x]. a b. 0. a 1 u a u + sin 2 2 a 4. (. a a b. 2. 0. a. =. 2. a. 2. 2. a b. 0. 9.

(27) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution. 2. 2. 2. 2. a b. a. 2. a b 2. a. 4 b. =. 4. 2. a a. ( a b ) 2. 2. 4. a 1 + sin 2. 2. 2. 2. a b a. 2. a bsin. 1. b+. =2. 2. 2. 2. a b a 2. a b 2. 2. 2. 30. The upper half of the torus is generated by rotating the curve (xR) +y =r , y>0 , about the y 2. 2. dy dx. dy axis. y =(xR) 1+ dx. (xR). =1+. 2. =. 2. 2. y +(xR) 2. y. =. r. 2. 2. r ( xR ). y. 2. . Thus,. R+r. S=. R+r. 2 2 x. 2. dy dx. 1+. Rr. dx=4. . rx 2. dx 2. r (xR). Rr. r. =. 4 r. . u+R 2. r u. r. du[u=xR] 2. r. =. 4 r. r. . udu 2. +4 Rr. r u. r. 2. . du 2. r u. r. 2. r. =. 4 r 0+8 Rr. . du 2. r u. 0. 1. [since the first integrand is odd and the second is even ] 2. r. = 8 Rr sin (u/r) =8 Rr 0. 31. The analogue of f x. *. i. S=lim. n. i=12. c f x. i. n . 1/2. 32. y=x. *. /. y =. 2. 2. =4 Rr. in the derivation of (4) is now c f x. *. i. 1+ f. /. x. *. i. 2. b. x= 2 c f (x). , so /. 1+ f (x). 2. dx .. a. 1 1/2 x 1+ y 2. ( / ) 2=1+1/4x , so by Exercise 31, 10.

(28) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution. 4. S= 2 ( 4 x ) 1+1/(4x) dx . Using a CAS, we get S=2 ln. (. 17+4 ) +. 0 2. 2. /. 2. ( 31 17+1 ) 80.6095 . 6. 2. 33. For the upper semicircle, f (x)= r x , f (x)=x/ r x . The surface area generated is r. S = 1. . r. (. 2. 2. 2 r r x. 2. ). 1+. x 2. 2. dx=4. r x. r. (. 2. 2. r r x. ). r 2. dx 2. r x. 0. r. =. 4. . r. 2. 2. 2. r. dx. r x. 0. r. 2. 2. /. For the lower semicircle, f (x)= r x and f (x)=. x. , so S =4. 2. 2. 2. r x. . r. 2. 2. +r. dx .. 2. r x. 0. Thus, the total area is r. S=S +S =8 1. 2. . r. 2. dx=8. 2. 2. r x. 0. 2. r sin. 1. x r. r. =8 r. 0. 2. 2. 2 2. =4 r .. 1 2 d and let the intersecting planes be y=c and y=c+h , where 4 1 1 2 2 1 2 d c dh . The sphere intersects the xy plane in the circle x +y = d . From this equation, 2 2 4 dx dx y we get x +y=0 , so = . The desired surface area is dy dy x 2. 2. 2. 34. Take the sphere x +y +z =. 11.

(29) Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution. S = 2 xds=2 c+h. = 2. c. c+h. . c+h. 2. x. 1+(dx/dy) dy=2. c. . x. 2. c+h. 2. 1+y /x dy=2. c. 1 d dy= d 2. . 2. 2. x +y dy. c. c+h. dy= dh c. 35. In the derivation of (4), we computed a typical contribution to the surface area to be y +y i1 i P P , the area of a frustum of a cone. When f (x) is not necessarily positive, the 2 i1 i 2. ( ). approximations y = f x f x i. ( ). y= f x i. i. i. f x. *. i1. 2. i. i1. ( ). and y = f x. i. i1. P P 2 f x i1 i. ( ). and y = f x. i. y +y 2. *. i1. *. 1+ f. i. b. as before, we obtain S= 2 f (x). i1. f x *. . Thus,. i. /. 1+ f (x). x. 2. *. i. 2. must be replaced by. i. f x /. *. x . Continuing with the rest of the derivation. dx .. a. /. /. 36. Since g(x)= f (x)+c , we have g (x)= f (x) . Thus, b. S = g. 2 g(x). 1+ g (x). 2. a. b. =. /. 2 f (x) a. b. dx= 2 f (x)+c. 2. /. 1+ f (x). dx. a. /. 1+ f (x). 2. b. dx+2 c a. /. 1+ f (x). 2. dx=S +2 cL f. 12.

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