M1120 Class 6 Dan Barbasch September 11, 2011 - PowerPoint PPT Presentation

M1120 Class 6 Dan Barbasch September 11, 2011 http://www.math.cornell.edu/ ˜ web1120/index.html Dan Barbasch () M1120 Class 6 September 11, 2011 1 / 17

Surface Area Dan Barbasch () M1120 Class 6 September 11, 2011 2 / 17

Surface Area Dan Barbasch () M1120 Class 6 September 11, 2011 2 / 17

Surface Area Central to surface area of a surface of revolution is the idea of approximating a little rotated piece of the surface by such a frustrum. Before rotation, the piece of curve has length ds . Dan Barbasch () M1120 Class 6 September 11, 2011 2 / 17

Surface Area After rotation, the area dA of the rotated piece is: dA = 2 π radius ds . � b So the total area is A = 2 π r ds . a Dan Barbasch () M1120 Class 6 September 11, 2011 2 / 17

Surface Area The surface area of a region obtained by rotating the curve y = f ( x ) ≥ 0 between x = a and x = b about the x − axis is � b � 1 + f ′ ( x ) 2 dx . A = 2 π f ( x ) a If the curve is x = g ( y ) , with c ≤ y ≤ d , then the formula is � d � 1 + g ′ ( y ) 2 dy . A = 2 π y c If the curve is given in parametric form, ( x ( t ) , y ( t )) for t 0 ≤ t ≤ t 1 , the formula is � t 1 � x ′ ( t ) 2 + y ′ ( t ) 2 dt . A = 2 π y ( t ) t 0 REMARK: You have to plug in the appropriate r which depends on the axis of rotation. In the formulas above we rotate about hte x − axis so the radius is always | y | . Dan Barbasch () M1120 Class 6 September 11, 2011 2 / 17

Problems from Section 6.4 Problem 18: Find the surface area of the region obtained by rotating x = 1 3 1 2 − y 2 about the y -axis for 1 ≤ y ≤ 3 3 y Problem 26: Dan Barbasch () M1120 Class 6 September 11, 2011 3 / 17

Problems from Section 6.4 Problem 30: Find the surface area (without bottom ) of the 90 ft NWS Dome Dan Barbasch () M1120 Class 6 September 11, 2011 3 / 17

Hints to the Exercises: Problem 18: The graph is Dan Barbasch () M1120 Class 6 September 11, 2011 4 / 17

Hints to the Exercises: 1 + ( r / h ) 2 dx . � Problem 26: Since y = ( r / h ) x , dy = ( r / h ) dx , so ds = So the area is � h � h � � 1 + ( r / h ) 2 dx =2 π 1 + ( r / h ) 2 dx = A = 2 π ( r / h ) x y 0 0 � r 2 + h 2 . = π r Verify the arithmetic. Dan Barbasch () M1120 Class 6 September 11, 2011 4 / 17

Hints to the Exercises: R 2 − y 2 for − r ≤ y ≤ R � Hint to Problem 30: Rotate the curve x = about the y − axis. In this problem R = 45 and r = 22 . 5 A better way is to use polar coordinates: x ( t ) = R cos t , y ( t ) = R sin t . In this problem, R = 45 . To generate the surface of the dome, we rotate the portion of the circle on the right of the axis. The bounds of the angle t are − π/ 6 ≤ t ≤ π/ 2 . The radius is r = x = R cos t . The formulas are � � R 2 cos 2 t + R 2 sin 2 t dt = R dt , x ′ ( t ) 2 + y ′ ( t ) 2 dt = ds = � π/ 2 � π/ 2 2 π R cos t · Rdt = 2 π R 2 A = cos t dt = − π/ 6 − π/ 6 2 π R 2 (sin π/ 2 − sin( − π/ 6)) = 3 π · 45 2 (Check the arithmetic carefully!) Dan Barbasch () M1120 Class 6 September 11, 2011 4 / 17

Integration by Parts � � Basic Formula: u dv = uv − vdu . d ( uv ) = udv + vdu ⇔ udv = d ( uv ) − vdu ⇔ � � � udv = d ( uv ) − vdu . Dan Barbasch () M1120 Class 6 September 11, 2011 5 / 17

x sec 2 x dx . � Dan Barbasch () M1120 Class 6 September 11, 2011 6 / 17

x sec 2 x dx . � Example x sec 2 x dx set u = x and dv = sec 2 x dx . Then � For dv = sec 2 x dx u = x du = dx v = tan x . The integration by parts formula gives � � x sec 2 x dx = x tan x − tan x dx = x tan x − ln | cos x | + C . Dan Barbasch () M1120 Class 6 September 11, 2011 7 / 17

Further Examples � (a) x sin x dx . xe 3 x dx . � (b) x 4 ln x dx . � (c) Dan Barbasch () M1120 Class 6 September 11, 2011 8 / 17

x 4 ln x dx � Dan Barbasch () M1120 Class 6 September 11, 2011 9 / 17

Example (c) Set dv = x 4 dx u = ln x v = x 5 du = 1 x dx 5 . So � 1 5 dx = x 5 ln x x 4 dx = x 5 ln x x 4 ln x dx = x 5 x 5 − x 5 − 1 � � 5 ln x − 25+ C . 5 5 5 x Dan Barbasch () M1120 Class 6 September 11, 2011 10 / 17

Harder Examples e ax sin ( bx ) dx . � (d) � (e) ln x dx . Dan Barbasch () M1120 Class 6 September 11, 2011 11 / 17

e x sin x dx � Dan Barbasch () M1120 Class 6 September 11, 2011 12 / 17

e 2 x cos x dx � u = e 2 x dv = cos x dx du = 2 e 2 x dx v = sin x . Then � � � e 2 x cos x dx = e 2 x sin x − (sin x )(2 e 2 x dx ) = e 2 x sin x − 2 e 2 x sin x dx . For the second integral, u = e 2 x dv = sin x dx du = 2 e 2 x dx v = − cos x . So � � e 2 x sin x dx = − e 2 x cos x − ( − cos x )(2 e 2 x dx ) = � = − e 2 x cos x + 2 e 2 x cos x dx . Dan Barbasch () M1120 Class 6 September 11, 2011 13 / 17

e 2 x cos x dx � Plugging into the first equation, � � � � e 2 x cos x dx = e 2 x sin x − 2 − e 2 x cos x + 2 e 2 x cos x dx = � e 2 x sin x + 2 e 2 x cos x e 2 x cos x dx . � � = − 4 This is an equation, the integral we want appears on both sides. So we move it to the left and solve: � e 2 x cos x dx = e 2 x sin x + 2 e 2 x cos x (1 + 4) e 2 x cos x dx = e 2 x sin x + 2 e 2 x cos x � + C . 5 Dan Barbasch () M1120 Class 6 September 11, 2011 13 / 17

Reduction formulas The expression to be integrated depends on some integer n . We write the integral in terms of similar integrals, but for smaller n . sec n x dx . � (f) x n e x dx . � (g) Dan Barbasch () M1120 Class 6 September 11, 2011 14 / 17

Example (f) Dan Barbasch () M1120 Class 6 September 11, 2011 15 / 17

Example (f) u = sec n − 2 x dv = sec 2 x dx du = ( n − 2) sec n − 3 x sec x tan x dx v = tan x . � � sec n x dx = sec n − 2 x tan x − ( n − 2) sec n − 2 x tan 2 x dx = � = sec n − 2 x tan x − ( n − 2) sec n − 2 x − 1 + sec 2 x � � dx = � � = sec n − 2 x tan x + ( n − 2) sec n − 2 x dx − ( n − 2) sec n x dx sec n x dx to the other side of the equation: � Now move the integral � � sec 2 x dx = sec n − 2 x tan x + ( n − 2) sec n − 2 x dx (1 + n − 2) sec n x dx = sec n − 2 x tan x � + n − 2 � sec n − 2 x dx . n − 1 n − 1 Dan Barbasch () M1120 Class 6 September 11, 2011 16 / 17

Exercises for next time (ln x ) 2 dx . � (h) sin √ x dx . � (i) sec 3 x dx � (j) For (j) you use the recursion formula from the previous page to express the � integral in terms of sec x dx . For (i) you make a change of variables, and then use integration by parts. For (h) try u = (ln x ) 2 dv = dx . Dan Barbasch () M1120 Class 6 September 11, 2011 17 / 17