A = (cosh x sinh x) dx = (sinh x cosh x) = sinh1 cosh1 sinh 0 + cosh 0 =

Transcription

1 Calculus 7 Review Consider the region between curves y= cosh, y= sinh, =, =.. Find the area of the region. e + e e e Solution. Recall that cosh = and sinh =, whence sinh cosh. Therefore the area is given by the integral A = (cosh sinh ) d = (sinh cosh ) = sinh cosh sinh + cosh = e e e+ e e = + = =.63. e e. Find the volume of the solid of revolution when the region is revolved about the -ais. Solution. We will compute the volume using the washers method and the identity cosh sinh =. V = π (cosh sinh ) d = π d = π.

2 3. Find the volume of the solid of revolution when the region is revolved about the y-ais. Solution. We will use the cylindrical shells method. V = π (cosh sinh ) d. We will integrate by parts taking u = and dv = (cosh sinh ) d. Then du = d and v= sinh cosh whence V = π(sinh cosh ) π (sinh cosh ) d= = π(sinh cosh) + π (cosh sinh ) d. The last integral we have already computed in Problem and therefore e e Vy = π + π = π.66 e e e 4. Find the coordinates of the geometric center of the region. Solution. By the Pappus s Centroid Theorem we have c Vy e = =.4, and π A e V e yc = =.8. π A ( e ) 5. Find the area of the surface of revolution when the region is revolved about the -ais. Solution. The area in question consists of four parts. (a) The area of the surface generated by the rotation of the segment of the y-ais from to about the -ais. This surface is a disk with the radius and its area is equal toπ. (b) The area of the surface generated by the rotation of the segment of the vertical line = from sinh to cosh about the -ais. This surface is an annulus with the radii sinh and cosh and its area is equal toπ(cosh sinh ) = π. (c) The area of the surface generated by the rotation of the curve y = cosh about the - ais. This area is given by the formula

3 dy S πy d π cosh sinh d d = + = + = = π cosh cosh d= π cosh d= π (cosh + ) d= 4 π e e e + 4e = sinh + π = π π. + = 4 4e (d) The area of the surface generated by the rotation of the curve y = sinh about the -ais. This area is given by the formula S = π sinh + cosh d. Letu = cosh, then du = sinh dand therefore cosh formula u cosh S = π + u du = π + u + ln( u + + u = = π cosh cosh ln( ) ln(cosh cosh ). Adding all four areas from (a) (d) we get that the surface area in question is approimately Find the area of the surface of revolution when the parametric curve = cosh t, y= sinh t, t, is revolved about the -ais and about the y-ais. Solution. (a) To find the surface area in case when the curve is revolved about the -ais we use the formula d dy S = π y( t) + = π sinh t sinh t+ cosh tdt = π sinh t cosh t dt. dt dt Taking u = cosh twe have du = sinh tdt and S = π u du. Let v= uthen cosh ( ) cosh formula 39 cosh π S = π v dv = v v ln( v + v = π + = π + cosh+ cosh (cosh cosh ) ln 5.7 (b) If the curve is revolved about the y-ais we have

4 d dy Sy = π ( t) + = π cosh t sinh t+ cosh tdt = π cosh t sinh t+ dt. dt dt Taking u = sinh twe have du = cosh tdt and S y = π u + du. Let v= uthen sinh sinh formula sinh π Sy = π v dv v v ln( v v + = = π = π Find the arc length of the curve y= 3,. sinh sinh ln( sinh sinh ). Solution. According to the formula for arc length we have dy L= + d= + (3 ln3) d= + 9 ln 3 d. d Letu = + 9 ln 3. Then u = + 9 ln 3whence 3 9 (ln9)(ln 3) (ln 3)9 udu = d = dand, because 9 du du have d = = 3 (ln 3)9 (ln 3)( u ). Therefore + 9ln 3 + 9ln 3 u du L = = + du = ln 3 u ln 3 u u+ + ln 3 + ln 3 =, we u ln 3 + 9ln 3 + ln 3 = + 9 ln 3 + ln 3 + ln ln.5. ln 3 + 9ln 3+ + ln A force of 3 N is required to maintain a spring stretched from its natural length of cm to a length of 5 cm. How much work is done in stretching the spring from cm to cm? Solution. According to the Hooke s law the force required to hold the spring stretched meters beyond its natural length is f () = k. When the spring is stretched from cm to 5 cm, the amount stretched is 3 cm =.3 m. This means that f (.3) = 3, so.3k = 3,k = 3 =. Thus f () = and the work done in stretching the.3 spring from cm to cm is

5 .8 W = d = 5.8 = 3.J 9. A tank full of water has the shape of a paraboloid of revolution; that is, its shape is obtained by rotating a parabola about a vertical ais. If its height is 4 ft and the radius at the top is 4 ft, find the work required to pump the water out of the tank. Solution. The equation of the revolved parabola is y = a. Because f (4) = 4 we have y = 4 and = y. Therefore the area of cross-section at height y is A(y) = 4π y. Considering a thin slice of water from the height of y to y + Δy we see that the work required to raise the water in this slice to the top of the tank is approimately ΔW = 6.5 4π y(4 y)δy. Finally, W = 5π 4 y(4 y)dy = 8π lb-ft. 3