spring from 1 cm to 2 cm is given by

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Stanley Dickerson
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1 Problem [8 pts] Tre or Flse. Give brief explntion or exmple to jstify yor nswer. ) [ pts] Given solid generted by revolving region bot the line x, if we re sing the shell method to compte its volme, then we wold integrte with respect to x. Tre: The line x is prllel to the y xis. When yo se shells the vrible of integrtion is opposite to the xis of rottion, so we shold integrte over x if we wnt to se shells b) [ pts] Given spring tht obeys Hooke s Lw, the work reqired to stretch the spring from eqilibrim to cm is the sme s the work reqired to stretch the spring from cm to cm. Flse: Given fixed spring constnt k, the work reqired to stretch the spring from eqilibrim to cm is given by. kx k(.) k,. spring from cm to cm is given by. Menwhile, the work reqired to stretch the c) [ pts] sin θ cos θ dθ ( 8 ) d where sin θ.. kx k((.) (.) k,. Ths the work streching the spring from cm to cm is greter. Tre: sin θ cos θ dθ sin θ( sin θ) cos θ dθ Let sin θ. Then d cos θ dθ. So or integrl is eql to ( 8 ) d, s desired. d) [ pts] dx ln cos x + C cos x Flse: cos x dx sec x dx ln sec x + tn x + C ln cos x + C Alterntively one cold differentite ln cos x + C to show the sttement ws flse. e) [ pts] x dx represents the re of the region bonded by the grph of y x nd the x-xis over the intervl [, ]. Flse: f) [ pts] π/ x dx x. sec x dx is the length of the crve y ln(sec x) on the intervl [, π ]. Tre: Arc length of the crve y(x) on the intervl [, π/] is given by π + (y (x)) dx. Ths rc length is given by π π + tn x dx sec x dx y (x) sec x tn x tn x. sec x π/ sec x dx.

2 Problem [ pts] Ares nd Volmes. Give the exct vles. Yo do not need to simplify yor finl nswer. ) [ pts] Find the re of the region bonded by the grphs x y+ nd x y y 7. Soltion: Let s eqte the two eqtions to identify their intersection pts. y + y y 7 y y 8 (y + )(y ) Ths or region is contined in the intervl [, ]. By evlting ech crve t the pt y, we see tht the crve x y + hs lrger x vles on the intervl [, ] thn the crve y y 7. Ths Are y + (y y 7) dy y + y + 8 dy y + y + 8y ( ( ) + ( ) ) + 8 ( ) b) [ pts] Consider the region R bonded by the grph of y sin (x) nd the x-xis over the intervl [, π ]. Find the volme of the solid whose bse is the region R nd whose cross sections perpendiclr to the x-xis re sqres. Soltion: On the intervl [, π ] sin(x). Ths for fixed x the side length of the sqre cross section perpendiclr to the x xis throgh tht point is given by sin(x) nd or forml for cross sectionl re t the point x is given by CSA(x) sin x. over the bonds of or object. Ths V olme π π [ x sin x dx ( cos x) dx ] π sin x [ π sin π ( sin )] π By the generlized slicing method volme is given by integrting cross sectionl re

3 Problem [ pts] A tnk is shped like prbolic bowl. It is formed by revolving the grph of y x for x (in meters) bot the y-xis. The tnk is filled with wter to height of meters. How mch work is reqired to pmp ll of the wter to n exit pipe t the top of the tnk? [Note: The density of wter is kg/m.] Soltion: Work for pmping liqid ot of tnk is given by the eqtion W ork b ρ g CSA(y) D(y) dy. Here [, b] is the intervl where the liqid exists. ρ is the density of the liqid. g is the grvittionl constnt. CSA(y) is the re of the horizontl cross section t the height y. D(y) is the distnce the cross section t height y hs to trvel. In or cse the tnk is filled p to m so or bonds of integrtion re nd b. ρ nd g 9.8. The tnk is formed s rottionl solid with xis of rottion the y xis. Ths the cross sections re circles nd by solving the reltion y x for y we find the rdis s fnction of height: y r(y). Ths CSA(y) π y. The height of the tnk is meters. Ths the distnce the yth cross section hs to trvel to pmp ot of the tnk is y. Ths W ork Plg in x into the eqtion y x 9.8 πy ( y) dy 98π y y dy 98π ] [8y y 98π ] [8() 978 J

4 Problem [ pts] Evlte the following integrls. Show yor work. ) [ pts] x 8 ln x dx Soltion: We will solve the integrl by integrting by prts. Let ln(x) nd dv x 8 dx. Then d x9 dx nd v, nd or integrl becomes: x 9 x 8 ln(x) dx x9 x 9 9 ln(x) 9 x dx x9 9 ln(x) 9 x 8 dx x9 9 ln(x) 9 x9 9 + C w b) [ pts] w + dw Soltion: Let w +. Then w nd dw d. At w nd t w. Ths w w + dw ( ) d + d d d d ln ( ln ) ln

5 Problem 5 [ pts] Consider the region nder the grph y where > is constnt. x + over the intervl x, ) [ pts] Set p n integrl (or integrls) tht represents the volme of the solid generted by revolving the region bot the y-xis. Soltion: The xis of rottion is the y-xis nd or vrible of integrtion is x ths we shold se shells. Ths or volme is fond by the integrl π x x + dx if we wnted to se disk method we wold need to integrte over y. b) [8 pts] Evlte yor integrl(s) in prt to find the volme of the solid. Note: Yor nswer will possibly contin the constnt. Soltion: Let x +. Then d x dx. At x, nd t x, +. Ths π x x + dx π + d π ln + π ln + π ln π ln +.

Math 2260 Written HW #8 Solutions

Math 2260 Written HW #8 Solutions Mth 60 Written HW #8 Soltions. Sppose nd b re two fied positive nmbers. Find the re enclosed by the ellipse + y b =. [Hint: yo might wnt to find the re of the hlf of the ellipse bove the -is nd then doble