Math 262 Exam 1 - Practice Problems. 1. Find the area between the given curves:

Transcription

1 Mat 6 Exam - Practice Problems. Find te area between te given curves: (a) = x + and = x First notice tat tese curves intersect wen x + = x, or wen x x+ =. Tat is, wen (x )(x ) =, or wen x = and x =. Next, notice tat x x + on [,. Tus te area between tese curves is given b: (x ) (x + ) dx = x + x dx = x + x x = 6 (b) = x and = x on [, First notice tat tese curves intersect wen x = x, or wen x +x =. Tat is, wen (x+)(x ) =, or wen x = and x =. Next, notice tat x x on [, wile x x on [,. Tus te area between tese curves is given b: ( x) (x ) dx + (x ) ( x) dx = = x x + x + x + x x =. x x + dx + x + x dx (c) = x, =, + x = 6, and = Tis integral is a bit easier if we slice te area orizontall and integrate wit respect to, so, solving for x in terms of, we ave x = and x = 6, wit 6 on [,. Terefore, te area of tis region is given b: (6 ) () d = 6 d = 6 = 4 = 8 (d) x =, x = 4 We will find te area b integrating wit respect to. Notice tat = 4 wen = ±. Also, 4 on [,. Terefore, te area of tis region is given b: 4 d = 4 =.. A potter jar as circular cross sections or radius 4 + sin( x ) inces for x. Sketc te jar and ten compute its volume. r = 4 r = 4 + \ V = = (4 + sin( x ) ) dx = [6x 4cos( x ) + x 4 sin(x ) ( x 6 + 8sin ) ( + sin x ) dx = ( ( x cos( x 6 + 8sin + ) = [6 4( ) + [ 4 + = ) ) dx

2 . Let R be te region bounded b = x, =, and x = (a) Sketc R and ten find its area. x A = x dx = 4 x4 = 4. (b) Find te volume of te solid formed b rotating R about te x-axis. Integrating wit respect to x and using disks, we see V = ( x ) dx = x 6 dx = 7 x7 = 7. (c) Find te volume of te solid formed b rotating R about te -axis. Integrating wit respect to x and using clindrical sells, we see: V = x ( x ) dx = x 4 dx = 5 x5 = Let R be te region bounded b = ax, =, and te -axis (were a and are positive constants). Compute te volume of te solid formed b revolving tis region about te -axis. Sow tat our answer is equal to alf te volume of a clinder of eigt and radius r = a. Sketc a picture illustrating te relationsip between tese two volumes. If we solve for x, = ax becomes x = and we can represent te volume of tis solid of revolution b slicing a orizontall into disks: ( ) V = d = d = a a a = a. Next, recall tat te formula for te volume of a clinder is: V = r = ( a ) = ( ) a = a, so we can see tat our answer is equal to alf te volume of tis clinder. x

3 5. Te great pramid at Gize is 5 feet ig rising from a square base wit side lengt 75 feet. Use integration to find te volume of tis pramid. We will find te volume of tis pramid b noticing tat slicing it orizontall ields square cross sections. Moreover, if we tink of te eigt of our slices wit 5, ten te side lengt of te square cross sections is given b s() = 75. Terefore, te volume of te great pramid of Gize is: 5 ( V = 75 ) ( )( d = ) (75 ) 5 = [ () (75) = 9,75,ft. 9 Notice tat tis agrees wit te result of te geometric formula V = b = (75) (5) = 9,75,ft. 6. Find te volume of te solid formed b revolving te region bounded b = x 4 and = 4 x about te line x =. Notice tat tese two curves meet wen x = ±. Terefore, te volume of tis solid of revolution is given b: V = ( x)[(4 x ) (x 4) dx = = x4 4 x 4x + 6x = ( x)(8 x ) dx = [ (8 6 + ) (8 + 6 ) x 4x 8x + 6 dx = [64 64 = Find te volume of te solid formed over te region bounded b = x 4 and = 4 x if its cross sections perpendicular to te -axis are semicircles. First, since te semicircles run orizontall, we sould integrate wit respect to. B smmetr wit respect to te x-axis, we can integrate over 4 and ten double our result. Notice tat te radius of eac semicircular cross section is given b r = 4. Terefore, te volume of tis region is given b: V = (4 ) d = [4 4 = [6 8 = Find te volume of te solid formed b revolving te region bounded b = x and te x-axis for x about te line x =. Based on te sape of tis region, it is best to use clindrical sells to find tis volume. We do so using te following integral: V = (x + )(x ) dx = x + x dx = [ 4 x4 + x = [( 4 + ) ( 4 ) = (4 ) = Find te volume of te solid formed b revolving te region bounded b x = and x = about te line =. Based on te sape of tis region, it is best to use clindrical sells to find tis volume. We do so using te following integral: V = ( + )( ) d = + + d = [ = [( ) ( ) = (4 4 ) = 6.. Find te volume of te solid formed b revolving te region bounded b x = and x = + about: First, notice tat tese two curves intersect wen = +, or wen =. Tat is wen ( )( + ) =. Terefore, te two points of intersection are (, ) and (4,). (a) te line x =. Notice tat on [, te region enclosed b tese curves is bounded above and below b te parabola x = wile on [,4 te region is bounded above b x = and below b x = +. Because of tis, in order to find te volume of tis solid, we must consider tese two subintervals separatel. Terefore, te volume of tis solid is given b:

4 V = (x + )( x) dx + = 4 x + x dx + 4[ 5 x 5 + x (x + )( x (x )) dx x + x + x + x + dx = + [ x + 5 x 5 + x + 4 x + x = 4[ [( ) ( ) = 5 [Note: It is actuall easier to do tis part using wasers see part (c) below. I just wanted to sow bot possibilities (b) te line =. Here we view x = + as te rigt function, x = as te left function bounding our region and we integrate using clindrical sells: V = ( + )[( + ) ( ) d = + + d = [ = [( ) ( ) = (c) te line x =. We could do tis similarl to part (a) above, but it is actuall simpler to do tis using wasers: V = (( + ) + ) ( + ) d = = [ = [( ) ( 5 ( ) ( ) d = ) = d. Set up te integral for te arc lengt of eac of te following curves on te given interval. You DO NOT need to evaluate te integral. (a) = x + x on [,5 Notice tat = x + and ( ) = 9x 4 + 6x + 5 Terefore, L = 9x4 + 6x + dx (b) = tan x on [, 4 = sec x and ( ) = sec 4 x 4 Terefore, L = sec4 x + dx (c) = x + on [, Notice tat = x (x + ) and ( ) = 4x Terefore, L = + (x + ) 4 dx 4x (x + ) 4 (d) x = Notice tat tis curve is an ellipse centered at te origin wit major axis of lengt 8 and minor axis of lengt 6. Solving for, we obtain: 44 9x 9 = x 6, so = 9 9x 6 or = ± 6 If we take te positive alf of tis equation, we can find te arc lengt of te ellipse and double it in order to obtain te arc lengt of te entire ellipse. Ten = ( ) 44 9x ( 8x ) 6 6 Terefore, L = + 76x 44 9x dx 4 = 9x 6 ( 44 9x 6 ) and ( ) = 8x 56 ( 44 9x 6 ) = 76x 44 9x

5 . Set up te integral for te surface area of eac of te following on te given interval. You DO NOT need to evaluate te integral. (a) = sin x on [, revolved about te x-axis. S = r ds = (sin x) + cos x dx (b) = x on [, revolved about te x-axis. S = r ds = (x ) + 9x 4 dx (c) = x on [, revolved about te -axis. Since we are revolving te region about te -axis, we need to solve te equation or our curve for x: + = x, so x = ( + ). Terefore, x = ( + ) Ten S = 7 r ds = 7 ( + ) + 9 ( + ) 4 d. A force of pounds stretces a spring inces. Find te work done in stretcing tis spring inces beond its natural lengt (give our answer in ft-lbs). Recall tat B Hooke s Law, te force needed to stretc of compress a spring is given b F(x) = kx. Here, F( ) = k ( 6) =, so k = 6. 4 ( ) Terefore, W = 6x dx = x = = 5 ft. lbs A clindrical tank is resting on te ground wit its axis vertical; it as a radius of 5 feet and a eigt of feet. Find te amount of work done in filling tis tank wit water pumped in from ground level (use ρ = 6.5/ft for te densit of water). W = F d = 5 (6.5) d = 5(6.5) = 5(6.5)(5) = 785ft. lbs. 5. A water tank is in te sape of a rigt circular cone of altitude feet and base radius 5 feet, wit its vertex one te ground (tink of an ice cream cone wit its point facing down). If te tank is alf full, find te work done in pumping all of te water out te top of te tank. If we slice te center of te cone verticall, we obtain a rigt triangular cross section wit eigt feet and widt 5 feet. Using similar triangles, if we move up te cone to a eigt, ten te orizontal cross section at eigt as radius r satisfing = r 5, so r =. Terefore, te total work in pumping all of te water out of te top of te cone is given b: 5 W = ( 5 5 ) (6.5)( ) d = (6.5) 4 d = (6.5)[ = (6.5)[ = A cain of densit 5 kilograms per meter is anging down te side of a building tat is meters tall. How muc work is done if tis cain is used to lift a 5 kilogram mass feet up te side of te building? First recall tat te downward force of gravit in a kg mass is 9.8 m sec. Next, ft ()(.48 m ft ) =.48meters. Terefore, te work done in using tis cain to lift a 5kg mass is given b te integral: W =.48 (9.8)(5 + (5)( )) d = (9.8) = (9.8)[ = 75, Joules Find te center of mass of te following sstems of point masses: 5 d = (9.8)[ 5.48 (a) A 5 kg mass 5 units left of te x-axis, a kg mass unit left of te x-axis and a kg mass units rigt of te x-axis. m = = wile M = (5)( 5) + ()( ) + ()() = 4 Terefore, x = 4 =.4

6 (b) A 4 kg mass at te point (-, 5), a 5 kg mass at te point (, -) and a lb mass at te point (5, 7). Notice tat lb = ()(.456kg/lb) = 4.56kg m = = 69.56kg M = (4)( ) + (5)() + (4.56)(5) = 7.68 wile M x = (4)(5) + (5)( ) + (4.56)(7) = 6.75 Terefore, x = M m = =.54 wile = = Compute bot te mass and te center of mass of a steel rod wit densit ρ(x) = x 6 for x Te mass of tis bar is given b: m = ρ(x) dx = x 6 dx = x 6 x = 8 6 = 5kg. 6 6 M = xρ(x) dx = x x 6 dx = x 6 8 x = 54 = 4. Terefore, x = M m = 4 5 =.8 meters from te left end of te rod. 9. Find te centroid of te region R bounded b = 4 x and te x-axis. First notice tat, b smmetr, x =. Also, trougout tis problem, we will take ρ =. Ten m = Next, M x = ρ(4 x ) dx = = = Terefore, = 4 x dx = [4x x (4 x + ())(ρ)((4 x ) ()) dx = 56 5 = 8 5 Hence te center of mass is at (, 8 5 ) = [8 8 = 6 8x + x 4 dx = 6x 8 x + 5 x5. Find te centroid of te isosceles triangle formed b te points (,), (6,) and (,9) Tis problem is muc easier to do is we tink of sifting te coordinate sstem for tis triangle left b units, centering te triangle on te -axis and making its points (,), (,) and (,9) First, notice tat if we assume ρ =, ten m = A ρ = b = (6)(9) = 7 For our sifted triangle, we see b smmetr tat x =. Next, we find te equation for te line containing (,) and (,9): It as slope: 9 Ten M x = = =, so te equation is = x + 9 (f(x) + g(x))(ρ)(f(x) g(x)) dx = 9x 54x + 8 dx = x 7x + 8x = = 8. Terefore, = 8 7 = Hence te center of mass of te sifted triangle is at (,). ( x ())( x + 9 ()) dx Sifting tis rigt units, te center of mass of te original triangle is at (,).