6 APPLICATIONS OF INTEGRATION

Transcription

1 6 APPLICATIONS OF INTEGRATION 6. Areas Between Curves. A. A x x y y (y T y B) dx (x R x L ) dy + (). A (9 x ) (x +) dx ( x x ) dx x x x (x x ) x dx (x x ) dx x x 6 y (y ) dy (y / y +)dy y/ y + y () 7. The curves intersect when x x x x x(x ) x or. A (x x ) dx x x 6 9. First find the points of intersection: x + x + x + x + x + (x +) (x +) (x +) (x +)[ (x +)] (x + )( x) x or. So A x x + + dx (x +)/ (x +) 6 ( )

2 CHAPTER 6 APPLICATIONS OF INTEGRATION. A x x dx x/ x. x x 6 x x 9 x ±,so A ( x ) (x 6) dx x dx [by symmetry] x x [( ) ] (6)7. The curves intersect when cosx sec x cos x cos x cos x x π for <x< π.bysymmetry, A π/ ( cos x sec x) dx sinx tan x π/ 6 7. x x x x x x x(x ) x or,so A x x dx + 9 x x dx x/ x x x/ 9

3 SECTION 6. AREAS BETWEEN CURVES 9. y +y y y ±,so A ( + y ) y dy ( y ) dy [by symmetry] y y. The curves intersect when y y y y y ±. A ( y ) (y ) dy ( y ) dy ( y ) dy y y. Notice that cos x sinx sinx cos x sinx cos x cos x cos x ( sin x ) sinx or cos x x π 6 or π. A π/6 (cos x sin x) dx + sin x + cos x π/6 π/ π/6 (sin x cos x) dx + cos x sin x π/ π/ From the graph, we see that the curves intersect at x, x π,andx π. By symmetry, A π cos x x π/ dx π sin x x + π x π/ π + π π cos x x dx π π/ π + π π cos x + x dx π

4 6 CHAPTER 6 APPLICATIONS OF INTEGRATION 7. Graph the three functions y /x, y x,andy x; then determine the points of intersection: (, ), (, ),and,. A x x dx + 7 xdx+ x x dx 7 6 x + x 6 x x x dx An equation of the line through (, ) and (, ) is y x; through (, ) and (, 6) is y 6x; through (, ) and (, 6) is y x +. A x + x + ( 6x) dx + dx + x + 6 dx x + x dx (x +)dx + x + dx x + x + x + x + [( +) ] +. The curves intersect when sin x cosx (on [,π/]) sin x sin x sin x +sinx ( sin x )(sin x +) sin x x π 6. A π/ π/6 sin x cos x dx (cos x sin x) dx + π/ π/6 (sin x cos x) dx sin x +cosx π/6 + cos x sin x π/ π/6 + πx. Let f(x) cos The shaded area is given by ( + ) + ( ) πx sin and x. A f(x) dx M f + f + f + f 7.67

5 SECTION 6. AREAS BETWEEN CURVES 7. From the graph, we see that the curves intersect at x and x a.96,with x sin(x ) >x on (,a). So the area A of the region bounded by the curves is A a x sin(x ) x dx cos(x ) x a cos(a ) a From the graph, we see that the curves intersect at x a.,x b., and x c.6, with x x +> x x on (a, b) and x x >x x + on (b, c). So the area of the region bounded by the curves is A b a b a (x x +) (x x) c dx + (x x) (x x +) dx (x x x +)dx + c b b ( x +x + x ) dx x x x +x b a + x + x + x x c b. 9. As the figure illustrates, the curves y x and y x 6x +x enclose a four-part region symmetric about the origin (since x 6x +x and x are odd functions of x). The curves intersect at values of x where x 6x +x x;thatis,where x(x 6x +). That happens at x and where x 6 ± 6 ± 6;thatis,atx + 6, 6,, 6,and + 6.Theexactareais + 6 (x 6x +x) x + 6 dx x 6x +x dx 6 (x 6x + 6 +x) dx + ( x +6x x) dx CAS 6 9 6

6 CHAPTER 6 APPLICATIONS OF INTEGRATION. second 6 hour, so s 6 h. With the given data, we can take n to use the Midpoint Rule. t /6,so distance Kelly distance Chris /6 v K dt /6 v C dt /6 (v K v C ) dt. Let h(x) denote the height of the wing at x cm from the left end. A M M [(vk vc)() + (vk vc)() + (vk vc)() +(v K v C )(7) + (v K v C )(9)] [( ) + ( 6) + (7 6) + (6 7) + (9 6)] (+6+9++) () mile, or 7 feet [h() + h(6) + h() + h() + h()] ( ) (.) cm. We know that the area under curve A between t and t x is x va(t) dt sa(x),whereva(t) is the velocity of car A and s A is its displacement. Similarly, the area under curve B between t and t x is x v B(t) dt s B (x). (a)afteroneminute,theareaundercurvea is greater than the area under curve B. So car A is ahead after one minute. (b) The area of the shaded region has numerical value s A() s B(), which is the distance by which A is ahead of B after minute. (c) After two minutes, car B is traveling faster than car A and has gained some ground, but the area under curve A from t to t is still greater than the corresponding area for curve B, so car A is still ahead. (d) From the graph, it appears that the area between curves A and B for t (when car A is going faster), which corresponds to the distance by which car A is ahead, seems to be about squares. Therefore, the cars will be side by side at the time x where the area between the curves for t x (when car B is going faster) is the same as the area for t. From the graph, it appears that this time is x.. So the cars are side by side when t. minutes. 7. To graph this function, we must first express it as a combination of explicit functions of y;namely,y ±x x +. We can see from the graph that the loop extends from x to x, and that by symmetry, the area we seek is just twice the area under the top half of the curve on this interval, the equation of the top half being y x x +. So the area is A x x + dx. We substitute u x +,sodu dx and the limits change to and,andweget A [(u ) u ] du (u/ u / ) du u/ u /

7 SECTION 6. AREAS BETWEEN CURVES 9 9. By the symmetry of the problem, we consider only the first quadrant, where y x x y. We are looking for a number b such that b ydy b b ydy y / y / b b / / b / b / b / b /... We first assume that c>, sincec can be replaced by c in both equations without changing the graphs, and if c the curves do not enclose a region. We see from the graph that the enclosed area A lies between x c and x c, and by symmetry, it is equal to four times the area in the first quadrant. The enclosed area is A c (c x ) dx c x x c c c c c So A 76 c 76 c 6 c 6 6. Note that c 6 is another solution, since the graphs are the same.. A x dx ln x + x x ln + (ln + ) ln.9. The curves intersect when tan x sin x (on [ π/,π/]) sin x sinx cos x sinx cos x sin x sin x ( cos x ) sin x or cos x x or x ± π. A π/ π/ π/ ( sin x tan x) dx ( sin x tan x) dx [by symmetry] cosx ln sec x π/ [( ln ) ( )] ( ln ) ln

8 CHAPTER 6 APPLICATIONS OF INTEGRATION 6. Volumes. A cross-section is a disk with radius x, so its area is A(x) π x. V π A(x) dx x + x dx π x x + x π + π π π x dx +. A cross-section is a disk with radius /x, so its area is A(x) π(/x). V A(x) dx π dx x π x dx π x π ( ) π. A cross-section is a disk with radius y, so its area is A(y) π. y V 9 A(y) dy 9 π y 9 π() 6π π 9 y dy π ydy 7. A cross-section is a washer (annulus) with inner radius x and outer radius x, so its area is A(x) π(x) π(x ) π(x x 6 ). V A(x) dx π(x x 6 ) dx π x x7 π 7 7 π

9 SECTION 6. VOLUMES 9. A cross-section is a washer with inner radius y and outer radius y, so its area is A(y) π(y) π(y ) π(y y ). V A(y) dy π (y y ) dy π y y π 6 π. A cross-section is a washer with inner radius x and outer radius x, so its area is A(x) π( x) π x π ( x + x ) x + x π x + x + x. V A(x) dx π x + x + x dx π x + x + x/ π + π 6. A cross-section is a washer with inner radius ( + sec x) secxand outer radius, so its area is A(x) π (sec x) π( sec x). V π/ π π/ π/ A(x) dx π x tan x π π π/ π/ π( sec x) dx ( sec x) dx [by symmetry] π/ π π. V π( y ) dy π ( y + y ) dy π y y + y π 6 π π( y ) dy

10 CHAPTER 6 APPLICATIONS OF INTEGRATION 7. y x x y for x. The outer radius is the distance from x to x y and the inner radius is the distance from x to x y. V π y ( ) y ( ) dy π π π y + y + y y dy π y + y/ y y π + 9 π y + (y +) dy y + y y y dy 9. R about OA (the line y ): V. R about AB (the line x ): V A(y) dy π + π A(x) dx π(x ) dx π x 6 dx π 7 x7 π 7 π y dy π ( y / + y / ) dy π y y/ + y/. R about OA (the line y ): V A(x) dx π() π x dx π ( x) dx π x x π π. R about AB (the line x ): V A(y) dy π() π( y ) dy π ( y + y ) dy π (y y ) dy π y y π 7 π 7. R about OA (the line y ): V A(x) dx π x π(x ) dx π (x x 6 ) dx π x x7 π 7 7 π. Note: Let R R + R + R.IfwerotateR about any of the segments OA, OC, AB,orBC, we obtain a right circular cylinder of height and radius. Its volume is πr h π() π. As a check for Exercises 9,, and 7, we can add the answers, and that sum must equal π. Thus, π + π + π π π. 9. R about AB (the line x ): V A(y) dy π( y ) π y dy π ( y + y ) ( y / + y / ) dy π ( y + y +y / y / ) dy π y + y + y/ y/ π + + Note: See the note in Exercise 7. For Exercises,, and 9, we have π + 7π + π ++ π π. π

11 SECTION 6. VOLUMES. V π π/ ( tan x) dx. V π π π π ( ) ( sin x) dx ( sin x) dx. V π [ ( )] y + ( ) dy π +y + dy 7. y +x cos x and y x + x +intersect at x a. and x b.. V π b a [( + x cos x) (x + x +) ]dx.7 9. V π π CAS π sin x ( ) [ ( )] dx

12 CHAPTER 6 APPLICATIONS OF INTEGRATION. π π/ cos xdxdescribes the volume of the solid obtained by rotating the region R (x, y) x π, y cos x. π of the xy-plane about the x-axis. (y y ) dy π (y ) (y ) dy describes the volume of the solid obtained by rotating the region R (x, y) y,y x y of the xy-plane about the y-axis.. There are subintervals over the -cm length, so we ll use n / for the Midpoint Rule. V A(x) dx M [A(.) + A(.) + A(7.) + A(.) + A(.)] ( ) 7 cm 7. (a) V π [f(x)] dx π [f()] +[f()] +[f(7)] +[f(9)] π (.) +(.) +(.) +(.) 96 units (b) V π (outer radius) (inner radius) dy π (9.9) (.) + (9.7) (.) + (9.) (.6) + (.7) (6.) units 9. We ll form a right circular cone with height h and base radius r by revolving the line y r x about the x-axis. h V π h π r h h r h x dx π h πr h r h h x dx π r h x Another solution: Revolve x r y + r about the y-axis. h h V π r dy h h y + r r π h y r h y + r dy r h π h y r h y + r y π r h r h + r h πr h Or use substitution with u r r h y and du r dy to get h π u h r r du π h r u π h r r r πr h.

13 SECTION 6. VOLUMES. x + y r x r y V π π r r h r y dy π r r r y y r (r h) r r h (r h) π r (r h) r (r h) π r (r h) r r rh + h π r (r h) r +rh h π r r r h + rh +r h +rh h π rh h πh (r h), or, equivalently, πh r h. For a cross-section at height y, we see from similar triangles that α/ b/ h y h,soα b y h Similarly, for cross-sections having b as their base and β replacing α, β b y.so h V h h A(y) dy b y h h b y y h + y h b y b y dy h h dy h b yh y + dy h h b h h + h b h [ Bh where B is the area of the base, as with any pyramid.]. A cross-section at height z is a triangle similar to the base, so we ll multiply the legs of the base triangle, and, by a proportionality factor of ( z)/.thus,thetriangleatheightz has area A(z) z z 6 z,so V A(z) dz 6 z dz 6 u cm u ( du) u z/, du dz. 7. If l is a leg of the isosceles right triangle and y is the hypotenuse, then l + l (y) l y l y. V A(x) dx A(x) dx (l)(l) dx ( x ) dx 9 (6 9x ) dx 9 x x 9 y dx

14 6 CHAPTER 6 APPLICATIONS OF INTEGRATION 9. The cross-section of the base corresponding to the coordinate x has length y x. The corresponding square with side s has area A(x) s ( x) x + x. Therefore, V A(x) dx ( x + x ) dx x x + x + Or: ( x) dx u ( du) [u x] u 6. The cross-section of the base b corresponding to the coordinate x has length x. The height h also has length x, so the corresponding isosceles triangle has area A(x) bh ( x ). Therefore, V ( x ) dx ( x + x ) dx [by symmetry] x x + x + 6. (a) The torus is obtainedbyrotatingthecircle(x R) + y r about the y-axis. Solving for x, we see that the right half of the circle is given by x R + r y f(y) and the left half by x R r y g(y). So V π r r [f(y)] [g(y)] dy π r R +R r y + r y π r R r y dy πr r r y dy R R r y + r y dy (b) Observe that the integral represents a quarter of the area of a circle with radius r, so πr r r y dy πr πr π r R. 6. (a) Volume(S ) h A(z) dz Volume(S ) since the cross-sectional area A(z) at height z is the same for both solids. (b) By Cavalieri s Principle, the volume of the cylinder in the figure is the same as that of a right circular cylinder with radius r and height h,thatis,πr h.

15 SECTION 6. VOLUMES The volume is obtained by rotating the area common to two circles of radius r,as shown. The volume of the right half is V right π r/ y dx π r/ r r + x dx π r x r + x r/ π r r r πr So by symmetry, the total volume is twice this, or πr. Another solution: We observe that the volume is the twice the volume of a cap of a sphere, so we can use the formula from Exercise with h r: V πh (r h) π r r r πr. 69. Take the x-axis to be the axis of the cylindrical hole of radius r. A quarter of the cross-section through y, perpendicular to the y-axis, is the rectangle shown. Using the Pythagorean Theorem twice, we see that the dimensions of this rectangle are x R y and z r y,so A(y) xz r y R y,and V r r A(y) dy r r r y R y dy r r y R y dy 7. (a) The radius of the barrel is the same at each end by symmetry, since the function y R cx is even. Since the barrel is obtained by rotating the graph of the function y about the x-axis, this radius is equal to the value of y at x h,whichisr c h R d r. (b) The barrel is symmetric about the y-axis, so its volume is twice the volume of that part of the barrel for x>. Also,the barrel is a volume of rotation, so V h/ πy dx π h/ π R h Rch + 6 c h R cx dx π R x Rcx + c x h/ Trying to make this look more like the expression we want, we rewrite it as V πh R + R Rch + c h. But R Rch + c h R ch c h (R d) ch r d. Substituting this back into V,weseethatV πh R + r d, as required.

16 CHAPTER 6 APPLICATIONS OF INTEGRATION 6. Volumes by Cylindrical Shells. If we were to use the washer method, we would first have to locate the local maximum point (a, b) of y x(x ) using the methods of Chapter. Then we would have to solve the equation y x(x ) for x in terms of y to obtain the functions x g (y) and x g (y) showninthefirst figure. This step would be difficult because it involves the cubic formula. Finally we would find the volume using V π b [g (y)] [g (y)] dy. Using shells, we find that a typical approximating shell has radius x, so its circumference is πx. Its height is y,thatis, x(x ). So the total volume is. V V πx dx π x πx x(x ) dx π dx π x π( ) π x x + x dx π x x + x π. V πx( x ) dx π (x x ) dx π x x π( ) π 7. The curves intersect when (x ) x x +7 x 6x +6x x +7 x x +9 (x x +) (x )(x ),sox or. V π π x x x +7 (x ) dx π x(x x +7 x +6x 6) dx x( x +x 9) dx π( ) (x x +x) dx 6π x x + x 6π π π 6π

17 SECTION 6. VOLUMESBYCYLINDRICALSHELLS 9 9. V πy( + y ) dy π (y + y ) dy π y + y π ( + ) + π π. V π π y( y ) dy y / dy π y7/ 7 6π 7 (7/ ) 6π 7 (7 ) 76 7 π. Theheightoftheshellis +(y ) (y ) y y + y +y. V π y( y +y ) dy π ( y +y y) dy π y + y y π +6 7 π 6 π +. The shell has radius x, circumference π( x), and height x. V π( x)x dx π (x x ) dx π x 6 x6 π 6 π 7 7 π

18 CHAPTER 6 APPLICATIONS OF INTEGRATION 7. The shell has radius x, circumference π(x ), and height (x x ) x +x. V π(x )( x +x ) dx π ( x +x 7x +)dx π x + x 7 x +x π π π 9. The shell has radius y, circumference π( y), and height y y x x y. V π( y)( y/ ) dy π ( y y/ + y / ) dy π y y y/ + 7 y7/ π + 7 π π. V π π πx sin xdx. V π[x ( )] sin π x x dx. V π π( y) sin ydy

19 SECTION 6. VOLUMESBYCYLINDRICALSHELLS 7. V πx +x dx. Letf(x) x +x. Then the Midpoint Rule with n gives f(x) dx [f(.) + f(.) + f(.) + f(.7) + f(.9)].(.99) Multiplying by π gives V πx dx π x(x ) dx. The solid is obtained by rotating the region y x, x about the y-axis using cylindrical shells.. π( y)( y ) dy. The solid is obtained by rotating the region bounded by (i) x y, x,andy or (ii) x y, x,andy about the line y using cylindrical shells.. From the graph, the curves intersect at x and at x a.,with. V π π/ CAS π π x sin x sin x dx x + x x > on the interval (,a). So the volume of the solid obtained by rotating the region about the y-axis is V π a [x(x + x x )] dx π a (x + x x ) dx π x + x 6 x6 a. 7. Use shells: V πx( x +6x ) dx π ( x +6x x) dx π x +x x π[( 6 + 6) ( +6 6)] π() π 9. Use shells: V π[x ( )][ (x x +9)]dx π (x +)( x +x ) dx π ( x +x + x ) dx π x + x + x x π π 6 6 π

20 CHAPTER 6 APPLICATIONS OF INTEGRATION. Use disks: x +(y ) x ± (y ) V π (y ) dy π (y y ) dy π y y π π. Use shells: V r πx r x dx π r (r x ) / ( x) dx r π (r x ) /. V π π( r ) πr r πh x r + x x hr x + h dx πh r r πh r 6 πr h x r + x dx 6. Work. W Fd mgd ()(9.)(.) J. W b a 9 f(x) dx dx ( + x) du [u +x, du dx] u u + 9ft-lb. The force function is given by F (x) (in newtons) and the work (in joules) is the area under the curve, given by F (x) dx F (x) dx + F (x) dx ()() + ()() J. 7. f(x) kx k [ inches foot], so k lb/ft and f(x) x. Now6 inches foot, so W / xdx x / ft-lb. 9. (a) If. kx dx J, then kx. k(.).7k and k N/m. Thus, the work needed to stretch the spring from cm to cm is. xdx x /. 9 9 / 9. J. (b) f(x) kx,so 7 9 x and x m. cm

21 . Thedistancefrom cm to cm is. m, so with f(x) kx,wegetw. kx dx k x. Now W.. kx dx k x.. k k.thus,w W. SECTION 6. WORK k. In Exercises, n is the number of subintervals of length x, andx i isasamplepointintheith subinterval [x i,x i].. (a) The portion of the rope from x ft to (x + x) ft below the top of the building weighs x lb and must be lifted x i ft, so its contribution to the total work is x i x ft-lb. The total work is W lim n n i x i x xdx x 6ft-lb Notice that the exact height of the building does not matter (as long as it is more than ft). (b) When half the rope is pulled to the top of the building, the work to lift the top half of the rope is W that is W xdx x 6 ft-lb. The bottom half of the rope is lifted ft and the work needed to accomplish dx is W W + W ft-lb. x 6 ft-lb. The total work done in pulling half the rope to the top of the building. Theworkneededtoliftthecableis lim n n i x i x xdx x, ft-lb.theworkneededtolift the coal is lb ft, ft-lb. Thus, the total work required is, +, 6, ft-lb. 7. At a height of x meters ( x ), the mass of the rope is (. kg/m)( x m) (9.6.x) kg and the mass of the water is 6 kg/m ( x m) (6 x) kg. The mass of the bucket is kg, so the total mass is (9.6.x)+(6 x)+ (.6.x) kg, and hence, the total force is 9.(.6.x) N. The work needed to lift the bucket x m through the ith subinterval of [, ] is 9.(.6.x i ) x, so the total work is W lim n n i 9.(.6.x i ) x (9.)(.6.x) dx 9..6x.9x 9.(9.6) 7 J 9. A slice ofwater x m thick and lying at a depth of x i m(where x i )hasvolume( x) m,amassof x kg, weighs about (9.)( x) 9,6 x N, and thus requires about 9,6x i x J of work for its removal. So W lim n n i 9,6x i x / 9,6xdx 9x / J.. A rectangular slice of water x m thick and lying x m above the bottom has width x mandvolumex x m.itweighs about (9. )(x x) N, and must be lifted ( x) m by the pump, so the work needed is about (9. )( x)(x x) J. The total work required is W (9. )( x)xdx(9. ) (x x ) dx (9. ) x x (9. )( 7) (9. )()..6 6 J

22 CHAPTER 6 APPLICATIONS OF INTEGRATION. Let x measure depth (in feet) below the spout at the top of the tank. A horizontal disk-shaped slice of water x ft thick and lying at coordinate x has radius (6 x) ft () andvolumeπr x π 9 (6 6 x) x ft.itweighs about (6.) 9π 6 (6 x) x lb and must be lifted x ft by the pump, so the work needed to pump it out is about (6.)x 9π 6 (6 x) x ft-lb. The total work required is W 9π (6.)x (6 6 x) dx (6.) 9π 6 (6.) 9π 6 (6.) 9π 6 x(6 x + x ) dx (6x x + x ) dx (6.) 9π 6 x x + x, 6,π. ft-lb d () From similar triangles, x. So r +d + ( x) () (6 x) + ( x). If only.7 J of work is done, then only the water above a certain level (call it h) will be pumped out. So we use the same formula as in Exercise, except that the work is fixed, and we are trying to find the lower limit of integration:.7 h (9. )( x)xdx 9. x x h.7 9. h h h h +.Tofind the solution of this equation, we plot h h +between h and h. We see that the equation is satisfied for h..sothedepthofwaterremaininginthetankisabout. m. 7. V πr x,sov is a function of x and P canalsoberegardedasafunctionofx. IfV πr x and V πr x,then W 9. W x x F (x) dx V x x πr P (V (x)) dx V P (V ) dv by the Substitution Rule. b a F (r) dr b a x G m m dr Gm r m r x P (V (x)) dv (x) [Let V (x) πr x,sodv (x) πr dx.] b a Gm m a b 6. Average Value of a Function. f ave b b a f(x) dx (x a x ) dx x x. g ave b b a g(x) dx a. f ave t +t dt 6 u / du 6 xdx 7 x/ (6 ) 6 u du 6 u / (6/ ) [u +t, du tdt]

23 7. h ave π π cos x sin xdx π u ( du) [u cosx, du sin xdx] π u du π u du [by Theorem..6(a)] u π π SECTION 6. AVERAGE VALUE OF A FUNCTION 9. (a) f ave (x ) dx (x ) (c) 9 ( ) ( + ) 9 (b) f(c) f ave (c ) c ± c or. (a) f ave π π π π ( sin x sin x) dx cosx + cos x π + + π (c) (b) f(c) f ave sinc sin c π c. or c.. f is continuous on [, ], so by the Mean Value Theorem for Integrals there exists a number c in [, ] such that f(x) dx f(c)( ) f(c); that is, there is a number c such that f(c).. f ave f(x) dx M 7. Let t and t correspond to 9 AM and 9 PM, respectively. 9. ρ ave x + dx. V ave V (t) dt T ave [f() + f() + f()] ( ) + sin πt dt t cos πt π F 9 F + π + π + π (x +) / dx x + 9 6kg/m π cos πt dt π cos πt dt π t sin πt [( ) ]. L π π π. Let F (x) x f(t) dt for x in [a, b]. ThenF is continuous on [a, b] and differentiable on (a, b), sobythemeanvalue a Theorem there is a number c in (a, b) such that F (b) F (a) F (c)(b a). ButF (x) f(x) by the Fundamental Theorem of Calculus. Therefore, b f(t) dt f(c)(b a). a

24 6 CHAPTER 6 APPLICATIONS OF INTEGRATION 6 Review. (a) See Section 6., Figure and Equations 6.. and 6... (b) Instead of using top minus bottom and integrating from left to right, we use right minus left and integrate from bottom to top. See Figures and in Section 6... The numerical value of the area represents the number of meters by which Sue is ahead of Kathy after minute.. (a) See the discussion in Section 6., near Figures and, ending in the Definition of Volume. (b) See the discussion between Examples and 6 in Section 6.. If the cross-section is a disk, find the radius in terms of x or y and use A π(radius). If the cross-section is a washer, find the inner radius r in and outer radius r out and use A π r out π r in.. (a) V πrh r (circumference)(height)(thickness). (b) For a typical shell, find the circumference and height in terms of x or y and calculate V b (circumference)(height)(dx or dy),wherea and b are the limits on x or y. a (c) Sometimes slicing produces washers or disks whose radii are difficult (or impossible) to find explicitly. On other 6 occasions, the cylindrical shell method leads to an easier integral than slicing does. f(x) dx represents the amount of work done. Its units are newton-meters, or joules. 6. (a) The average value of a function f on an interval [a, b] is f ave b a b a f(x) dx. (b) The Mean Value Theorem for Integrals says that there is a number c at which the value of f is exactly equal to the average value of the function, that is, f(c) f ave. For a geometric interpretation of the Mean Value Theorem for Integrals, see Figure in Section 6. and the discussion that accompanies it.. The curves intersect when x x x x x x(x ) x or. A (x x ) x dx (x x ) dx x x 6. If x,then x x, and the graphs intersect when x x x + x (x )(x +) x or,but <. By symmetry, we can double the area from x to x. A / ( x ) x dx / ( x x +)dx x x + x / + 7 7

25 . A sin πx π cos πx (x x) dx x + x CHAPTER 6 REVIEW 7 π + π + + π 7. Using washers with inner radius x and outer radius x,wehave V π (x) (x ) dx π (x x ) dx π x x π π 6 π 9. V π (9 y ) ( ) [ ( )] dy π ( y ) dy π ( y + y ) dy π (99 y + y ) dy π 99y y + y π π. The graph of x y a is a hyperbola with right and left branches. Solving for y gives us y x a y ± x a. We ll use shells and the height of each shell is x a x a x a. The volume is V a+h a πx x a dx. Toevaluate,letu x a, so du xdxand xdx du. Whenx a, u,andwhenx a + h, u (a + h) a a +ah + h a ah + h. ah+h ah+h Thus, V π u du π u/ π ah + h /.. A shell has radius π x, circumference π π x, and height cos x. y cos x intersects y when cos x cos x ± [ x π/] x ± π. V π/ π/ π π x cos x dx

26 CHAPTER 6 APPLICATIONS OF INTEGRATION. (a) A cross-section is a washer with inner radius x and outer radius x. V π (x) (x ) dx π(x x ) dx π x x π π (b) A cross-section is a washer with inner radius y and outer radius y. V π y y dy π(y y ) dy π y y π π 6 (c) A cross-section is a washer with inner radius x and outer radius x. V π ( x ) ( x) dx π(x x +x) dx π x x +x π + π 7. (a) Using the Midpoint Rule on [, ] with f(x) tan(x ) and n,weestimate A tan(x ) dx tan +tan +tan +tan 7 (b) Using the Midpoint Rule on [, ] with f(x) π tan (x ) (fordisks)andn,weestimate (.). V tan f(x) dx π +tan +tan +tan 7 π (.).7 9. π/ πx cos xdx π/ (πx)cosxdx The solid is obtained by rotating the region R (x, y) x π, y cos x about the y-axis.. π π( sin x) dx The solid is obtained by rotating the region R {(x, y) x π, y sin x} about the x-axis.. Takethebasetobethediskx + y 9. ThenV A(x) dx,wherea(x ) is the area of the isosceles right triangle whose hypotenuse lies along the line x x in the xy-plane. The length of the hypotenuse is 9 x and the length of each leg is 9 x. A(x) 9 x 9 x,so V A(x) dx (9 x ) dx 9x x (7 9) 6. Equilateral triangles with sides measuring x meters have height x sin 6 x. Therefore, A(x) x x 6 x. V A(x) dx x dx x m. 7. f(x) kx N k( ) cm k N/cm N/m. cm cm. m W. kx dx. xdx x. (.). N m. J.

27 CHAPTER 6 REVIEW 9 9. (a) The parabola has equation y ax with vertex at the origin and passing through (, ). a a y x x y x y. Each circular disk has radius y and is moved y ft. W π y 6.( y) dy π y( y) dy π y y π 6 π 7 ft-lb (b) In part (a) we knew the final water level () but not the amount of work done. Here we use the same equation, except with the work fixed, and the lower limit of integration (that is, the final water level call it h) unknown: W π y y 6 6 h π h h h 6h +. We graph the function f(h) π h 6h + π on the interval [, ] to see where it is. From the graph, f(h) for h.. So the depth of water remaining is about. ft.. lim h f ave lim h (x + h) x x+h x F (x + h) F (x) f(t) dt lim,wheref(x) x f(t) dt. But we recognize this h h a limit as being F (x) by the definition of a derivative. Therefore, lim h f ave F (x) f(x) by FTC.

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29 PROBLEMS PLUS. (a) The area under the graph of f from to t is equal to t f(x) dx, so the requirement is that t f(x) dx t for all t. We differentiate both sides of this equation with respect to t (with the help of FTC) to get f(t) t. This function is positive and continuous, as required. (b) The volume generated from x to x b is b π[f(x)] dx. Hence, we are given that b b π[f(x)] dx for all b>. Differentiating both sides of this equation with respect to b using the Fundamental Theorem of Calculus gives b π[f(b)] f(b) b/π,sincef is positive. Therefore, f(x) x/π.. Let a and b be the x-coordinates of the points where the line intersects the curve. From the figure, R R a c x 7x dx b a x 7x c dx cx x + 7 x a x 7 x cx b a ac a + 7 a b 7 b bc a 7 a ac b 7 b bc b 7 b b b 7b b 7 b b +7b b b b b So for b>, b 6 b. Thus, c b 9 7b (a) V πh (r h/) πh (r h). See the solution to Exercise 6... (b) The smaller segment has height h x andsobypart(a)itsvolumeis V π( x) [() ( x)] π(x ) (x +). This volume must be of the total volume of the sphere, which is π().so π(x ) (x +) π (x x +)(x +) x x + x 9x +. Using Newton s method with f(x) x 9x +, f (x) 9x 9, weget x n+ x n x n 9x n +.Takingx 9x,wegetx.,andx.6 x, so, correct to four decimal n 9 places, x.6. (c) With r. and s.7, the equation x rx +r s becomes x (.)x +(.) (.7) x x + x x +. We use Newton s method with f(x) x x +, f (x) x x,sox n+ x n x n x n + x n x n.takex.. Thenx.6667,andx.676 x. So to four decimal places the depth is.676 m.

30 CHAPTER 6 PROBLEMS PLUS (d) (i) From part (a) with r in., the volume of water in the bowl is V πh (r h) πh ( h) πh πh. We are given that dv dt. in /s and we want to find dh dt when h.now dv dt dh dt. π( ). in/s. π dh πh dt dh πh dt,sodh dt..whenh,wehave π(h h ) (ii) From part (a), the volume of water required to fill the bowl from the instant that the water is in. deep is V π() π() ( ) π 6 this volume by the rate: Time 7π/. 7π 7 s 6. min. 7. We are given that the rate of change of the volume of water is dv dt 7 π π.tofind the time required to fill the bowl we divide ka(x),wherek is some positive constant and A(x) is the area of the surface when the water has depth x. Now we are concerned with the rate of change of the depth of the water with respect to time, that is, dx dv. But by the Chain Rule, dt dt dv dx dx,sothefirst equation can be written dt dv dx dx dt ka(x) (). Also, we know that the total volume of water up to a depth x is V (x) x A(s) ds,wherea(s) is the area of a cross-section of the water at a depth s. Differentiating this equation with respect to x,wegetdv/dx A(x). Substituting this into equation,wegeta(x)(dx/dt) ka(x) dx/dt k, a constant. 9. We must find expressions for the areas A and B, and then set them equal and see what this says about the curve C. If P a, a,thenareaa is just a (x x ) dx a x dx a.tofind area B, weusey as the variable of integration. So we find the equation of the middle curve as a function of y: y x x y/, sinceweare concerned with the first quadrant only. We can express area B as a a a y/ C(y) dy (y/)/ C(y) dy a a C(y) dy where C(y) is the function with graph C. SettingA B,weget a a a C(y) dy a C(y) dy a. Now we differentiate this equation with respect to a using the Chain Rule and the Fundamental Theorem: C(a )(a) a C(y) y/,wherey a.nowwecansolvefory: x y/ x 9 (y/) y 6 9 x.. (a) Stacking disks along the y-axis gives us V h π [f(y)] dy. (b) Using the Chain Rule, dv dt dv dh dh dt π [f(h)] dh dt. (c) ka h π[f(h)] dh dt.setdh dt C: π[f(h)] C ka h [f(h)] ka ka h f(h) πc πc h/ ;that ka is, f(y) πc y/. The advantage of having dh C is that the markings on the container are equally spaced. dt

31 CHAPTER 6 PROBLEMS PLUS. The cubic polynomial passes through the origin, so let its equation be y px + qx + rx. The curves intersect when px + qx + rx x px +(q )x + rx.calltheleftsidef(x). Sincef(a) f(b), another form of f is f(x)px(x a)(x b) px[x (a + b)x + ab] p[x (a + b)x + abx] Since the two areas are equal, we must have a f(x) dx b f(x) dx a [F (x)] a [F (x)]a b F (a) F () F (a) F (b) F () F (b),wheref is an antiderivative of f. Now F (x) f(x) dx p[x (a + b)x + abx] dx p x (a + b)x + abx + C,so F () F (b) C p b (a + b)b + ab + C p b (a + b)b + ab b (a + b)+6a [multiply by /(pb ), b 6 ] b a b +6a b a. Hence, b is twice the value of a.. We assume that P lies in the region of positive x. Sincey x is an odd function, this assumption will not affect the result of the calculation. Let P a, a. The slope of the tangent to the curve y x at P is a,andso the equation of the tangent is y a a (x a) y a x a. We solve this simultaneously with y x to find the other point of intersection: x a x a (x a) (x +a).soq a, a is the other point of intersection. The equation of the tangent at Q is y ( a )a [x ( a)] y a x +6a. By symmetry, this tangent will intersect the curve again at x ( a) a.thecurveliesabovethefirst tangent, and below the second, so we are looking for a relationship between A a a x (a x a ) dx and B a a (a x +6a ) x dx. We calculate A x a x +a x a a a ( 6a ) 7 a,and B 6a x +6a x x a a 96a ( a ) a. We see that B 6A A.Thisisbecauseour calculation of area B was essentially the same as that of area A,witha replaced by a, so if we replace a with a in our expression for A,weget 7 ( a) a B.

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