Problem Set Solutions: Applications of Integration

SOLUTIONS TO 18.01 EXERCISES Unit 4. Applications of integration 4A. Areas between curves. 1 � 4A-1 a) 1/2 �1 (3x − 1 − 2x2 )dx = (3/2)x2 − x − (2/3)x3 �1/2 = 1/24 b) x3 = ax =⇒ x = ±a or x = 0. There are two enclosed pieces (−a < x < 0 and 0 < x < a) with the same area by symmetry. Thus the total area is: � √a � √a 2 (ax − x3 )dx = ax2 − (1/2)x4 �0 = a2 /2 0 5/2 a3/2 (1,2) 1 1 a1/2 (1/2,1/2) a1/2 1 2 a 3/2 1a 1c 1b 1d c) x + 1/x = 5/2 =⇒ x2 + 1 = 5x/2 =⇒ x = 2 or 1/2. Therefore, the area is � 2 1/2 � d) 0 1 �2 [5/2 − (x + 1/x)]dx = 5x/2 − x2 /2 − ln x�1/2 = 15/8 − 2 ln 2 �1 (y − y 2 )dy = y 2 /2 − y 3 /3�0 = 1/6 4A-2 First way (dx): � 1 � 1 �1 (1 − x2 )dx = 2 (1 − x2 )dx = 2x − 2x3 /3�0 = 4/3 −1 √ 0 Second way (dy): (x = ± 1 − y) � 1 � �1 � 2 1 − ydy = (4/3)(1 − y)3/2 � = 4/3 0 0 1 E. Solutions to 18.01 Exercises 4. Applications of integration y = 1-x 2 -1 1 1 x = - 1-y x = 1-y 4A-3 4 − x2 = 3x =⇒ x = 1 or − 4. First way (dx): � 1 �1 (4 − x2 − 3x)dx = 4x − x3 /3 − 3x2 /2�−4 = 125/6 −4 Second way (dy): Lower section has area � 3 �3 � � (y/3 + 4 − ydy = y 2 /6 − (2/3)(4 − y)3/2 � −12 −12 = 117/6 4 (1,3) x=- 4-y x= 4-y x= y 3 (-4,-12) Upper section has area � 4 � �4 � 2 4 − ydy = −(4/3)(4 − y)3/2 � = 4/3 3 3 (See picture for limits of integration.) Note that 117/6 + 4/3 = 125/6. 4A-4 sin x = cos x =⇒ x = π/4 + kπ. So the area is � 5π/4 √ 5π/4 (sin x − cos x)dx = (− cos x − sin x)|π/4 = 2 2 π/4 2 4. Applications of integration E. Solutions to 18.01 Exercises 5π/4 π/4 4B. Volumes by slicing; volumes of revolution 1 � πy 2 dx = 4B-1 a) � −1 1 π(1 − x2 )2 dx = 2π −1 1 � (1 − 2x2 + x4 )dx 0 �1 = 2π(x − 2x3 /3 + x5 /5)�0 = 16π/15 b) �a πy 2 dx = −a �a −a π(a2 − x2 )2 dx = 2π �a 0 (a4 − 2a2 x2 + x4 )dx �a = 2π(a4 x − 2a2 x3 /3 + x5 /5)�0 = 16πa5 /15 1 � πx2 dx = π/3 c) 0 a � πx2 dx = πa3 /3 d) 0 2 � π(2x − x2 )2 dx = e) 0 � 0 2 �2 π(4x2 − 4x3 + x4 )dx = π(4x3 /3 − x4 + x5 /5)�0 = 16π/15 (Why (e) the same as (a)? Complete the square and translate.) 1b 1d 1f 1g 1h b −a ∆x a y = a2− x 2 (for 1a, set a = 1) f) � 2a 0 0 a 0 y = 2ax − x2 y=x (for 1c, set a = 1) π(2ax − x2 )2 dx = � 2a 0 a y= ax (for 1e, set a = 1) π(4a2 x2 − 4ax3 + x4 )dx 3 0 a y 2= b 2(1 − x 2/a 2) E. Solutions to 18.01 Exercises 4. Applications of integration �2 = π(4a2 x3 /3 − ax4 + x5 /5)�0 = 16πa5 /15 (Why is (f) the same as (b)? Complete the square and translate.) a � axdx = πa3 /2 g) 0 a � πy 2 dx = h) 0 � a � 0 �a πb2 (1 − x2 /a2 )dx = πb2 (x − x3 /3a2 )�0 = 2πb2 a/3 1 π(1 − y)dy = π/2 4B-2 a) � 01 c) π(1 − y 2 )dy = 2π/3 a2 �0 a d) 0 e) x2 − 2x + y = 0 =⇒ x = 1 ± � � π(a2 − y)dy = πa4 /2 b) π(a2 − y 2 )dy = 2πa3 /3 √0 1 − y. Using the method of washers: 1 π[(1 + � 1 − y)2 − (1 − � 1 − y)2 ]dy = 1 � 4π 0 � 1 − ydy 0 �1 � = −(8/3)π(1 − y)3/2 � = 8π/3 0 (In contrast with 1(e) and 1(a), rotation around the y-axis makes the solid in 2(e) different from 2(a).) a 2 2b a,2fa2) 2d (a,a) a 2g 2h b ∆y x = a2 − y (for 2a, set a = 1) x=y x = a + a 2− y (for 2c, set a = 1) f) x2 − 2ax + y = 0 =⇒ x = a ± � a2 a 0 0 a 2 x = y /a x2 = a2(1 − y2 /b2 ) (for 2e, set a = 1) � a2 − y. Using the method of washers: � � � 2 2 2 2 π[(a + a − y) − (a − a − y) ]dy = 0 a2 4πa � a2 − ydy 0 �1 � = −(8/3)πa(a2 − y)3/2 � = 8πa4 /3 0 4 4. Applications of integration E. Solutions to 18.01 Exercises g) Using washers: � a �a π(a2 − (y 2 /a)2 )dy = π(a2 y − y 5 /5a2 )�0 = 4πa3 /5. 0 h) � b −b πx2 dy = 2π b � 0 �b a2 (1 − y 2 /b2 )dy = 2π(a2 y − a2 y 3 /3b2 )�0 = 4πa2 b/3 (The answer in 2(h) is double the answer in 1(h), with a and b reversed. Can you see why?) 4B-3 Put the pyramid upside-down. By similar triangles, the base of the smaller bottom pyramid has sides of length (z/h)L and (z/h)M . The base of the big pyramid has area b = LM ; the base of the smaller pyramid forms a cross-sectional slice, and has area M L h z (z/h)L · (z/h)M = (z/h)2 LM = (z/h)2 b Therefore, the volume is � 0 h �h (z/h)2 bdz = bz 3 /3h2 �0 = bh/3 4B-4 The slice perpendicular to the xz-plane are right triangles with base of length x and height z = 2x. Therefore the area of a slice is x2 . The volume is y 2 x z 1 2x x x 1 x side view of wedge along y-axis � 1 x2 dy = top view of wedge along z-axis � side view of slice along y-axis 1 (1 − y 2 )dy = 4/3 −1 −1 √ 4B-5 One side can be described by y = 3x for 0 ≤ x ≤ a/2. Therefore, the volume is � a/2 � 2 2 πy dx2 0 0 a/2 √ π( 3x)2 dx = πa3 /4 5 E. Solutions to 18.01 Exercises 4. Applications of integration y = 3x a/2 4B-6 If the hypotenuse of an isoceles right triangle has length √ h, then its area is√h2 /4. The endpoints of the slice in the xy-plane are y = ± a2 − x2 , so h = 2 a2 − x2 . In all the volume is � a � a 2 (h /4)dx = (a2 − x2 )dx = 4a3 /3 −a −a 4B-7 Solving for x in y = (x − 1)2 and y = (x + 1)2 gives the values a2-x 2 a 2 a2-x 2 x slice top view x=1± √ y and x = −1 ± √ y The hard part is deciding which sign of the square root representing the endpoints of the square. x = - 1+ y 1 x =1 - y -1 1 2 (x- y) Method 1: The point (0, 1) has to be on the two curves. Plug in y = 1 and x = 0 √ to see that the square root must have the opposite sign from 1: x = 1 − y and √ x = −1 + y. √ Method 2: Look at the picture. x = 1 + y is the wrong choice because it is the √ right half of the parabola with vertex (1, 0). We want the left half: x = 1 − y. √ Similarly, we want x = −1 + y, the right half of the parabola with vertex (−1, 0). √ √ Hence, the side of the square is the interval −1 + y ≤ x ≤ 1 − y, whose length √ is 2(1 − y), and the � 1 � 1 √ √ 2 Volume = (2(1 − y) dy = 4 (1 − 2 y + y)dy = 2/3 . 0 6 0 4. Applications of integration E. Solutions to 18.01 Exercises 4C. Volumes by shells 4C-1 a) � b+a Shells: b+a � � 4πx a2 − (x − b)2 dx (2πx)(2y)dx = b−a b−a � (x − b)2 = a2 − y 2 =⇒ x = b ± a2 − y 2 � a � a � � 2 2 Washers: π(x2 − x1 )dy = π((b + a2 − y 2 )2 − (b − a2 − y 2 )2 )dy −a −a � a � =π 4b a2 − y 2 dy b) −a y = a2 - (x - b) 2 x = b - a 2- y 2 a a b-a x = b + a2 - y 2 b+a b-a -a b+a -a 2 y = - a 2- (x - b} Washers Shells a � � c) a2 − y 2 dy = πa2 /2, because it’s the area of a semicircle of radius a. −a Thus (b) =⇒ Volume of torus = 2π 2 a2 b d) z = x − b, dz = dx � b+a 4πx � a2 − (x − b)2 dx = � a � � 4π(z + b) a2 − z 2 dz = −a b−a a 4πb � a2 − z 2 dz −a because the part of the integrand with the factor z is odd, and so it integrates to 0. � 4C-2 1 1 � 2πx3 dx = π/2 2πxydx = 0 0 1 1 1 y = x2 4C−2 (shells) y= x 4C−3a (shells) 7 x = y2 4C−3b (discs) E. Solutions to 18.01 Exercises 1 � 4C-3 4. Applications of integration 0 � Disks: 1 πx2 dy = πy 4 dy = π/5 1 � y 0 � 1 − ydy 0 a2 � x)dx = π/5 1 � 2πy(2x)dy = 4π b) √ 0 1 � 2πx(1 − 0 0 4C-4 a) 1 � 2πx(1 − y)dx = Shells: � 2πy(2x)dy = 4π 0 a2 y � a2 − ydy 0 1 � 2πy(1 − y)dy c) 0 a � 2πy(a − y)dy d) 0 √ e) x2 − 2x + y = 0 =⇒ x = 1 ± 1 − y. � � � The interval 1 − 1 − y ≤ x ≤ 1 + 1 − y has length 2 1 − y � 1 � 1 � � =⇒ V = 2πy(2 1 − y)dy = 4π y 1 − ydy 0 0 � f) x2 − 2ax + y = 0 =⇒ x = a ± a2 − y. � � The interval a − a2 − y ≤ x ≤ a + a2 − y � a2 � � 2 =⇒ V = 2πy(2 a − ydy = 4π 0 4b � has length 2 a2 − y a2 y 4f 4d 4g 2 (a, a ) g) 0 x = a + a 2- y x = a - a2 - y a 2πy(a − y 2 /a)dy 8 b 0 0 a x=y 4h (a,a) a � a2 − ydy 0 a2 x = a 2 - y (right) x = - a 2- y (left) � x = y2/a a 2 x = a 1 - y /b 2 4. Applications of integration b � h) b � 2πy(a2 (1 − y 2 /b2 )dy 2πyxdy = 0 E. Solutions to 18.01 Exercises 0 (Why is the lower limit of integration 0 rather than −b?) 1 � c) 0 d) 0 2 � 0 � 2πx(2x − x2 )dx f) 0 5b 2πx2 dx 0 2 � 2πxydx = a � 2πxydx = 0 e) 2πx2 dx 0 a � 2πx(a2 − x2 )dx 1 � 2πxydx = 0 a � b) 1 � 2πx(1 − x2 )dx 4C-5 a) 2a 0 5d 2a � 2πx(ax − 2πxydx = 0 5f b 5g 5h a -a ∆x a a � g) a y = ax √ 2πx axdx 0 a � a � 2πxydx = 0 h) y = 2ax - x 2 (for 5e, set a =1 ) y=x for 5c, set a = 1 ) y=a-x (for 5a, set a = 1 ) x2 )dx 0 a 0 2 2 a � 2πx(2b2 (1 − x2 /a2 ))dx 2πx(2y)dx = 0 0 (Why did y get doubled this time?) 4C-6 � b b � 2πx(2y)dx = a 2πx(2 � b2 − x2 )dx a �b � = −(4/3)π(b2 − x2 )3/2 � = (4π/3)(b2 − a2 )3/2 a 4D. Average value 4D-1 Cross-sectional area at x is = πy 2 = π · (x2 )2 = πx4 . Therefore, 9 b y = b 1 - x 2/a 2 y = -b 1 - x 2/a 2 E. Solutions to 18.01 Exercises 4. Applications of integration y= b2- x 2 Shells a b 2 y=- b - x 2 base of removed cylinder average cross-sectional area = 1 4D-2 Average = a � 2a a 1 2 2 � πx4 dx = 0 �2 πx5 �� 16π = . 10 �0 5 �2a � � � 1 1 1 2a ln 2 dx = ln x�� = (ln 2a − ln a) = ln = . x a a a a a a 4D-3 Let s(t) be the distance function; then the velocity is v(t) = s� (t) Average value of velocity = 1 b−a � b s� (t)dt = a s(b) − s(a) by FT1 b−a = average velocity over time interval [a,b] 4D-4 By symmetry, we can restrict P to the upper semicircle. By the law of cosines, we have average of |P Q|2 = 1 π � |P Q|2 = 12 + 12 − 2 cos θ. Thus π (2 − 2 cos θ)dθ = 0 1 π [2θ − 2 sin θ]0 = 2 π (This is the value of |P Q|2 when θ = π/2, so the answer is reasonable.)) P 1 θ 1 Q � 1 x 4D-5 By hypothesis, g(x) = f (t)dt To express f (x) in terms of g(x), x 0 multiply thourgh by x and apply the Sec. Fund. Thm: 10 4. Applications of integration � E. Solutions to 18.01 Exercises x f (t)dt = xg(x) ⇒ f (x) = g(x) + xg � (x) , by FT2.. 0 1 4D-6 Average value of A(t) = T T � A0 ert dt = 0 1 A0 rt T A0 rT e |0 = (e − 1) T r rT (rT )2 , so we get 2 2 A0 (rT ) rT (rT + ) = A0 (1 + ). A(t) ≈ rT 2 2 If rT is small, we can approximate: erT ≈ 1 + rT + (If T ≈ 0, at the end of T years the interest added will be A0 rT ; thus the average is approximately what the account grows to in T /2 years, which seems reasonable.) 4D-7 1 b � b x2 dx = b2 /3 0 4D-8 The average on each side is the same as the average over all four sides. Thus the average distance is � 1 a/2 � 2 x + (a/2)2 dx a −a/2 x2+ (a/2) 2 a/2 x Can’t be evaluated by a formula until Unit 5. The average of the square of the distance is � � 1 a/2 2 2 a/2 2 (x + (a/2)2 )dx = (x + (a/2)2 )dx = a2 /3 a −a/2 a 0 �π/a � π/a � 1 1 4D-9 sin ax dx − cos(ax)�� = 2/π π/a 0 π 0 4D’. Work 4D’-1 According to Hooke’s law, we have F = kx, where F is the force, x is the displacement (i.e., the added length), and k is the Hooke’s law constant for the 11 E. Solutions to 18.01 Exercises 4. Applications of integration spring. To find k, substitute into Hooke’s law: 2, 000 = k · (1/2) ⇒ k = 4000. To find the work W , we have � 6 � 6 �6 W = F dx = 4000x dx = 2000x2 0 = 72, 000 inch-pounds = 6, 000 foot-pounds. 0 0 4D’-2 Let W (h) = weight of pail and paint at height h. 1 h, since the pulling and leakage W (0) = 12, W (30) = 10 ⇒ W (h) = 12 − 15 both occur at a constant rate. �30 � 30 � 30 h h2 work = W (h) dh = (12 − ) dh = 12h − = 330 ft-lbs. 15 30 0 0 0 4D’-3 Think of the hose as divided into many equal little infinitesimal pieces, of length dh, each of which must be hauled up to the top of the building. The piece at distance h from the top end has weight 2 dh; to haul it up to the top requires 2h dh ft-lbs. Adding these up, � 50 �50 total work = 2h dh = h2 0 = 2500ft-lbs. 0 g m1 m2 . x2 � � � � 1 1 g m1 m2 n − 1 = −g m1 m2 − = . nd d d n 4D’-4 If they are x units apart, the gravitational force between them is � nd g m1 m2 g m1 m2 work = dx = − 2 x x d g m1 m2 The limit as n → ∞ is . d �nd d 4E. Parametric equations 4E-1 y − x = t2 , y − 2x = −t. Therefore, y − x = (y − 2x)2 =⇒ y 2 − 4xy + 4x2 − y + x = 0 (parabola) 4E-2 x2 = t2 + 2 + 1/t2 and y 2 = t2 − 2 + 1/t2 . Subtract, getting the hyperbola x2 − y 2 = 4 4E-3 (x − 1)2 + (y − 4)2 = sin2 θ + cos2 t = 1 (circle) 4E-4 1 + tan2 t = sec2 t =⇒ 1 + x2 = y 2 (hyperbola) � 4E-5 x = sin 2t = 2 sin t cos t = ±2 1 − y 2 y. This gives x2 = 4y 2 − 4y 4 . 12 4. Applications of integration E. Solutions to 18.01 Exercises 4E-6 y � = 2x, so t = 2x and y = t2 /4 x = t/2, 4E-7 Implicit differentiation gives 2x + 2yy � = 0, so that y � = −x/y. So the parameter is t = −x/y. Substitute x = −ty in x2 + y 2 = a2 to get t2 y 2 + y 2 = a2 =⇒ y 2 = a2 /(1 + t2 ) Thus a −at y=√ , x= √ 2 1+t 1 + t2 For −∞ < t < ∞, this parametrization traverses the upper semicircle y > 0 (going clockwise). One can also get the lower semicircle (also clockwise) by taking the negative square root when solving for y, −a at y=√ , x= √ 2 1 + t2 1+t 4E-8 The tip Q of the hour hand is given in terms of the angle θ by Q = (cos θ, sin θ) (units are meters). Next we express θ in terms of the time parameter t (hours). We have � θ= π/2, t = 0 π/3, t = 1 =⇒ θ − π = 2 π 3 − � θ decreases linearly with t π 2 · (t − 0) . Thus we get θ = 1−0 π 2 − π6 t. Q P θ Finally, for the snail’s position P , we have P = (t cos θ, t sin θ) , where t increases from 0 to 1. So, x = t cos( π π π − t) = t sin t, 2 6 6 y = t sin( π2 − π6 t) = t cos π6 t 4F. Arclength � � √ � 2 4F-1 a) ds = 1 + (y ) dx = 26dx. Arclength = 13 0 1 √ 26dx = √ 26. E. Solutions to 18.01 Exercises 4. Applications of integration � � b) ds = 1 + (y � )2 dx = 1 + (9/4)xdx. � 1� �1 � Arclength = 1 + (9/4)xdx = (8/27)(1 + 9x/4)3/2 � = (8/27)((13/4)3/2 − 1) 0 0 � c) y � = −x−1/3 (1 − x2/3 )1/2 = − x−2/3 − 1. Therefore, ds = x−1/3 dx, and � 1 �1 � Arclength = x−1/3 dx = (3/2)x2/3 � = 3/2 0 0 � d) y � = x(2 + x2 )1/2 . Therefore, ds = 1 + 2x2 + x4 dx = (1 + x2 )dx and � 2 �2 Arclength = (1 + x2 )dx = x + x3 /3�1 = 10/3 1 � −x x 4F-2 y = (e − e ydx. Thus, )/2, so the hint says 1 + (y � )2 = y 2 and ds = b � Arclength = (1/2) 0 � 4F-3 y = 2x, � 1+ (y � )2 � 1 + (y � )2 dx = �b (ex + e−x )dx = (1/2)(ex − e−x )�0 = (eb − e−b )/2 √ = 1 + 4x2 . Hence, arclength = � b � 1 + 4x2 dx. 0 4F-4 ds = √ (dx/dt)2 + (dy/dt)2 dt = 4t2 + 9t4 dt. Therefore, � 2� � 2 2 4 Arclength = 4t + 9t dt = (4 + 9t2 )1/2 tdt � 0 0 �2 2 3/2 � = (1/27)(4 + 9t ) � = (403/2 − 8)/27 0 2 2 4F-5 dx/dt = 1 − 1/t , dy/dt = 1 + 1/t . Thus � � ds = (dx/dt)2 + (dy/dt)2 dt = 2 + 2/t4 dt � 2� 2 + 2/t4 dt Arclength = and 1 4F-6 a) dx/dt = 1 − cos t, dy/dt = sin t. � √ ds/dt = (dx/dt)2 + (dy/dt)2 = 2 − 2 cos t (speed of the point) Forward motion (dx/dt) is largest for t an odd multiple of π (cos t = −1). Forward motion is smallest for t an even multiple of π (cos t = 1). (continued →) Remark: The largest forward motion is when the point is at the top of the wheel and the smallest is when the point is at the bottom (since y = 1 − cos t.) � b) 0 2π √ 2 − 2 cos tdt = � 0 2π 2π 2 sin(t/2)dt = −4 cos(t/2)|0 = 8 14 4. Applications of integration 2π � E. Solutions to 18.01 Exercises � a2 sin2 t + b2 cos2 tdt 4F-7 0 4F-8 dx/dt = et (cos t − sin t), dy/dt = et (cos t + sin t). � � √ ds = e2t (cos t − sin t)2 + e2t (cos t + sin t)2 dt = et 2 cos2 t + 2 sin2 tdt = 2et dt Therefore, the arclength is 10 � √ 2et dt = √ 10 2(e − 1) 0 4G. Surface Area 4G-1 The curve y = √ R2 − x2 for a ≤ x ≤ b is revolved around the x-axis. √ Since we have y � = −x/ R2 − x2 , we get y= R2- x 2 a ds = b � � � 1 + (y � )2 dx = 1 + x2 /(R2 − x2 )dx = R2 /(R2 − x2 )dx = (R/y)dx Therefore, the area element is dA = 2πyds = 2πRdx and the area is � b 2πRdx = 2πR(b − a) √ 4G-2 Limits are 0 ≤ x ≤ 1/2. ds = 5dx, so √ √ � dA = 2πyds = 2π(1 − 2x) 5dx =⇒ A = 2π 5 a 1/2 (1 − 2x)dx = √ 5π/2 0 4G-3 Limits are 0 ≤ y ≤ 1. x = (1 − y)/2, dx/dy = −1/2. Thus ds = � 1 + (dx/dy)2 dy = � 5/4dy; √ √ √ �1 dA = 2πyds = π(1 − y)( 5/2)dx =⇒ A = ( 5π/2) 0 (1 − y)dy = 5π/4 15 E. Solutions to 18.01 Exercises 4. Applications of integration 4G−3 y = 1 − 2x x = (1 − y)/2 1 4G−2 1/2 � 4G-4 A = 4 � 2πx2 2πyds = � 1 + 4x2 dx 0 4G-5 x = � √ √ y, dx/dy = −1/2 y, and ds = 1 + 1/4ydy � � 2 √ � A = 2πxds = 2π y 1 + 1/4ydy 0 2 � = 2π � y + 1/4dy 0 �2 � = (4π/3)(y + 1/4)3/2 � = (4π/3)((9/4)3/2 − (1/4)3/2 ) 0 = 13π/3 4G-6 y = (a2/3 − x2/3 )3/2 =⇒ y � = −x−1/3 (a2/3 − x2/3 )1/2 . Hence � ds = 1 + x−2/3 (a2/3 − x2/3 )dx = a1/3 x−1/3 dx Therefore, (using symmetry on the interval −a ≤ x ≤ a) -a a y = (a2/3- x 2/3) � � 3/2 a 2π(a2/3 − x2/3 )3/2 a1/3 x−1/3 dx � �a = (4π)(2/5)(−3/2)a1/3 (a2/3 − x2/3 )5/2 � A= 2πyds = 2 0 0 = (12π/5)a2 � 4G-7 a) Top half: y = a2 − (x − b)2 , y � = (b − x)/y. Hence, � � ds = 1 + (b − x)2 /y 2 dx = (y 2 + (b − x)2 )/y 2 dx = (a/y)dx Since we are only covering the top half we double the integral for area: � � b+a xdx � A = 2πxds = 4πa 2 a − (x − b)2 b−a b) We need to rotate two curves x2 = b + 16 � a2 − y 2 4. Applications of integration E. Solutions to 18.01 Exercises y = a 2- (x-b) a 2 a b y = - a 2- (x-b) 2 upper and lower surfaces are symmetrical and equal and x1 = b − � a2 − y 2 around the y-axis. The value � dx2 /dy = −(dx1 /dy) = −y/ a2 − y 2 So in both cases, ds = � � 1 + y 2 /(a2 − y 2 )dy = (a/ a2 − y 2 )dy The integral is x = b + a 2- y 2 x = b - a -2 y 2 inner and outer surfaces are not symmetrical and not equal � A= � 2πx2 ds + � 2πx1 ds = But x1 + x2 = 2b, so � a A = 4πab −a a ady 2π(x1 + x2 ) � a2 − y 2 −a dy � 2 a − y2 c) Substitute y = a sin θ, dy = a cos θdθ to get � π/2 � π/2 a cos θdθ A = 4πab = 4πab dθ = 4π 2 ab a cos θ −π/2 −π/2 4H. Polar coordinate graphs 4H-1 We give the polar coordinates in the form (r, θ): 17 E. Solutions to 18.01 Exercises 4. Applications of integration √ a) (3, √ π/2) b) (2, π) c) (2, π/3) d) (2 2, 3π/4) e) ( 2, −π/4 or 7π/4) f) (2,√ −π/2 or 3π/2) g) (2, −π/6 or 11π/6) h) (2 2, −3π/4 or 5π/4) 4H-2 a) (i) (x − a)2 + y 2 = a2 ⇒ x2 − 2ax + y 2 = 0 ⇒ r2 − 2ar cos θ = 0 ⇒ r = 2a cos θ. (ii) ∠OP Q = 90o , since it is an angle inscribed in a semicircle. In the right triangle OPQ, |OP | = |OQ| cos θ, i.e., r = 2a cos θ. b) (i) Analogous to 4H-2a(i); ans: r = 2a sin θ. (ii) analogous to 4H-2a(ii); note that ∠OQP = θ, since both angles are comple­ ments of ∠P OQ. c) (i) OQP is a right triangle, |OP | = r, and ∠P OQ = α − θ. The polar equation is r cos(α − θ) = a, or in expanded form, r(cos α cos θ + sin α sin θ) = a , or finally x y + = 1, A B a since from the right triangles OAQ and OBQ, we have cos α = , sin α = A a cos BOQ = . B d) Since |OQ| = sin θ, we have: if P is above the x-axis, sin θ > 0, OP | = |OQ| − |QR|, or r = a − a sin θ; if P is below the x-axis, sin θ < 0, OP | = |OQ| + |QR|, or r = a + a| sin θ| = a − a sin θ. Thus the equation is r = a(1 − sin θ). e) Briefly, when P = (0, 0), |P Q||P R| = a · a = a2 , the constant. Using the law of cosines, |P R|2 = r2 + a2 − 2ar cos θ; |P Q|2 = r2 + a2 − 2ar cos(π − θ) = r2 + a2 + 2ar cos θ Therefore |P Q|2 |P R|2 = (r2 + a2 )2 − (2ar cos θ)2 = (a2 )2 which simplifies to r2 = 2a2 cos 2θ. 4H-3 a) r = sec θ =⇒ r cos θ = 1 =⇒ x = 1 b) r = 2a cos θ =⇒ r2 = r · 2a cos θ = 2ax =⇒ x2 + y 2 = 2ax c) r = (a + b cos θ) (This figure is a cardiod for a = b, a limaçon with a loop for 0 < a < b, and a limaçon without a loop for a > b > 0.) 18 4. Applications of integration E. Solutions to 18.01 Exercises � r2 = ar + br · cos θ = ar + bx =⇒ x2 + y 2 = a x2 + y 2 + bx b=|c| parabola y r r θ θ a r r θ 2a x b>|c| ellipse r θ θ hyperbola b<|c| 1 limacon ab 8c r = a/(b + c cos θ) 8d =⇒ r(b + c cos θ) = a =⇒ rb = a − cx =⇒ rb + cx = a 2 2 r b = a2 − 2acx + c2 x2 =⇒ =⇒ a2 − 2acx + (c2 − b2 )x2 − b2 y 2 = 0 (e) r = a sin(2θ) r = a cos 2 θ =⇒ r = 2a sin θ cos θ = 2axy/r2 =⇒ r3 = 2axy r2 = a2sin 2θ 2 r2 = a cos 2θ r = a sin 2θ r = a cos(2θ) = a(2 cos2 θ − 1) = a( f) (x2 + y 2 )3/2 = 2axy =⇒ 2x2 − 1) =⇒ (x2 + y 2 )3/2 = + y2 x2 a(x2 − y 2 ) g) r2 = a2 sin(2θ) = 2a2 sin θ cos θ = 2a2 2axy h) r2 = a2 cos(2θ) = a2 ( i) xy =⇒ r4 = 2a2 xy =⇒ (x2 +y 2 )2 = r2 2x2 − 1) =⇒ (x2 + y 2 )2 = a2 (x2 − y 2 ) + y2 x2 r = eaθ =⇒ ln r = aθ =⇒ ln � x2 + y 2 = a tan−1 y x 4I. Area and arclength in polar coordinates 19 E. Solutions to 18.01 Exercises 4I-1 � 4. Applications of integration (dr/dθ)2 + r2 dθ a) sec2 θdθ b) 2adθ c) � a2 + b2 + 2ab cos θdθ d) √ a b2 + c2 + 2bc cos θ dθ (b + c cos θ)2 � e) a 4 cos2 (2θ) + sin2 (2θ)dθ � f) a 4 sin2 (2θ) + cos2 (2θ)dθ g) Use implicit differentiation: 2rr� = 2a2 cos(2θ) =⇒ r� = a2 cos(2θ)/r =⇒ (r� )2 = a2 cos2 (2θ)/ sin(2θ) Hence, using a common denominator and cos2 + sin2 = 1, � a ds = a2 cos2 (2θ)/ sin(2θ) + a2 sin(2θ)dθ = � dθ sin(2θ) h) This is similar to (g): ds = � i) a cos(2θ) dθ � 1 + a2 eaθ dθ 4I-2 dA = (r2 /2)dθ. The main difficulty is to decide on the endpoints of integra­ tion. Endpoints are successive times when r = 0. cos(3θ) = 0 =⇒ 3θ = π/2 + kπ =⇒ θ = π/6 + kπ/3, � Thus, π/6 (a2 cos2 (3θ)/2)dθ = a2 A= −π/6 k an integer. π/6 � cos2 (3θ)dθ. 0 (Stop here in Unit 4. Evaluated in Unit 5.) � 4I-3 A = 2 π � (r /2)dθ = 0 �π (e6θ /2)dθ = (1/12)e6θ �0 = (e6π − 1)/12 20 4. Applications of integration E. Solutions to 18.01 Exercises θ = π/6 θ =−π/6 three-leaf rose three empty sectors e3π 1 4I-4 Endpoints are successive time when r = 0. � Thus, A = sin(2θ) = 0 =⇒ 2θ = kπ, k an integer. � π/2 �π/2 (r2 /2)dθ = (a2 /2) sin(2θ)dθ = −(a2 /4) cos(2θ)�0 = a2 /2. 0 4I-5 r = 2a cos θ, ds = 2adθ, −π/2 < θ < π/2. (The range was chosen carefully so that r > 0.) Total length of the circle is 2πa. Since the upper and lower semicircles are symmetric, it suffices to calculate the average over the upper semicircle: 2a r = 2a cos θ �π/2 � π/2 � 1 4a 4a 2a cos θ(2a)dθ = sin θ�� = πa 0 π π 0 4I-6 a) Since the upper and lower halves of the cardiod are symmetric, it suffices to calculate the average distance to the x-axis just for a point on the upper half. We have r = a(1 − cos θ), and the distance to the x-axis is r sin θ, so 1 π � 0 π 1 r sin θdθ = π � π a(1 − cos θ) sin θdθ = 0 21 �π a 2a � (1 − cos θ)2 � = 2π π 0 E. Solutions to 18.01 Exercises 4. Applications of integration P r 0 Q (b) � (dr/dθ)2 + r2 dθ = a (1 − cos θ)2 + sin2 θdθ √ = a 2 − 2 cos θdθ = 2a sin(θ/2)dθ, using the half angle formula. � 2π 2π arclength = 2a sin(2θ)dθ = −4a cos(θ/2)|0 = 8a ds = � 0 For the average, don’t use the half-angle version of the formula for ds, and use the interval −π < θ < π, where sin θ is odd: � π � π √ √ 1 1 Average = |r sin θ|a 2 − 2 cos θdθ = | sin θ| 2a2 (1 − cos θ)3/2 dθ 8a −π 8a −π �π √ � π √ � 2a 2a 4 � = (1 − cos θ)3/2 sin θdθ = (1 − cos θ)5/2 � = a � 4 10 5 0 0 4I-7 dx = −a sin θdθ. So the semicircle y > 0 has area � a � 0 � 2 ydx = a sin θ(−a sin θ)dθ = a −a π But � 0 π π sin2 θdθ 0 1 sin θdθ = 2 2 � π (1 − cos(2θ)dθ = π/2 0 So the area is πa2 /2 as it should be for a semicircle. Arclength: ds2 = dx2 + dy 2 =⇒ (ds)2 = (−a sin θdθ)2 + (a cos θdθ)2 = a2 (sin2 dθ + cos2 dθ)(dθ)2 =⇒ ds = adθ (obvious from picture). ds dθ θ a -a � 2π � ds = adθ = 2πa 0 22 4. Applications of integration E. Solutions to 18.01 Exercises 4J. Other applications 4J-1 Divide the water in the hole into n equal circular discs of thickness Δy. � �2 1 Volume of each disc: π Δy 2 π Energy to raise the disc of water at depth yi to surface: kyi Δy. 4 Adding up the energies for the different discs, and passing to the limit, E = lim n � π n→∞ 1 4 � 100 kyi Δy = 0 π πk y 2 ky dy = 4 4 2 �100 = 0 πk104 . 8 4J-2 Divide the hour into n equal small time intervals Δt. At time ti , i = 1, . . . , n, there are x0 e−kti grams of material, producing approx­ imately rx0 e−kti Δt radiation units over the time interval [ti , ti + Δt]. Adding and passing to the limit, R = lim n→∞ n � r x0 e −kti � Δt = 60 r x0 e 0 1 −kt e−kt dt = r x0 −k �60 = 0 � r x0 � 1 − e−60k . k 4J-3 Divide up the pool into n thin concentric cylindrical shells, of radius ri , i = 1, . . . , n, and thickness Δr. The volume of the i-th shell is approximately 2π ri D Δr. k The amount of chemical in the i-th shell is approximately 2π ri D Δr. 1 + ri2 Adding, and passing to the limit, n � � R k r 2π r D Δr = 2πkD dr i 2 n→∞ 1 + r 1 + r2 0 i 1 �R 2 = πkD ln(1 + r ) = πkD ln(1 + R2 ) gms. A = lim 0 4J-4 Divide the time interval into n equal small intervals of length Δt by the points ti , i = 1, . . . , n. The approximate number of heating units required to maintain the temperature at 75o over the time interval [ti , ti + Δt]: is � � �� πti · k Δt. 75 − 10 6 − cos 12 23 E. Solutions to 18.01 Exercises 4. Applications of integration Adding over the time intervals and passing to the limit: � �� n � � πti · k Δt total heat = lim 75 − 10 6 − cos n→∞ 12 1 � �� � 24 � πt = k 75 − 10 6 − cos dt 12 0 � � �24 � 24 � πt 120 πt = k 15 + 10 cos dt = k 15t + sin = 360k. 12 π 12 0 0 4J-5 Divide the month into n equal intervals of length Δt by the points ti , i = 1, . . . , n. Over the time interval [ti .ti + Δt], the number of units produced is about (10 + ti ) Δt. The cost of holding these in inventory until the end of the month is c(30 − ti )(10 + ti ) Δt. Adding and passing to the limit, n � total cost = lim c(30 − ti )(10 + ti ) Δt n→∞ � = 0 30 1 � �30 t3 c(30 − t)(10 + t) dt = c 300t + 10t2 − = 9000c. 3 0 4J-6 Divide the water in the tank into thin horizontal slices of width� dy. If the slice is at height y above the center of the tank, its radius is r2 − y 2 . This formula for the radius of the slice is correct even if y < 0 – i.e., the slice is below the center of the tank – as long as −r < y < r, so that there really is a slice at that height. Volume of water in the slice = π(r2 − y 2 ) dy Weight of water in the slice = πw(r2 − y 2 ) dy Work to lift this slice from the ground to the height h+y = πw(r2 −y 2 ) dy (h+y). � r Total work = πw(r2 − y 2 )(h + y) dy −r � r = πw (r2 h + r2 y − hy 2 − y 3 ) −r � �r r2 y2 hy 3 y4 = πw r2 hy + − − . 2 3 4 −r In this last line, the even powers of y have the same value at −r and r, so contribute 0 when it is evaluated; we get therefore � �r � � y3 r3 4 = πwh r2 y − = 2πwh r3 − = πwhr3 . 3 −r 3 3 24