18.02SC Supplementary Problem Solutions: Double Integrals

3. Double Integrals 3A. Double integrals in rectangular coordinates 3A-1 a) Inner: 6x2 y + y 2 �1 = 12x2 ; y=−1 Outer: 4x3 �2 0 = 32 . �π = 2u + 21 π 2 cos u b) Inner: −u cos t + 12 t2 cos u t=0 �π/2 = ( 12 π)2 + 12 π 2 = 34 π 2 . Outer: u2 + 21 π 2 sin u 0 c) Inner: x2 y 2 �x2 √ x = x6 − x3 ; 1 7 7x Outer: �u √ √ d) Inner: v u2 + 4 = u u2 + 4; − 14 x4 1 2 3 (u Outer: 0 �1 0 = 1 7 + 4)3/2 − 1 4 �1 √ = 31 (5 5 − 8) 0 3 = − 28 y 3A-2 a) b) 2 �� (i) dy dx = R � 0 � −2 2 dy dx �� (ii) −x dx dy = R � 2 0 � 0 dx dy y=-x −y x i) The ends of R are at 0 and 2, since 2x − x2 = 0 has 0 and 2 as roots. �� � 2 � 2x−x2 dydx = dydx R 0 y=2x-x 2 0 ii) We solve y = 2x − x2 for x in terms of y: write the equation as √ 2 x −2x+y = 0 and solve for x by the quadratic formula, getting x = 1± 1 − y. Note also that the maximum point of the graph is (1, 1) (it lies midway between the two roots 0 and 2). We get �� c) �� (i) (ii) dy dx = R �� � dx dy = R √ 2 0 � � √ 2 � dxdy = R x dy dx + 0 � √4−y2 � 2 √ 2 1 0 � � √ 1 √ 1+ 1−y √ 1− 1−y dxdy, 4−x2 dy dx 0 y=x 2 2 dx dy 2 y 0 d) Hint: First you have to find the points where the two curves intersect, by solving simultaneously y 2 = x and y = x − 2 (eliminate x). �� �� The integral dy dx requires two pieces; dx dy only one. R 3A-3 a) �� R Inner: x(1 − 21 x) 2 � 1−x/2 x dy dx; �2 Outer: 21 x2 − 16 x3 = x dA = R � 0 0 0 4 2 1 − 8 6 = 23 . 2 2 b) E. 18.02 EXERCISES �� 1 � � 1−y 2 (2x + y ) dA = (2x + y 2 ) dx dy 0 0 �1 �1−y2 = 1 − y2 ; Outer: y − 31 y 3 = 23 . Inner: x2 + y 2 x 2 R 0 c) �� 1 � 0 1−y � y dx dy y dA = �1−y 0 y−1 Inner: xy = y[(1 − y) − (y − 1)] = 2y − 2y 2 R Outer: y 2 − 23 y 3 y−1 �� 3A-4 a) 2 sin x dA = R Inner: y sin2 x b) �� �cos x xy dA = R Inner: 1 2 2 xy �x x2 1 0 π/2 −π/2 � � 0 = 13 . cos x sin2 x dy dx 0 = cos x sin2 x 0 � � �1 1 3 Outer: � π/2 sin3 x −π/2 = 13 (1 − (−1)) = 32 . x (xy) dy dx. x2 = 12 (x3 − x5 ) Outer: � �1 1 x4 x6 1 1 1 = · − = . 2 4 6 0 2 12 24 c) The function x2 − y 2 is zero on the lines y = x and y = −x, and positive on the region R shown, lying between x = 0 and x = 1. Therefore �� � 1� x 2 2 (x2 − y 2 ) dy dx. Volume = (x − y ) dA = R 0 Inner: x2 y − 31 y 3 � 3A-5 a) b) c) � � 1 4 0 1 0 3A-6 � � 1 2 √ t 1 x1/3 0; 2 0 � 2 e �x = 34 x3 ; dy dx = x u e du dt = u � 1 2 0 � � Outer: u2 0 2 0 u y e −y 2 1 4 3x �1 0 � 1 2 0 4 2 � 2 e 0 −y 2 � 2 2 1 1 y dy = − e−y = (1 − e−4 ) 2 2 0 u eu du = (u − 1)eu � 12 0 =1− Q x2 dA; 0 Q 1 ≤1 1 + x4 + y 4 �� � 1� 1 �1 x x dA 1 .7 ln 2 2 b) ≤ dx dy = ln(1 + x ) = < . 2 2 2 2 2 2 0 R 1+x +y 0 0 1+x 3A-7 a) x4 + y 4 ≥ 0 ⇒ y=x 5a 1√ e 2 � 1 � u3 � 1 �1 1 1 u3 1 ln 2 4 du dx = dx du = du = ln(1 + u ) . = 4 4 4 1+u 1+u 4 4 0 0 0 0 1+u �� �� x2 dA, Q = first quadrant 2 ex dA, S = upper half of R; 4 �� 2 y=-x = 13 . dx dy = 0 e dt du = u S 0; � 1 −x −x −y 2 y=x 1/2 t=u 2 1/4 5b 1 x=u3 1 5c 3 3B. Double Integrals in polar coordinates 3B-1 2 1 a) In polar coordinates, the line x = −1 becomes r cos θ = −1, or r = − sec θ. We also need the polar angle of the intersection points; since the right triangle is a 30-60-90 triangle (it has one leg 1 and hypotenuse 2), the limits are (no integrand is given): �� � dr dθ = R 4π/3 2π/3 � 2 dr dθ. − sec θ c) We need the polar angle of the intersection points. To find it, we solve the two equations r = 23 and r = 1 − cos θ simultanously. Eliminating r, we get 23 = 1 − cos θ, from which θ = 2π/3 and 4π/3. Thus the limits are (no integrand is given): �� dr dθ = R � 4π/3 2π/3 � -2 3/2 -2 1−cos θ dr dθ. 3/2 a R d) The circle has polar equation r = 2a cos θ. The line y = a has polar equation r sin θ = a, or r = a csc θ. Thus the limits are (no integrand): �� dr dθ = R r= sin 2 θ � π/2 π/4 a � a csc θ 2a cos θ a � b) c) d) � � π/2 0 π/4 0 π/2 � � a 0 0 � sin 2θ 0 r dr dθ = r sec θ tan2 θ · r dr dθ = 0 � π/2 0 2d �π/2 1 1 sin 2θ dθ = − cos 2θ = − (−1 − 1) = 1. 2 2 0 �a π 1 π r 2 dr dθ = · ln(1 + r ) = ln(1 + a2 ). 1 + r2 2 2 4 0 1 2 � π/4 tan2 θ sec2 θ dθ = 0 �π/4 1 1 = . tan3 θ 6 6 0 sin θ r √ dr dθ 1 − r2 0 0 �sin θ √ Inner: − 1 − r2 = 1 − cos θ � � 1 2c 2b π/2 r=sin θ 1 x=1 r=sec θ r=a 2a 3B-2 a) a dr dθ. 0 Outer: θ − sin θ 3B-3 a) the hemisphere is the graph of z = � �π/2 0 = π/2 − 1. a2 − x 2 − y 2 = √ a2 − r2 , so we get 4 E. 18.02 EXERCISES �� � R b) � a2 − r2 dA = π/2 0 � � 2π 0 � a 0 � �a 1 1 2 a2 − r2 r dr dθ = 2π · − (a2 − r2 )3/2 = 2π · a3 = πa3 . 3 3 3 0 a (r cos θ)(r sin θ)r dr dθ = 0 � a r3 dr 0 � π/2 sin θ cos θ dθ = 0 a4 1 a4 · = . 4 2 8 c) In order to be able to use the integral formulas at the beginning of 3B, we use symmetry about the y-axis to compute the volume of just the right side, and double the answer. � π/2 � π/2 � 2 sin θ �� � r=2sin θ y 1 r r dr dθ = 2 x2 + y 2 dA = 2 (2 sin θ)3 dθ 3 1 0 0 0 R 32 8 2 · = , by the integral formula at the beginning of 3B. 3 3 9 � π/2 � π/2 � √cos θ 1 1 π π 2 r r dr dθ = 2 d) 2 cos2 θ dθ = 2 · · = . 4 4 4 8 0 0 0 =2· top view x r 2=cos θ 3C. Applications of Double Integration Placing the figure so its legs are on the positive x- and y-axes, � a � a−x �a−x 1 1 �a 1 4 = x2 (a − x); Outer: x3 a − x4 = x2 dy dx Inner: x2 y a) M.I. = a . 3 4 12 0 0 0 0 �� �� �� 1 1 1 4 a 2 2 2 y=a-x a + a 4 = a4 . b) (x + y ) dA = x dA + y 2 dA = 12 12 6 R R R 3C-1 a c) Divide the triangle symmetrically into two smaller triangles, their legs are √ ; 2 a4 1 � a �4 √ = Using the result of part (a), M.I. of R about hypotenuse = 2 · 12 24 2 3C-2 In both cases, x̄ is clear by symmetry; we only need ȳ. � π �� dA = sin x dx = 2 a) Mass is R� � 0 � π � sin x � 1 π 2 π π y dy dx = y-moment is y dA = sin x dx = ; therefore ȳ = . 2 4 8 0 0 0 R �� π b) Mass is Using the formulas at the beginning of 3B, y dA = , by part (a). 4 R �� � π/2 � π � sin x sin3 x 1 2 4 y-moment is y 2 dy dx = 2 y 2 dA = dx = 2 · · = , 3 3 3 9 R 0 0 0 16 4 4 . Therefore ȳ = · = 9 π 9π a a a 2 5 3C-3 Place the segment either horizontally or vertically, so the diameter is respectively on the x or y axis. Find the moment of half the segment and double the answer. (a) (Horizontally, using rectangular coordinates) � b 0 � √ a2 −x2 y dy dx = c � b 0 2 2 2 ans: 2 3 b . 3 b c Note that a − c = b . 1 2 1� 1 x 3 �b = b3 ; (a − x2 − c2 ) dx = b2 x − 2 2 3 0 3 x2+y 2=a 2 a a r=a (b) (Vertically, using polar coordinates). Note that x = c becomes r = c sec θ. � α� a �a = 31 (a3 cos θ − c3 sec2 θ) (r cos θ) r dr dθ Inner: 31 r3 cos θ Moment = c sec θ c sec θ 0 � �α 2 Outer: 31 a3 sin θ − c3 tan θ = 13 (a2 b − c2 b) = 31 b3 ; ans: b3 . 3 0 a b α a c 3C-4 Place the sector so its vertex is at the origin and its axis of symmetry lies along the positive x-axis. By symmetry, the center of mass lies on the x-axis, so we only need find x̄. 2α = a2 α. Since δ = 1, the area and mass of the disc are the same: πa2 · 2π � α� a �a r cos θ · r dr dθ Inner: 32 r3 cos θ ; x-moment: 2 0 Outer: 2 3 3a 0 0 sin θ �α 0 = 32 a3 sin α x̄ = a α 2 3 3a sin α 2 sin α = ·a· . a2 α 3 α 3C-5 By symmetry, we use just the upper half of the loop and double the answer. The upper half lies between θ = 0 and θ = π/4. � π/4 � π/4 � a√cos 2θ 1 4 r2 r dr dθ = 2 a cos2 2θ dθ 2 4 0 0 0 � π/2 a4 a4 1 π πa4 r 2=cos 2θ Putting u = 2θ, the above = cos2 u du = · · = . 2·2 0 4 2 2 16 3D. Changing Variables � � ∂(x, y) ∂(u, v) �� 1 −3 �� 1 = 7; =� = . 2 1� ∂(x, y) ∂(u, v) 7 � � �� 1 7 4u x − 3y dx dy = dv du 7 0 1 v R 2x + y �7 �4 49 ln 4 1 49 ln 4 1 2 ; Ans: = 7 ln 2 Inner: u ln v = u ln 4; Outer: 2 u ln 4 = 2 7 2 1 0 3D-1 u=0 Let u = x − 3y, v = 2x + y; R u=7 v= 1 v= 4 6 E. 18.02 EXERCISES 3D-2 Let u = x + y, v = x − y. Then ∂(u, v) = 2; ∂(x, y) ∂(x, y) 1 = . ∂(u, v) 2 To get the uv-equation of the bottom of the triangular region: y = 0 ⇒ u = x, v = x ⇒ u = v. � � �� � � x−y v 1 2 u cos cos dv du dx dy = x + y 2 u R 0 0 v Inner: u sin u 3D-3 �u = u sin 1 1 2 2u Outer: 0 v =0 u =2 R u =v �2 sin 1 = 2 sin 1 u =0 Ans: sin 1 0 Let u = x, v = 2y; � �1 ∂(x, y) = �� 0 ∂(u, v) � 1 0 �� 1 � = 2 2 Letting R be the elliptical region whose boundary is x2 + 4y 2 = 16 in xy-coordinates, and u2 + v 2 = 16 in uv-coordinates (a circular disc), we have �� �� 1 (16 − u2 − v 2 ) dv du (16 − x2 − 4y 2 ) dy dx = 2 R R = 1 2 � 2π 0 � 4 0 � �4 r2 r4 (16 − r2 ) r dr dθ = π 16 − = 64π. 2 4 0 � � ∂(u, v) �� 1 1 �� = −5; = 3D-4 Let u = x + y, v = 2x − 3y; then ∂(x, y) � 2 −3 � We next express the boundary of the region R in uv-coordinates. For the x-axis, we have y = 0, so u = x, v = 2x, giving v = 2u. For the y-axis, we have x = 0, so u = y, v = −3y, giving v = −3u. ∂(x, y) 1 = . ∂(u, v) 5 v=0 v=2 u v=4 It is best to integrate first over the lines shown, v = c; this means v is held constant, i.e., we are integrating first with respect to u. This gives �� v2 3 u Inner: 15 �v/2 −v/3 R v/2 du dv . 5 −v/3 0 � �4 � � v6 1 1 46 1 1 Outer: = . + + 6 · 15 8 27 0 6 · 15 8 27 (2x − 3y)2 (x + y)2 dx dy = � � v 2 3 1 −1 = v − 15 8 27 4 v=- 3u � � v 2 u2 Let u = xy, v = y/x; in the other direction this gives y 2 = uv, x2 = u/v. � � ∂(u, v) �� ∂(x, y) 1 y x �� 2y = We have =� = 2v; = ; this gives −y/x2 1/x � ∂(x, y) x ∂(u, v) 2v � �� � 3 � 2� u 1 + uv dv du. (x2 + y 2 ) dx dy = v 2v R 0 1 3D-5 Inner: 3D-8 −u u + v 2v 2 �2 1 � � 1 1 1 3u = ; = u − +1+ − 4 2 2 4 Outer: a) y = x2 ; therefore u = x3 , v = x, which gives u = v 3 . 3 2 u 8 �3 0 = 27 . 8 (v=2) y=2x (v=1) y=x R u =0 xy=3 (u=3) 7 b) We get u v + uv = 1, or u = 2 ; (cf. 3D-5) v v +1