3. Double Integrals Solutions

3. Double Integrals Solutions - 3. Double Integrals 3A....

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3. Double Integrals 3A. Double integrals in rectangular coordinates 1 2 a) Inner: 6x2y + y2 ] y=-1 = 12xZ; Outer: 4x3] = 32 . 0 b) Inner: -u cos t + it2cos u I I0 = 2u + in2cos u Outer: u2 + in2 sinu = + $n2 = an2. x2 1 = x6 - x3; Outer: +x7 - = + - a = 28 1 d) Inner: dm] o = udm; Outer: $(u2 + 4)3/2]o = $(5& - 8) b) i) The ends of R are at 0 and 2, since 2x - = 0 has 0 and 2 as roots. 2x-x2 dydx JJ, dydx = 1 Jo ii) We solve y = 22 - x2 for x in terms of y: write the equation as x2 - 22 + y = 0 and solve for x by the quadratic formula, getting x = 1 f fi. Note also that the maximum point of the graph is (1,l) (it lies midway between the two roots 0 and 2). We get d) Hint: First you have to find the points where the two curves intersect, by solving simultaneously y2 = x and y = x - 2 (eliminate x). The integral JL dy dx requires two pieces; dx dy only one. JL 1-x/2 3A-3 a) JJ,X~A=$J, xdydx; Inner: x(1- ix) Outer: ix2 -
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2 E. 18.02 EXERCISES 1 Inner: x2 + Y2x I = 1-y2; outer: y - ty310 = 5 :-y2 C) JJR~~A=~~J,:~~~~~~ 1-Y 1 Inner: xy I = y[(l - y) - (y - I)] = 29 - 2y2 Outer: yZ - iy3I0 = ). y-1 3A-4 a) JL sin2 x dA = sin2 x dy dx cos x Inner: y sin2 x] = cos x sin2 x Outer: sin3 x -~/2"/~ = )(I - (- 1)) = 3. 0 I 1 x4 l11 1 Inner: txy2] = t(x3 - x5) - - g) = - . - - - - x 2 2 4 2 12 24' c) The function x2 - y2 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 volume = JJA(x~ - y2)d~ = L1 ~:(XZ - y2) d~ dx.
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3. Double Integrals Solutions - 3. Double Integrals 3A....

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