dbl_intgrls_sol

dbl_intgrls_sol - MIT OpenCourseWare http:/ocw.mit.edu...

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MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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3. Double Integrals 3A. Double integrals in rectangular coordinates I' 2 a) Inner: 6x2y + y2 = 12x2; Outer: 4x3] = 32 y=-1 0 b) Inner: -u cos t + it2 cos u I I0 = 2u + i~~ cos u Outer: u2 + !p2 sinu = (i~)~ + $T~ = ;r2. x2 1 = x6 - 1 - x3; Outer: +xi 1 = 7 - i - 28 1 d) Inner: dm] o = udm; Outer: k(u2 + 4)3/2] o = f (5& - 8) b) i) The ends of R are at 0 and 2, since 22 - = 0 has 0 and 2 as roots. 2x-x2 ii) We solve y = 22 - x2 for x in terms of y: write the equation as - 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). dy dx requires two pieces; 1-x/2 3A-3 a) JJ,.~A=JI(~JI( xdydx; 2 In: (1 - x) Outer: $x2 - $x3IO = $ - 9 = $.
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2 E. 18.02 EXERCISES I 1 Inner: x2 + y2x = 1-y2; outer: y - :-y2 ty310 = $. C) SJ,Y~A = JolJ,'Jyydxdy I 1 -Y 1 Inner: xy = y[(l- y) - (y - 1)] = 29 - 2y2 Outer: y2 - $Y3] = y-1 i. 3A-4 a) sin2 x dA = sin2 x dy dx cos Z Inner: y sin2 x] = cos x sin2 x Outer: t sin3 x I -r/2"I2 = ) (1 2 - (- 1)) = I. 0 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.
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dbl_intgrls_sol - MIT OpenCourseWare http:/ocw.mit.edu...

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