ps7sol - 18.02 Fall 2008 Problem Set 7 Part B Solutions 1...

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Unformatted text preview: 18.02 Fall 2008 Problem Set 7, Part B Solutions 1. Average distance = /2 1 Area r dA = 1 a2 2 0 1 2 2r 0 a r r dr d = sin 2 0 1 1 2 r3 a2 3 a 0 = 2 a. 3 sin 2 2. a) Area = 0 /2 0 1 2 (1 r dr d. Inner: = 1 2 sin2 2. 8. Outer: 1 2 0 1 - cos 4) d = 4 - 1 16 sin 4 /2 0 = b) By symmetry the centroid must be on the diagonal line y = x, so calculating x is enough. x= 1 Area 1 3 3r 8 3 8 3 /2 0 0 sin 2 0 sin 2 r cos r dr d. cos = 1 3 1 sin3 2 cos = 3 (2 sin cos )3 cos = 8 3 8 3 Inner: = Outer: sin3 cos4 sin cos4 (1 - cos2 ) = 1 - 5 cos5 + 1 cos7 7 0 /2 sin (cos4 - cos6 ). 8 1 1 3 5-7 = = 16 105 . Therefore x = y = 8 16 128 = . 105 105 3. Rotate so that one vertex is at (1, 0). The other two vertices are at (cos 1 , sin 1 ) and (cos 2 , sin 2 ). The assumption that any point on the circle is as likely as any other and that the choice of 1 is independent of the choice of 2 means that average is the average with respect to d1 d2 . When 1 < 2 the area of the triangle is given by f (1 , 2 ) = 1 2 cos 1 - 1 sin 1 cos 2 - 1 sin 2 = 1 [(cos 1 - 1) sin 2 - sin 1 (cos 2 - 1)] 2 1 = [sin 1 + sin(2 - 1 ) - sin 2 ]. 2 (This formula can also be obtained by splitting the triangle into three smaller triangles having O as a common vertex.) If you integrate f over the whole square 0 1 2, 0 2 2, you get zero, the wrong answer. This is because the determinant is negative for 1 > 2 , and the correct formula for the area is |f (1 , 2 )|. One can deal with this problem in two equivalent ways. The first is to break the integral into two pieces, and integrate over 1 > 2 and 1 < 2 . By symmetry these two integrals will give the same answer. The second way is to average over the half of the parameter space 0 1 2 2 for which the formula for the area is f . By symmetry the average is the same over each half, so it suffices to consider one half. The area of the region 0 1 2 2 is (2)2 /2 = 2 2 , and the average value is 1 f= 2 2 = = 1 4 2 1 4 2 2 0 2 0 2 0 2 1 1 (sin 1 + sin(2 - 1 ) - sin 2 ) d2 d1 2 2 1 (2 - 1 ) sin 1 + - cos(2 - 1 ) + cos 2 (2 - 1 ) sin 1 + 2 - 2 cos 1 d1 = 1 d1 2 1 1 - 4 2 1 sin 1 d1 . 0 2 2 Integrating by parts, (- cos )d = -2, so the average 1 3 3 3/4 1 area is f = + = . (The ratio of maximum to average is = 3/2 2.72.) 2 2 3/2 0 0 sin d = [- cos ]2 - 0 4. Let u = x + 4y - 5, v = 5x + 3y + 8. Then Therefore du dv = 17 dx dy, and so dx dy = (u, v) ux = vx (x, y) uy vy = 1 du dv. 17 1 du dv = . 17 17 1 4 = -17. 5 3 dx dy = (x+4y-5)2 +(5x+3y+8)2 <1 u2 +v 2 <1 (The area of the unit disk is .) 5. u = xy, v = y/x: so uv = y 2 and u/v = x2 , which gives x2 + y 2 = uv + u/v. (u, v) 2y ux uy y x = = 2v. The Jacobian is = = vx vy -y/x2 1/x (x, y) x x 1 2y dx dy, and dx dy = du dv = du dv. Thus du dv = x 2y 2v Limits of integration: 0 < xy < 1, 1 < x < 2. In uv-coordinates, the first inequality becomes 0 < u < 1; and the second one becomes 1 < x2 = u/v < 4, or equivalently v < u < 4v, which means that v < u and v > 1 u. So 4 1 u (x2 + y 2 ) dx dy = R 0 1 u/4 u u/4 1 uv + u v 1 dv du 2v = 0 u u + 2 dv du 2 2v u u/4 = 0 1 u uv - 2 2v du du 1 0 = 0 1 1 2 1 1 2 u - u -2 - 2 2 8 3 2 3 1 3 du = u3 + u u + 8 2 8 2 x0 0 = 0 = 13 1 3 + = . 8 2 8 x0 0 6. a) C1 F dr = C1 -c x dx + y dy = x2 + y 2 -c x dx c = - ln(x2 + 1) 2+1 x 2 c = - ln(x2 + 1). 0 2 C2 b) Since the field is directed radially inward everywhere, F T = 0, and so 2 F dr = 0. (or by calculation: x = a cos t, y = a sin t, 0 -c a2 cos t(- sin t) + a2 sin t cos t dt = 0) a2 1 c) Parametrizing e.g. as x = 1 - t, y = t: 1 C3 F dr = 0 1 0 -c (1 - t)(-dt) + t dt (1 - t)2 + t2 = -c 0 2t - 1 c dt = - ln(2t2 - 2t + 1) 2t2 - 2t + 1 2 c = - (ln 1 - ln 1) = 0. 2 2 ...
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This note was uploaded on 04/28/2009 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

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