18.02: Practice Exam 3A 1. Let (¯ x, y) ¯ be the center of mass of the triangle, with vertices at (−2, 0), (0, 1), (2, 0) and uniform density � = 1. a) Write an integral formula for ȳ. Do not evaluate the integral(s), but write explicitly the integrand and limits of integration. b) Find x̄. 2. Find the polar moment of inertia of the unit disk with density equal to the distance from the y-axis. 3. Let F = (ax2 y + y 3 + 1)i + (2x3 + bxy 2 + 2)j be a vector ﬁeld, where a and b are constants. a) Find the values of a and b for which F is conservative. b) For these values of a and b, ﬁnd f (x, y) such that F = �f . � c) Still using the values of a and b from part (a), compute F · dr along C the curve C such that x = et cos t, y = et sin t, 0 � t � �. � 4. For F = yx3 i + y 2 j ﬁnd C F · dr on the portion of the curve y = x2 from (0, 0) to (1, 1). 5. Consider the region R in the ﬁrst quadrant bounded by the curves y = x 2 , y = x2 /5, xy = 2, and xy = 4. a) Compute dxdy in terms of dudv if u = x 2 /y and v = xy. b) Find a double integral for the area of R in uv coordinates and evaluate it. 6. a) Let C be a simple closed curve going counterclockwise around a region � R. Let M = M (x, y). Express M dx as a double integral over R. C � b) Find M so that M dx is the mass of R with density �(x, y) = (x + y) 2 . C 7. Consider the region R enclosed by the x-axis, x = 1 and y = x 3 . Travelling in a counterclockwise direction along the boundary C or R, call C1 the portion of C that goes from (0, 0) to (0, 1), C 2 the portion that goes from (1, 0) to (1, 1) and C3 the portion that goes from (1, 1) to (0, 0). a) Find the total work of F = (1 + y 2 )i around the boundary C of R, in a counterclockwise direction. b) Calculate the work of F along C1 and C2 . c) Use parts (a) and (b) to ﬁnd the work along the third side C 3 . 1 2 18.02 Practice Exam 3A Solutions 1. a) Area of triangle is base times height = 2, so ȳ = 1 � 2−2y 1 2 � � �/2 0 y dxdy 2y−2 b) By symmetry x̄ = 0 2. � = |x| = r| cos �|. I0 = � 2� 0 � � � 1 2 r |r cos �|rdrd� = 4 0 � r 2 �rdrd� = D �/2 0 � 1 4 r cos �drd� = 4 0 0 1 4 cos �d� = 5 5 3. a) Nx = 6x2 + by 2 , My = ax2 + 3y 2 . Nx = My provided a = 6 and b = 3. b) fx = 6x2 y + y 3 + 1 =� f = 2x3 y + xy 3 + x + c(y). Therefore, fy = 2x3 +3xy 2 +c� (y). Setting this equal to N , we have 2x 3 +3xy 2 +c� (y) = 2x3 + 3xy 2 + 2 so c� (y) = 2 and c = 2y (+constant). In all, f = 2x3 y + xy 3 + x + 2y c) C starts at (1, 0) and ends at (−e� , 0), so (+constant) � F · dr = f (−e� , 0) − f (1, 0) = C −e� − 1. � � 1 � 1 4. yx3 dx + y 2 dy = x2 x3 dx + (x2 )2 (2xdx) = 3x5 dx = 1/2 C 0 0 � � � � � ux uy � � 2x/y −x2 /y 2 � �=� � = 3x2 /y. Therefore, 5. a) �� vx vy � � y x dudv = (3x2 /y)dxdy = 3u dxdy =� dxdy = 1 dudv 3u � 4 1 1 2 b) dudv = ln 5 dv = ln 5 3 2 1 3u 2 3 � � � 6. a) M dx = −My dA � 4� 5 C R 1 b) We want M such that −My = (x + y)2 . Use M = − (x + y)3 3 �� 7. a) For F, My = 2y and Nx = 0, hence C F · dr = R −2y dA = �1 � 1 � x3 1 6 0 0 −2y dydx = 0 −x dx = − 7 . b) For the work through C1 , we have F · i = 1 + y 2 = 1 + 0 = 1. The length of C1 is 1, so the total work through C1 is 1. The work through C2 is zero because F · j = 0. � � � � 1 8 c) F · dr = F · dr − F · dr − F · dr = − − 1 − 0 = − 7 7 C2 C3 C C1

18.02: Exam 3 Practice Solutions

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