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Practice Test 4a solutions

# Practice Test 4a solutions - Solutions for Practice Test IV...

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Solutions for Practice Test IV A for Math 1501, Calculus I Problem 1: A two dimensional domain is bounded by the positive x -axis, the positive y axis and the curve x + y = a . (a) Sketch this domain. Solution Solving for y as a function of x , we see y = ( a - x ) 2 . Evidently y ( a ) = 0 and y (0) = a . Also, y ( a/ 4) = a/ 4 so ( a, 0) , (0 , a ) and ( a/ 4 , a/ 4) are all points on the graph. Connecting the dots (or using your graphing calculator), you find: 0 0.2 0.4 0.6 0.8 1 y 0.2 0.4 0.6 0.8 1 x We have used a = 1 to draw the graph in Maple: Just change the length units so a = 1 in the new units. Note that had the equation been x 2 + y 2 = a 2 , we would have a quadrant of a circle. Had it been x + y = a , it would have been a quadrant of a diamond, with the diagonal part of the boundary being the line segment connecting ( a, 0) and (0 , a ). When the power is 1 / 2, the curve “curves in” as opposed to “curving out” as it does when the power is 2. (b) Compute its center of mass. (Assume that the mass density is uniform). Solution First, the region is symmetric about the line y = x . You see this from the fact that x and y enter the equations defining the region in a symmetric way, or from the picture. This means that ¯ x = ¯ y

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so we only have to calculate one of these to find the center of mass. Always think about symmetry before beginning the computation of a centroid. The area A is given by A = Z a 0 ( a - x ) 2 d x = a 2 6 . The numerator in ¯ x is Z a 0 x ( a - x ) 2 d x = a 2 30 . Hence ¯ x = a 5 . Therefore, the coordinates of the centroid are a 5 , a 5 . (c) Rotating this shape about the x = y axis yields a solid. Compute the volume of that solid. Solution This is best done with the shell method. In this case it is a little tricky since the axis of rotation is on a 45 angle instead of being one of the axes. The formula is Vol = 2 π Z r 1 r 0 rh ( r )d r where r is the distance of the shell or “pipe section” from the axis of rotation, and h ( r ) is the length of the shell or “pipe section” at radius r .
The long diagonal line is the axis of rotation. The short parallel segment generates the shell at radius r when rotated about the axis. The perpendicular segment shows the radius. Since the hypotenuse of a 45 triangle is 2 times as long as the shorter sides, we see that if x

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