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Practice Test 4b solutions

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

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Solutions for Practice Test IV B for Math 1501, Calculus I Problem 1: A solid is generated by rotating a square of side length a about its diagonal. Compute its volume. Solution: There are many ways to do this. Here is a diagram of the square. We have chosen to orient the square so that both the x axis and the y axis run along diagonals. (What could be more simple?) We have shown a “slice” running through x in the diagram. If we rotate the figure about the x axis, this segment generates a disk of volume πr 2 ( x )d x where r ( x ) is the radius of the disk at x . Clearly r ( x ) is the y coordinate of the point at the top of the segment generating the disk, and this is y = a 2 - | x | since y = a 2 - x is the equation of the line that bounds the square in the upper right quadrant, and y = a 2 + x is the equation of the line that bounds the square in the upper left quadrant. Note that x ranges over - a 2 x a 2 . Hence the volume is π Z a/ 2 - a/ 2 a 2 - x 2 d x = π 2 6 a 3 .

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Second Solution: This time we use the shell method. If we rotate around the y axis, which is also a diagonal, we generate a “piece of pipe” or a “shell”. The volume of the shell is 2 πxh ( x )d x , where h ( x ) is the length of the “piece of pipe” or“shell”. This is given by h ( x ) = 2 a 2 - x . Now in this approach, x is the radius of rotation; that is, the distance of the shell from the axis of rotation. Distances are always positive, so both limits of integration will be positive . The smallest radius is 0, and the largest radius is a/ 2. Hence in doing the integral, x ranges over 0 x a 2 .
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