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Compute a composite solid by two methods

Region RR is bounded by y=2xy=2x, y=3/2y=3/2, the x-axis and y=(3x2−1)/x2y=\sqrt{(3x^2-1)/x^2}, as shown. Find the volume when RR rotates through 2π2\pi about the x-axis, correct to three decimal places.

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Region R is above the x-axis until x=1/square root 3, then above the curve square root of (3 minus 1/x squared). Its upper boundary is y=2x until x=3/4 and y=3/2 afterwards, ending at x=2/square root 3.
Region R is above the x-axis until x=1/square root 3, then above the curve square root of (3 minus 1/x squared). Its upper boundary is y=2x until x=3/4 and y=3/2 afterwards, ending at x=2/square root 3.
Region R is above the x-axis until x=1/square root 3, then above the curve square root of (3 minus 1/x squared). Its upper boundary is y=2x until x=3/4 and y=3/2 afterwards, ending at x=2/square root 3.

[2007/HCI/I/11(modified)]

CJC · Discussion · Question 8

Break the outer boundary at x=3/4 and subtract the hollow part only where the curved lower boundary exists.

Read the worked solution