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Let R be the region bounded by y=lnx, the x-axis, and the line x=e.
(a) Find the volume of the solid generated when R is rotated about the x-axis.
(b) Find the volume of the solid formed with R as base and such that cross sections perpendicular to the y-axis are squares.
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Last edited by Stanley_Marsh (2007-04-15 06:27:24)
Numbers are the essence of the Universe
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The formula for volume of solid of revolution is
For for region R from x = 1 to x = e, its
So integrate [ln(x)][sup]2[/sup]. Hint: Integrate [ln(x)][ln(x)] by parts.
For the second part, shift the curve by e units to the left, so R is now y = ln(x+e) from x = 1−e to x = 0. (The new graph cuts the y-axis at (0,1).) But you now need to integrate along the y-axis, so you must express x in terms of y: x = e[sup]y[/sup] − e.
For an element of increase from y to y+Δy along the y-axis, the element of volume sliced out has base area xΔy cross-sectional area x[sup]2[/sup]; ∴ the element of volume is ΔV = x[sup]2[/sup]Δy.
Last edited by JaneFairfax (2007-04-17 08:48:20)
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thanks
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