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**canny****Member**- Registered: 2007-04-14
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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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**Stanley_Marsh****Member**- Registered: 2006-12-13
- Posts: 345

*Last edited by Stanley_Marsh (2007-04-15 06:27:24)*

Numbers are the essence of the Universe

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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*+Δ*y* along the *y*-axis, the element of volume sliced out has base area xΔ*y* 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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**canny****Member**- Registered: 2007-04-14
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thanks

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