題目
4.
( 10 % ) (10\%) ( 10% ) Let
Ω \Omega Ω be a region (on the
x y xy x y -plane) enclosed by
x = ln y x=\sqrt{\ln y} x = ln y ,
x = 0 x=0 x = 0 and
y = e y=e y = e . Let
S S S be the solid obtained by revolving
Ω \Omega Ω about the
y y y -axis. Let
V V V be the volume of
S S S .
V = ( 6 ) ‾ V=\underline{\quad(6)\quad} V = ( 6 ) .
解答
解法一
思路
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由 x = 0 x=0 x = 0 x = ln y x=\sqrt{\ln y} x = ln y y = 1 y=1 y = 1
繞 y y y y y y ln y \sqrt{\ln y} ln y
答題過程
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曲線 x = ln y x=\sqrt{\ln y} x = ln y x = 0 x=0 x = 0 ln y = 0 \ln y=0 ln y = 0 y = 1 y=1 y = 1 1 ≤ y ≤ e 1\le y\le e 1 ≤ y ≤ e
r ( y ) = ln y . r(y)=\sqrt{\ln y}. r ( y ) = ln y . 因此體積為
V = π ∫ 1 e r ( y ) 2 d y = π ∫ 1 e ln y d y = π [ y ln y − y ] 1 e = π . \begin{align*}
V
=&\,\pi\int_1^e r(y)^2\,\mathrm{d}y\\[4mm]
=&\,\pi\int_1^e\ln y\,\mathrm{d}y\\[4mm]
=&\,\pi\big[y\ln y-y\big]_1^e\\[4mm]
=&\,\pi.
\end{align*} V = = = = π ∫ 1 e r ( y ) 2 d y π ∫ 1 e ln y d y π [ y ln y − y ] 1 e π . 故 V = π \boxed{V=\pi} V = π
