2021-09-15T23:02:42-04:00
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.
1. y=ln(x), y=1, y=2, x=0; about the y axis
2. y=e^(-x), y=1, x=2; about y=2
1
2021-09-17T00:17:20-0400
1.
y = ln x = > x = e y y=\ln x=>x=e^y y = ln x => x = e y
V = π ∫ 1 2 ( e y ) 2 d y = π [ e 2 y 2 ] 2 1 V=\pi\displaystyle\int_{1}^{2}(e^y)^2dy=\pi[\dfrac{e^{2y}}{2}]\begin{matrix}
2\\
1
\end{matrix} V = π ∫ 1 2 ( e y ) 2 d y = π [ 2 e 2 y ] 2 1
= π ( e 4 − e 2 ) 2 ( u n i t s 3 ) =\dfrac{\pi(e^4-e^2)}{2}({units}^3) = 2 π ( e 4 − e 2 ) ( u ni t s 3 )
2.
e − x = 1 = > x = 0 e^{-x}=1=>x=0 e − x = 1 => x = 0
V = π ∫ 0 2 [ ( 2 − e − x ) 2 − ( 2 − 1 ) 2 ] d x V=\pi\displaystyle\int_{0}^{2}[(2-e^{-x})^2-(2-1)^2]dx V = π ∫ 0 2 [( 2 − e − x ) 2 − ( 2 − 1 ) 2 ] d x
= π ∫ 0 2 [ 3 − 4 e − x + e − 2 x ] d x =\pi\displaystyle\int_{0}^{2}[3-4e^{-x}+e^{-2x}]dx = π ∫ 0 2 [ 3 − 4 e − x + e − 2 x ] d x
= π [ 3 x + 4 e − x − e − 2 x 2 ] 2 0 =\pi[3x+4e^{-x}-\dfrac{e^{-2x}}{2}]\begin{matrix}
2\\
0
\end{matrix} = π [ 3 x + 4 e − x − 2 e − 2 x ] 2 0
= π ( 6 + 4 e − 2 − e − 4 2 − 0 − 4 + 1 2 ) =\pi(6+4e^{-2}-\dfrac{e^{-4}}{2}-0-4+\dfrac{1}{2}) = π ( 6 + 4 e − 2 − 2 e − 4 − 0 − 4 + 2 1 )
= π ( 5 2 + 4 e − 2 − e − 4 2 ) ( u n i t s 3 ) =\pi(\dfrac{5}{2}+4e^{-2}-\dfrac{e^{-4}}{2})({units}^3) = π ( 2 5 + 4 e − 2 − 2 e − 4 ) ( u ni t s 3 )
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