题目
Problem
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Figure 1 shows a sketch of the curve with equation
y = 3 x 3 x 2 + 5 , x ≥ 0. \begin{align*}
y=&\,\sqrt{\frac{3x}{3x^2+5}},\\[4mm]
x\ge&\,0.
\end{align*} y = x ≥ 3 x 2 + 5 3 x , 0. The finite region R R R x x x x = 5 x=\sqrt{5} x = 5 x = 5 x=5 x = 5
The region R R R 360 ∘ 360^\circ 36 0 ∘ x x x
Use integration to find the exact volume of the solid generated. Give your answer in the form a ln b a\ln b a ln b a a a b b b
(5)
题目中文翻译
本题中你必须写出解题过程的所有步骤。
完全依赖计算器技术的解法不被接受。
图 1 给出了曲线的示意图，其方程为
y = 3 x 3 x 2 + 5 , x ≥ 0. \begin{align*}
y=&\,\sqrt{\frac{3x}{3x^2+5}},\\[4mm]
x\ge&\,0.
\end{align*} y = x ≥ 3 x 2 + 5 3 x , 0. 图 1 中阴影所示的有限区域 R R R x x x x = 5 x=\sqrt{5} x = 5 x = 5 x=5 x = 5
区域 R R R x x x 360 ∘ 360^\circ 36 0 ∘
用积分求所生成立体的精确体积。将答案写成 a ln b a\ln b a ln b a a a b b b
(5)
解答
解法一
思路
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区域绕 x x x
V = π ∫ y 2 d x . V=\pi\int y^2\,\mathrm{d}x. V = π ∫ y 2 d x . 代入 y 2 y^2 y 2 ln 4 \ln4 ln 4 2 ln 2 2\ln2 2 ln 2
答题过程
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Using the volume of revolution about the x x x
V = π ∫ 5 5 y 2 d x = π ∫ 5 5 3 x 3 x 2 + 5 d x . \begin{align*}
V=&\,\pi\int_{\sqrt5}^{5}y^2\,\mathrm{d}x\\[4mm]
=&\,\pi\int_{\sqrt5}^{5}
\frac{3x}{3x^2+5}\,\mathrm{d}x.
\end{align*} V = = π ∫ 5 5 y 2 d x π ∫ 5 5 3 x 2 + 5 3 x d x . Since
d d x ( 3 x 2 + 5 ) = 6 x , \frac{\mathrm{d}}{\mathrm{d}x}(3x^2+5)=6x, d x d ( 3 x 2 + 5 ) = 6 x , an antiderivative of 3 x 3 x 2 + 5 \frac{3x}{3x^2+5} 3 x 2 + 5 3 x
F ( x ) = 1 2 ln ( 3 x 2 + 5 ) . F(x)=\frac12\ln(3x^2+5). F ( x ) = 2 1 ln ( 3 x 2 + 5 ) . At the endpoints,
F ( 5 ) = 1 2 ln 80 , F(5)=\frac12\ln80, F ( 5 ) = 2 1 ln 80 , F ( 5 ) = 1 2 ln 20. F(\sqrt5)=\frac12\ln20. F ( 5 ) = 2 1 ln 20. Therefore,
V = π [ F ( 5 ) − F ( 5 ) ] = π 2 ( ln 80 − ln 20 ) = π 2 ln 4 = π ln 2. \begin{align*}
V=&\,\pi\big[F(5)-F(\sqrt5)\big]\\[4mm]
=&\,\frac{\pi}{2}(\ln80-\ln20)\\[4mm]
=&\,\frac{\pi}{2}\ln4\\[4mm]
=&\,\pi\ln2.
\end{align*} V = = = = π [ F ( 5 ) − F ( 5 ) ] 2 π ( ln 80 − ln 20 ) 2 π ln 4 π ln 2. Hence the exact volume is
V = π ln 2 , \boxed{V=\pi\ln2}, V = π ln 2 , where a = π a=\pi a = π b = 2 b=2 b = 2
解法二
思路
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明确设 u = 3 x 2 + 5 u=3x^2+5 u = 3 x 2 + 5 1 u \frac1u u 1 u = 3 x 2 u=3x^2 u = 3 x 2
答题过程
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The required volume is
V = π ∫ 5 5 3 x 3 x 2 + 5 d x . V=\pi\int_{\sqrt5}^{5}
\frac{3x}{3x^2+5}\,\mathrm{d}x. V = π ∫ 5 5 3 x 2 + 5 3 x d x . Let u = 3 x 2 + 5 u=3x^2+5 u = 3 x 2 + 5
d u = 6 x d x , \mathrm{d}u=6x\,\mathrm{d}x, d u = 6 x d x , so
3 x d x = 1 2 d u . 3x\,\mathrm{d}x=\frac12\,\mathrm{d}u. 3 x d x = 2 1 d u . When x = 5 x=\sqrt5 x = 5 u = 20 u=20 u = 20 x = 5 x=5 x = 5 u = 80 u=80 u = 80
V = π 2 ∫ 20 80 1 u d u = π 2 [ ln u ] 20 80 = π 2 ( ln 80 − ln 20 ) = π 2 ln 4 = π ln 2 . \begin{align*}
V=&\,\frac{\pi}{2}\int_{20}^{80}
\frac1u\,\mathrm{d}u\\[4mm]
=&\,\frac{\pi}{2}[\ln u]_{20}^{80}\\[4mm]
=&\,\frac{\pi}{2}(\ln80-\ln20)\\[4mm]
=&\,\frac{\pi}{2}\ln4\\[4mm]
=&\,\boxed{\pi\ln2}.
\end{align*} V = = = = = 2 π ∫ 20 80 u 1 d u 2 π [ ln u ] 20 80 2 π ( ln 80 − ln 20 ) 2 π ln 4 π ln 2 . Here a = π a=\pi a = π b = 2 b=2 b = 2
若改设 u = 3 x 2 u=3x^2 u = 3 x 2 d u = 6 x d x \mathrm{d}u=6x\,\mathrm{d}x d u = 6 x d x 15 15 15 75 75 75
V = π 2 ∫ 15 75 1 u + 5 d u = π 2 [ ln ( u + 5 ) ] 15 75 = π 2 ( ln 80 − ln 20 ) = π ln 2. \begin{align*}
V=&\,\frac{\pi}{2}\int_{15}^{75}
\frac1{u+5}\,\mathrm{d}u\\[4mm]
=&\,\frac{\pi}{2}[\ln(u+5)]_{15}^{75}\\[4mm]
=&\,\frac{\pi}{2}(\ln80-\ln20)\\[4mm]
=&\,\pi\ln2.
\end{align*} V = = = = 2 π ∫ 15 75 u + 5 1 d u 2 π [ ln ( u + 5 ) ] 15 75 2 π ( ln 80 − ln 20 ) π ln 2. 这两种换元方式只相差常数平移。
