題目 5. (10 points) Find the volume of the solid of revolution about the xxx-axis bounded by the curve f(x)=1xx2+1f(x)=\frac{1}{x\sqrt{x^2+1}}f(x)=xx2+11 with x∈[1,∞)x \in [1,\infty)x∈[1,∞). 解答 解法一 思路 展開 繞 xxx 軸旋轉，採圓盤法；體積是 π∫1∞f(x)2 dx\pi\int_1^\infty f(x)^2\,\mathrm{d}xπ∫1∞f(x)2dx。 將 1x2(x2+1)\frac1{x^2(x^2+1)}x2(x2+1)1 拆成兩個熟悉的瑕積分。 答題過程 V= π∫1∞1x2(x2+1) dx= π∫1∞(1x2−1x2+1) dx= π[−1x−arctanx]1∞= π(1−π4).\begin{align*} V =&\,\pi\int_1^\infty\frac1{x^2(x^2+1)}\,\mathrm{d}x\\[4mm] =&\,\pi\int_1^\infty\left(\frac1{x^2}-\frac1{x^2+1}\right)\,\mathrm{d}x\\[4mm] =&\,\pi\left[-\frac1x-\arctan x\right]_1^\infty\\[4mm] =&\,\boxed{\pi\left(1-\frac\pi4\right)}. \end{align*}V====π∫1∞x2(x2+1)1dxπ∫1∞(x21−x2+11)dxπ[−x1−arctanx]1∞π(1−4π).
