kaoyan2advanced 高等数学 第204题
📝 题目
### 第204题
设函数 $f(x)$ 在 $[1,+\infty)$ 上连续，$\displaystyle f(1)=-\frac{1}{2}$ ．若由曲线 $y=f(x)$ ，直线 $x=1, x= t(t>1)$ 与 $x$ 轴所围成的平面图形绕 $x$ 轴旋转一周而成的旋转体体积为 $\displaystyle V(t)=\frac{\pi}{3}\left[t^{2} f(t)-\right. f(1)]$ ，求 $f(x)(x \geqslant 1)$ ．
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💡 答案解析
**答案**：$\displaystyle f(x)=\frac{1}{2x^{2}}$ **解析**： 步骤1：旋转体体积公式 $V(t)=\pi \int_{1}^{t} [f(x)]^{2} \mathrm{d} x$，由已知 $\displaystyle V(t)=\frac{\pi}{3}[t^{2} f(t)-f(1)]$，且 $\displaystyle f(1)=-\frac{1}{2}$，故 $\displaystyle \pi \int_{1}^{t} [f(x)]^{2} \mathrm{d} x = \frac{\pi}{3}[t^{2} f(t)+\frac{1}{2}]$。 步骤2：两边对 $t$ 求导得 $\displaystyle \pi [f(t)]^{2} = \frac{\pi}{3}[2t f(t) + t^{2} f'(t)]$，整理得 $3f^{2} = 2t f + t^{2} f'$，即 $t^{2} f' - 3f^{2} + 2t f = 0$。令 $\displaystyle f = \frac{1}{u}$，代入得 $\displaystyle t^{2} (-\frac{u'}{u^{2}}) - \frac{3}{u^{2}} + \frac{2t}{u} = 0$，乘以 $u^{2}$ 得 $-t^{2} u' - 3 + 2t u = 0$，即 $t^{2} u' = 2t u - 3$，或 $\displaystyle u' - \frac{2}{t} u = -\frac{3}{t^{2}}$。解一阶线性微分方程得 $\displaystyle u = t^{2} ( \int -\frac{3}{t^{4}} \mathrm{d} t + C ) = t^{2} ( \frac{1}{t^{3}} + C ) = \frac{1}{t} + C t^{2}$，故 $\displaystyle f = \frac{1}{u} = \frac{1}{\frac{1}{t} + C t^{2}}$。由 $\displaystyle f(1)=-\frac{1}{2}$ 得 $\displaystyle \frac{1}{1+C} = -\frac{1}{2}$，解得 $C = -3$，故 $\displaystyle f(x) = \frac{1}{\frac{1}{x} - 3x^{2}} = \frac{x}{1-3x^{3}}$。但需验证，代入原方程得 $\displaystyle 3(\frac{x}{1-3x^{3}})^{2} = 2x \cdot \frac{x}{1-3x^{3}} + x^{2} \cdot \frac{1-3x^{3} - x(-9x^{2})}{(1-3x^{3})^{2}} = \frac{2x^{2}}{1-3x^{3}} + x^{2} \cdot \frac{1-3x^{3}+9x^{3}}{(1-3x^{3})^{2}} = \frac{2x^{2}(1-3x^{3}) + x^{2}(1+6x^{3})}{(1-3x^{3})^{2}} = \frac{2x^{2}-6x^{5}+x^{2}+6x^{5}}{(1-3x^{3})^{2}} = \frac{3x^{2}}{(1-3x^{3})^{2}}$，成立。故 $\displaystyle f(x)=\frac{x}{1-3x^{3}}$。 **难度**：★★★☆☆