下册 8.2 三重积分 第24题

\section*{解题过程：} （1）利用球面坐标变换， $$ F(t)=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\pi} \mathrm{d} \varphi \int_{0}^{1} f\left(r^{2}\right) r^{2} \sin \varphi \mathrm{~d} r=\left.2 \pi(-\cos \varphi)\right|_{0} ^{\pi} \int_{0}^{t} f\left(r^{2}\right) r^{2} \mathrm{~d} r=4 \pi \int_{0}^{t} f\left(r^{2}\right) r^{2} \mathrm{~d} r $$ 已知 $f$ 是可微函数，所以 $F(t)$ 可微，且有 $F^{\prime}(t)=4 \pi f\left(t^{2}\right) t^{2}$ ． $$ \begin{aligned} \lim _{t \rightarrow 0^{+}} t^{-5} F(t) & =\lim _{t \rightarrow 0^{+}} \frac{4 \pi \int_{0}^{t} f\left(r^{2}\right) r^{2} \mathrm{~d} r}{t^{5}}=4 \pi \lim _{t \rightarrow 0^{+}} \frac{f\left(t^{2}\right) t^{2}}{5 t^{4}} \\ & =\frac{4}{5} \pi \lim _{t \rightarrow 0^{+}} \frac{f\left(t^{2}\right)}{t^{2}}=\frac{4}{5} \pi \lim _{t \rightarrow 0^{+}} \frac{f\left(t^{2}\right)-f(0)}{t^{2}-0}=\frac{4}{5} f^{\prime}(0) \pi \end{aligned} $$ （2）作球面坐标变换，则积分区域变成 $0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant \varphi \leqslant \pi, 0 \leqslant r \leqslant t$ ．于是 $$ \begin{aligned} \lim _{t \rightarrow 0^{+}} \frac{1}{t^{4}} \iiint_{V} f\left(\sqrt{x^{2}+y^{2}+z^{2}}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z & =\lim _{t \rightarrow 0^{+}} \frac{1}{t^{4}} \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\pi} \mathrm{d} \varphi \int_{0}^{t} f(r) r^{2} \sin \varphi \mathrm{~d} r \\ & =\pi \lim _{t \rightarrow 0^{+}} \frac{4 \int_{0}^{t} f(r) r^{2} \mathrm{~d} r}{t^{4}}=\pi \lim _{t \rightarrow 0^{+}} \frac{f(t)}{t}=\pi \lim _{t \rightarrow 0^{+}} \frac{f(t)-f(0)}{t-0}=\pi f_{+}^{\prime}(0)=\pi \end{aligned} $$ （3）作球面坐标变换，积分区域变成 $0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant \varphi \leqslant \pi, 0 \leqslant r \leqslant t$ ． $$ \begin{aligned} & \iiint_{V} f\left(x^{2}+y^{2}+z^{2}, \sqrt{x^{2}+y^{2}+z^{2}}, \sin \left(x^{2}+y^{2}+z^{2}\right)\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z \\ & =\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\pi} \sin \varphi \mathrm{d} \varphi \int_{0}^{r} \rho^{2} f\left(\rho^{2}, \rho, \sin \rho\right) \mathrm{d} \rho=4 \pi \int_{0}^{r} \rho^{2} f\left(\rho^{2}, \rho, \sin \rho\right) \mathrm{d} \rho \end{aligned} $$ 所以 $$ \begin{aligned} & \lim _{r \rightarrow 0^{+}} \frac{1}{r^{3}} \iiint_{V} f\left(x^{2}+y^{2}+z^{2}, \sqrt{x^{2}+y^{2}+z^{2}}, \sin \left(x^{2}+y^{2}+z^{2}\right)\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z \\ & =\lim _{r \rightarrow 0^{+}} \frac{4 \pi \int_{0}^{r} \rho^{2} f\left(\rho^{2}, \rho, \sin \rho^{2}\right) \mathrm{d} \rho}{r^{3}}=4 \pi \lim _{r \rightarrow 0^{+}} \frac{r^{2} f\left(r^{2}, r, \sin r^{2}\right)}{3 r^{2}}=\frac{4 \pi}{3} f(0,0,0) . \end{aligned} $$ （4）作柱面坐标变换，则积分区域变成 $0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant r \leqslant t, r \leqslant z \leqslant t$ ．于是 $$ F(t)=\iiint_{\Omega} f\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{1} \mathrm{~d} r \int_{r}^{1} r f\left(r^{2}\right) \mathrm{d} z=2 \pi\left(t \int_{0}^{t} r f\left(r^{2}\right) \mathrm{d} r-\int_{0}^{t} r^{2} f\left(r^{2}\right) \mathrm{d} r\right) . $$ 则 $$ F^{\prime}(t)=2 \pi\left(\int_{0}^{t} r f\left(r^{2}\right) \mathrm{d} r+t \cdot t f\left(t^{2}\right)-t^{2} f\left(t^{2}\right)\right)=2 \pi \int_{0}^{t} r f\left(r^{2}\right) \mathrm{d} r . $$ （5）作柱面坐标变换 $T: x=r \cos \theta, y=r \sin \theta, z=z$ ，则积分区域变成 $$ 0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant r \leqslant t, 0 \leqslant z \leqslant 1 . $$ 于是 $$ F(t)=\iiint_{\Omega}\left(z^{2}+f\left(x^{2}+y^{2}\right)\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{t} r \mathrm{~d} r \int_{0}^{1}\left(z^{2}+f\left(r^{2}\right)\right) \mathrm{d} z=2 \pi \int_{0}^{t} r\left(\frac{1}{3}+f\left(r^{2}\right)\right) \mathrm{d} r . $$ 从而 $\displaystyle F^{\prime}(t)=2 \pi\left(\frac{1}{3}+f\left(t^{2}\right)\right) t$ 。 （6）如图8．183所示，由柱面坐标变换得 $$ F(t)=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{t} r \mathrm{~d} r \int_{0}^{h}\left(z^{2}+f\left(r^{2}\right)\right) \mathrm{d} z=2 \pi \int_{0}^{t}\left(\frac{h^{3}}{3}+h f\left(r^{2}\right)\right) r \mathrm{~d} r . $$ 从而 $$ \lim _{t \rightarrow 0^{+}} \frac{F(t)}{t^{2}}=\lim _{t \rightarrow 0^{+}} \frac{2 \pi t\left(\frac{h^{3}}{3}+h f\left(t^{2}\right)\right)}{2 t}=\pi \lim _{t \rightarrow 0^{+}}\left(\frac{h^{3}}{3}+h f\left(t^{2}\right)\right)=\pi\left(\frac{h^{3}}{3}+h f(0)\right) . $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-225.jpg?height=1272&width=961&top_left_y=6202&top_left_x=4593} \captionsetup{labelformat=empty} \caption{图8．183} \end{figure}