원기둥·구면좌표 삼중적분

In spherical coordinates, $x=\rho\sin\varphi\cos\theta$, $y=\rho\sin\varphi\sin\theta$, $z=\rho\cos\varphi$ so that

$$\begin{aligned} &\frac{\rho^2\sin^2\varphi\cos\theta\sin\theta +\rho^3\sin\varphi\sin\theta\,\cos^2\varphi +\rho^3\sin\varphi\,\cos^2\varphi} {\rho^2\sin^2\varphi+\rho^4\cos^4\varphi} \\ & =\frac{\sin^2\varphi\cos\theta\sin\theta +\rho\sin\varphi\sin\theta\,\cos^2\varphi +\rho\sin\varphi\cos\theta\,\cos^2\varphi} {\sin^2\varphi+\rho^2\cos^4\varphi} \end{aligned}$$

As $(x,y,z)\to (0,0,0)$, the radius $\rho\to 0$ and the second and third terms in the numerator and the second term in the denominator converge to $0$. But that leaves

$$\frac{\sin^2\varphi\cos\theta\sin\theta}{\sin^2\varphi} =\cos\theta\sin\theta$$

which takes many different values. In particular, if we send $(x,y,z)\to (0,0,0)$ along either the $x$– or $y$–axis, that is with $z=0$ and either $x=0$ or $y=0$, then

$$\frac{xy+yz^2+xz^2}{x^2+y^2+z^4}\bigg|_{\genfrac{}{}{0pt}{}{x=0\text{ or }y=0}{z=0}}=0$$

converges to $0$. But, if we send $(x,y,z)\to (0,0,0)$ along the line $y=x$, $z=0$

$$\frac{xy+yz^2+xz^2}{x^2+y^2+z^4}\bigg|_{\genfrac{}{}{0pt}{}{y=x}{z=0}}=\frac{x^2}{2x^2} =\frac{1}{2}$$

converges to $1/2$. So $\frac{xy+yz^2+xz^2}{x^2+y^2+z^4}$ does not approach a single value as $(x,y,z)\to(0,0,0)$ and the limit does not exist.