选项1：

只需颠倒x和y即可切换函数的轴。

x = np.linspace(np.pi/8, np.pi*40/5, n)
y = (x-2)**(1/3)

选项2：

这个过程有点复杂，你也可以通过求原函数的反函数来完成。
f(x) = y = x^3 + 2的倒数是f^{-1}(y) = (y - 2)^(1/3)。
我修改了你提供的代码。

import matplotlib.pyplot as plt
import numpy as np

n = 100

fig = plt.figure(figsize=(14, 7))
ax1 = fig.add_subplot(221)
ax2 = fig.add_subplot(222, projection='3d')
ax3 = fig.add_subplot(223)
ax4 = fig.add_subplot(224, projection='3d')
y = np.linspace(np.pi / 8, np.pi * 40 / 5, n)
x = (y - 2) ** (1 / 3)
t = np.linspace(0, np.pi * 2, n)

xn = np.outer(x, np.cos(t))
yn = np.outer(x, np.sin(t))
zn = np.zeros_like(xn)
for i in range(len(x)):
    zn[i:i + 1, :] = np.full_like(zn[0, :], y[i])

ax1.plot(x, y)
ax1.set_title("$f(x)$")
ax2.plot_surface(xn, yn, zn)
ax2.set_title("$f(x)$: Revolution around $y$")

# find the inverse of the function
x_inverse = y
y_inverse = np.power(x_inverse - 2, 1 / 3)
xn_inverse = np.outer(x_inverse, np.cos(t))
yn_inverse = np.outer(x_inverse, np.sin(t))
zn_inverse = np.zeros_like(xn_inverse)
for i in range(len(x_inverse)):
    zn_inverse[i:i + 1, :] = np.full_like(zn_inverse[0, :], y_inverse[i])

ax3.plot(x_inverse, y_inverse)
ax3.set_title("Inverse of $f(x)$")
ax4.plot_surface(xn_inverse, yn_inverse, zn_inverse)
ax4.set_title("$f(x)$: Revolution around $x$")

plt.tight_layout()
plt.show()