Consider a function $ f(x)\equal{}xe^{\minus{}x^3}$ defined on any real numbers.

(1) Examine the variation and convexity of $ f(x)$ to draw the garph of $ f(x)$.

(2) For a positive number $ C$, let $ D_1$ be the region bounded by $ y\equal{}f(x)$, the $ x$-axis and $ x\equal{}C$. Denote $ V_1(C)$ the volume obtained by rotation of $ D_1$ about the $ x$-axis. Find $ \lim_{C\rightarrow \infty} V_1(C)$.

(3) Let $ M$ be the maximum value of $ y\equal{}f(x)$ for $ x\geq 0$. Denote $ D_2$ the region bounded by $ y\equal{}f(x)$, the $ y$-axis and $ y\equal{}M$.
Find the volume $ V_2$ obtained by rotation of $ D_2$ about the $ y$-axis.
1. **Variation and Convexity of \( f(x) = xe^{-x^3} \)**

   To analyze the variation and convexity of \( f(x) \), we need to find the first and second derivatives of \( f(x) \).

   - **First Derivative:**
     \[
     f'(x) = \frac{d}{dx} \left( xe^{-x^3} \right)
     \]
     Using the product rule:
     \[
     f'(x) = e^{-x^3} + x \cdot \frac{d}{dx} \left( e^{-x^3} \right)
     \]
     \[
     f'(x) = e^{-x^3} + x \cdot \left( -3x^2 e^{-x^3} \right)
     \]
     \[
     f'(x) = e^{-x^3} (1 - 3x^3)
     \]

     Setting \( f'(x) = 0 \) to find critical points:
     \[
     e^{-x^3} (1 - 3x^3) = 0
     \]
     Since \( e^{-x^3} \neq 0 \) for all \( x \), we have:
     \[
     1 - 3x^3 = 0 \implies x^3 = \frac{1}{3} \implies x = \sqrt[3]{\frac{1}{3}}
     \]

     - \( f(x) \) is increasing on \( (-\infty, \sqrt[3]{\frac{1}{3}}] \)
     - \( f(x) \) is decreasing on \( [\sqrt[3]{\frac{1}{3}}, +\infty) \)

   - **Second Derivative:**
     \[
     f''(x) = \frac{d}{dx} \left( e^{-x^3} (1 - 3x^3) \right)
     \]
     Using the product rule again:
     \[
     f''(x) = \frac{d}{dx} \left( e^{-x^3} \right) (1 - 3x^3) + e^{-x^3} \frac{d}{dx} (1 - 3x^3)
     \]
     \[
     f''(x) = \left( -3x^2 e^{-x^3} \right) (1 - 3x^3) + e^{-x^3} (-9x^2)
     \]
     \[
     f''(x) = -3x^2 e^{-x^3} (1 - 3x^3) - 9x^2 e^{-x^3}
     \]
     \[
     f''(x) = -3x^2 e^{-x^3} (1 - 3x^3 + 3)
     \]
     \[
     f''(x) = -3x^2 e^{-x^3} (4 - 3x^3)
     \]

     Setting \( f''(x) = 0 \) to find inflection points:
     \[
     -3x^2 e^{-x^3} (4 - 3x^3) = 0
     \]
     Since \( -3x^2 e^{-x^3} \neq 0 \) for \( x \neq 0 \), we have:
     \[
     4 - 3x^3 = 0 \implies x^3 = \frac{4}{3} \implies x = \sqrt[3]{\frac{4}{3}}
     \]

     - \( f(x) \) is concave on \( (-\infty, \sqrt[3]{\frac{4}{3}}] \)
     - \( f(x) \) is convex on \( [\sqrt[3]{\frac{4}{3}}, +\infty) \)

2. **Volume \( V_1(C) \) by Rotation about the \( x \)-axis**

   The volume \( V_1(C) \) is given by:
   \[
   V_1(C) = \pi \int_0^C (f(x))^2 \, dx = \pi \int_0^C x^2 e^{-2x^3} \, dx
   \]

   Let \( u = x^3 \), then \( du = 3x^2 dx \) and \( dx = \frac{du}{3x^2} \):
   \[
   V_1(C) = \pi \int_0^{C^3} \frac{u^{2/3}}{3} e^{-2u} \, du
   \]
   \[
   V_1(C) = \frac{\pi}{3} \int_0^{C^3} u^{2/3} e^{-2u} \, du
   \]

   Using the substitution \( v = 2u \), \( dv = 2du \), and \( du = \frac{dv}{2} \):
   \[
   V_1(C) = \frac{\pi}{6} \int_0^{2C^3} \left( \frac{v}{2} \right)^{2/3} e^{-v} \, dv
   \]
   \[
   V_1(C) = \frac{\pi}{6 \cdot 2^{2/3}} \int_0^{2C^3} v^{2/3} e^{-v} \, dv
   \]

   The integral \( \int_0^{\infty} v^{2/3} e^{-v} \, dv \) is a Gamma function \( \Gamma \left( \frac{5}{3} \right) \):
   \[
   \Gamma \left( \frac{5}{3} \right) = \int_0^{\infty} t^{\frac{5}{3} - 1} e^{-t} \, dt = \int_0^{\infty} t^{2/3} e^{-t} \, dt
   \]

   Therefore:
   \[
   \lim_{C \to \infty} V_1(C) = \frac{\pi}{6 \cdot 2^{2/3}} \Gamma \left( \frac{5}{3} \right)
   \]

   Using the property of the Gamma function \( \Gamma \left( \frac{5}{3} \right) = \frac{4}{3} \Gamma \left( \frac{2}{3} \right) \):
   \[
   \Gamma \left( \frac{2}{3} \right) = \frac{\sqrt{\pi}}{\Gamma \left( \frac{1}{3} \right)}
   \]

   Hence:
   \[
   \lim_{C \to \infty} V_1(C) = \frac{\pi}{6} \left( 1 - e^{-2C^3} \right) = \frac{\pi}{6}
   \]

3. **Volume \( V_2 \) by Rotation about the \( y \)-axis**

   The maximum value \( M \) of \( f(x) \) for \( x \geq 0 \) occurs at \( x = \sqrt[3]{\frac{1}{3}} \):
   \[
   M = f \left( \sqrt[3]{\frac{1}{3}} \right) = \sqrt[3]{\frac{1}{3}} e^{- \left( \sqrt[3]{\frac{1}{3}} \right)^3} = \sqrt[3]{\frac{1}{3}} e^{- \frac{1}{3}}
   \]

   The volume \( V_2 \) is given by:
   \[
   V_2 = \pi \int_0^M \left( f^{-1}(y) \right)^2 \, dy
   \]

   Since \( f(x) \) is invertible on \( [0, \sqrt[3]{\frac{1}{3}}] \):
   \[
   V_2 = \pi \int_0^{\sqrt[3]{\frac{1}{3}}} x^2 e^{-x^3} (1 - 3x^3) \, dx
   \]

   Using integration by parts:
   \[
   V_2 = \pi \left( e^{-\frac{1}{3}} - \frac{2}{3} \right)
   \]

The final answer is \( \boxed{\frac{\pi}{6}} \) for \( V_1(C) \) and \( \boxed{\pi \left( e^{-\frac{1}{3}} - \frac{2}{3} \right)} \) for \( V_2 \).