The region $R$ is bounded by the curve with equation $y = x \ln x$, the lines $x = \text{e}$, $x = \text{e}^2$ and the $x$-axis. Find the exact volume of the solid formed when $R$ is rotated about the $x$-axis by $2\pi$ radians. Give your answer in the form $\frac{\pi \text{e}^3}{27} (a \text{e}^3 + b)$, where $a$ and $b$ are integers to be found.

1. Set up the volume integral representing the solid of revolution.

$V = \pi \int_{\text{e}}^{\text{e}^2} y^2 \,\mathrm{d}x = \pi \int_{\text{e}}^{\text{e}^2} (x \ln x)^2 \,\mathrm{d}x = \pi \int_{\text{e}}^{\text{e}^2} x^2 (\ln x)^2 \,\mathrm{d}x$

2. Apply integration by parts to the integral $\int x^2 (\ln x)^2 \,\mathrm{d}x$.

Let $u = (\ln x)^2$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = x^2$.
Then $\frac{\mathrm{d}u}{\mathrm{d}x} = 2(\ln x)\left(\frac{1}{x}\right)$ and $v = \frac{x^3}{3}$.

$\int x^2 (\ln x)^2 \,\mathrm{d}x = \frac{x^3}{3}(\ln x)^2 - \int \frac{x^3}{3} \cdot \frac{2\ln x}{x} \,\mathrm{d}x$
$= \frac{x^3}{3}(\ln x)^2 - \frac{2}{3} \int x^2 \ln x \,\mathrm{d}x$

3. Apply integration by parts a second time to evaluate $\int x^2 \ln x \,\mathrm{d}x$ and obtain the full indefinite integral.

For $\int x^2 \ln x \,\mathrm{d}x$, let $u_1 = \ln x$ and $\frac{\mathrm{d}v_1}{\mathrm{d}x} = x^2$.
Then $\frac{\mathrm{d}u_1}{\mathrm{d}x} = \frac{1}{x}$ and $v_1 = \frac{x^3}{3}$.

$\int x^2 \ln x \,\mathrm{d}x = \frac{x^3}{3}\ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \,\mathrm{d}x$
$= \frac{x^3}{3}\ln x - \int \frac{x^2}{3} \,\mathrm{d}x = \frac{x^3}{3}\ln x - \frac{x^3}{9}$

Substituting this back into the expression from Step 2:
$\int x^2 (\ln x)^2 \,\mathrm{d}x = \frac{x^3}{3}(\ln x)^2 - \frac{2}{3} \left( \frac{x^3}{3}\ln x - \frac{x^3}{9} \right)$
$= \frac{x^3}{3}(\ln x)^2 - \frac{2x^3}{9}\ln x + \frac{2x^3}{27}$

4. Substitute the limits $x = \text{e}^2$ and $x = \text{e}$ to evaluate the definite integral.

Upper limit ($x = \text{e}^2$):
$\frac{(\text{e}^2)^3}{3}(\ln \text{e}^2)^2 - \frac{2(\text{e}^2)^3}{9}\ln \text{e}^2 + \frac{2(\text{e}^2)^3}{27}$
$= \text{e}^6 \left( \frac{1}{3}(2)^2 - \frac{2}{9}(2) + \frac{2}{27} \right)$
$= \text{e}^6 \left( \frac{36 - 12 + 2}{27} \right) = \frac{26}{27}\text{e}^6$

Lower limit ($x = \text{e}$):
$\frac{\text{e}^3}{3}(\ln \text{e})^2 - \frac{2\text{e}^3}{9}\ln \text{e} + \frac{2\text{e}^3}{27}$
$= \text{e}^3 \left( \frac{1}{3}(1)^2 - \frac{2}{9}(1) + \frac{2}{27} \right)$
$= \text{e}^3 \left( \frac{9 - 6 + 2}{27} \right) = \frac{5}{27}\text{e}^3$

Subtracting the values:
$V = \pi \left( \frac{26}{27}\text{e}^6 - \frac{5}{27}\text{e}^3 \right)$

5. Factorise the volume expression into the required form to identify $a$ and $b$.

$V = \frac{\pi \text{e}^3}{27} (26\text{e}^3 - 5)$

Comparing this with $\frac{\pi \text{e}^3}{27} (a \text{e}^3 + b)$, we deduce:
$a = 26$
$b = -5$