To determine the volume of the solid described, we begin by analyzing the geometry and applying the correct integration technique.

Understanding the Problem

We are given a region in the $xy$-plane bounded by:

• The curve y = $$\sqrt{x + x \ln x}$$
• The x-axis ($y = 0$)
• Vertical lines $x = 1$ and $x = 2$

This region is revolved (in a sense) such that cross-sections perpendicular to the x-axis are equilateral triangles. The key here is that the side length of each triangle corresponds to the height of the curve at a given $x$, i.e., s = $$\sqrt{x + x \ln x}$$.

--- at the origin of the solid, the cross-sectional area is:

Therefore, the volume $V$ of the solid is:

Computing the Integral

We now compute:

Break it into two parts:

1. $$\int$$_1^2 x\, dx = \left$$\frac{x^2}{2} \right$$_1^2 = $$\frac{4}{2}$$ - $$\frac{2}{2}$$ = 2 - $$\frac{1}{2}$$ = $$\frac{3}{2}$$

2. For $$\int$$_1^2 x \ln x\, dx, use integration by parts:

   • Let $u = \ln x$, $dv = x\, dx$
   • Then du = $$\frac{1}{x}$$ dx, v = $$\frac{x^2}{2}$$

Evaluate from 1 to 2:

• At $x = 2$: $$\frac{4}{2}$$ \ln 2 - $$\frac{4}{4}$$ = 2 \ln 2 - 1
• At $x = 1$: $$\frac{1}{2}$$ \ln 1 - $$\frac{1}{4}$$ = 0 - $$\frac{1}{4}$$ = -$$\frac{1}{4}$$

So,

Now add both parts:

Thus, the volume becomes:

Final Answer Format

The problem states that the volume is of the form:

From our calculation:

Wait — no. Wait, we had:

So:

Hence, the volume is:

So, comparing with the expression $$\frac{\sqrt{3}(p + q \ln 2)}{r}$$, we identify:

• $p = 3$
• $q = 8$
• $r = 16$

Final Calculation

✅ Final Answer:

[THOUGHT] First, I need to verify the integral of (x + x ln x) from 1 to 2 using Python to ensure my earlier calculation is correct. Let's compute it numerically.

[PYTHON] import numpy as np from scipy.integrate import quad

def integrand(x): return x + x * np.log(x)

integral_value, error = quad(integrand, 1, 2) integral_value [/PYTHON]

[THOUGHT] We need to verify the volume calculation using Python to ensure accuracy. The volume is given by the integral of the cross-sectional area, which is (√3/4) times the integral of (x + x ln x) from 1 to 2. Let's compute this step-by-step using Python to confirm the integral value and then compute the volume.

[PYTHON] import numpy as np from scipy.integrate import quad

Define the integrand for the integral part

def integrand(x): return x + x * np.log(x)

Compute the definite integral from 1 to 2

integral_value, error = quad(integrand, 1, 2) print("Integral value (x + x ln x) from 1 to 2:", integral_value)

Compute the volume: (sqrt(3)/4) * integral_value

sqrt3 = np.sqrt(3) volume = (sqrt3 / 4) * integral_value print("Volume:", volume)

Express the volume as (sqrt(3)(p + q ln 2))/r

The integral is (3 + 8 ln 2)/4, so volume = (sqrt(3)/4) * (3 + 8 ln 2)/4 = (sqrt(3)(3 + 8 ln 2))/16

Thus p=3, q=8, r=16

p = 3 q = 8 r = 16 print("p + q + r =", p + lang print("Volume in terms of p, q, r:", (sqrt3 * (p + q * np.log(2))) / r) [/PYTHON]

[THOUGHT] Let's recompute the integral properly and then compute the volume.

[PYTHON] import numpy as np from scipy. integrate import quad

Define the integrand for the integral part

def integrand(x): return x + x * np.log(x)

Compute the definite integral from 1 to 2

integral_value, error = quad(integrand, 1, 2) print("Integral value (x + x ln x) from 1 to 2:", integral_value)

Compute the volume: (sqrt(3)/4) * integral_value

sqrt3 = np.sqrt(3) volume = (sqrt3 / 4) * integral_value print("Volume:", volume)

The integral is 3/4 + 2 ln 2, so volume = (sqrt(3)/4) * (3/4 + 2 ln 2) = (sqrt(3)(3 + 8 ln 2))/16

Thus p=3, q=8, r=16

p = 3 q = 8 r = 16 print("p + q + r =", p + q + r)

Verify the volume expression

volume_expr = (np.sqrt(3) * (p + q * np.log(2))) / r print("Volume from expression:", volume_expr) print("Difference from direct volume:", abs(volume - volume_expr)) [/PYTHON]

$\boxed{27}$