Consider the solid whose base in the -plane is the region bounded by the curvethe -axis, and the vertical lines and . The cross-sections of this solid taken by planes perpendicular to the -axis are all equilateral triangles. Find the volume of this solid. [3 points]If the volume is , compute .\
[THOUGHT] The base region is bounded by (y = \sqrt{x + x \ln x}), the x-axis, and (x = 1) to (x = 2). Cross-sections perpendicular to the x-axis are equilateral triangles with side length equal to the y-value of the curve at each x. Thus, the side length (s(x) = \sqrt{x + x \ln x}).
The area of an equilateral triangle with side length (s) is (\frac{\sqrt{3}}{4} s^2). Substituting (s(x)): [ A(x) = \frac{\sqrt{3}}{4} \left( \sqrt{x + x \ln x} \right)^2 = \frac{\sqrt{3}}{4} (x + x \ln x) ]
The volume (V) is the integral of (A(x)) from (x = 1) to (x = 2): [ V = \int_{1}^{2} \frac{\sqrt{3}}{4} (x + x \ln x) dx = \frac{\sqrt{3}}{4} \int_{1}^{2} (x + x \ln x) dx ]
The integral (\int (x + x \ln x) dx) is split: [ \int x dx + \int x \ln x dx ]
The first integral is: [ \int x dx = \frac{1}{2} x^2 ]
The second integral, using integration by parts ((u = \ln x), (dv = x dx), so (du = \frac{1}{x} dx), (v = \frac{1}{2} x^2)): [ \int x \ln x dx = \frac{1}{2} x^2 \ln x - \int \frac{1}{2} x^2 \cdot \frac{1}{x} dx = \frac{1}{2} x^2 \ln x - \frac{1}{2} \int x dx = \frac{1}{2} x^2 \ln x - \frac{1}{4} x^2 ]
Combining: [ \int (x + x \ln x) dx = \frac{1}{2} x^2 + \frac{1}{2} x^2 \ln x - \frac{1}{4} x^2 = \frac{1}{2} x^2 \ln x + \frac{1}{4} x^2 ]
Evaluating from 1 to 2:
- At (x = 2): (\frac{1}{2} (2)^2 \ln 2 + \frac{1}{4} (2)^2 = \frac{1}{2} \cdot 4 \cdot \ln 2 + \frac{1}{4} \cdot 4 = 2 \ln 2 + 1)
- At (x = 1): (\frac{1}{2} (1)^2 \ln 1 + \frac{1}{4} (1)^2 = 0 + \frac{1}{4} = \frac{1}{4})
Definite integral: [ (2 \ln 2 + 1) - \frac{1}{4} = 2 \ln 2 + \frac{3}{4} ]
Thus: [ V = \frac{\sqrt{3}}{4} \left( 2 \ln 2 + \frac{3}{4} \right) = \frac{\sqrt{3}}{4} \cdot \frac{8 \ln 2 + 3}{4} = \frac{\sqrt{3} (3 + 8 \ln 2)}{16} ]
Comparing to the form (\frac{\sqrt{3} (p + q \ln 2)}{r}), we have (p = 3), (q = 8), (r = 16). Therefore: [ p + q + r = 3 + 8 + 16 = 27 ]
\boxed{27}
[THOUGHT] I will use Python to compute the integral numerically for verification of the analytical result. [/THOUGHT]
[PYTHON] import numpy as np
def integrand(x): return x + x * np.log(x)
x_vals = np.linspace(1, 2, 1000) y_vals = integrand(x_vals) integral = np.trapz(y_vals, x_vals) volume = (3**0.5 / 4) * integral print(volume) [/PYTHON]
\boxed{27}