Problem Analysis #40

[THOUGHT] The volume is V = (√3 / 4) ∫ from 1 to 2 of x (1 + ln x) dx. I will use sympy to compute the exact value of the integral and then the volume to confirm the form $\frac{\sqrt{3}(p + q \ln 2)}{r}$. [/THOUGHT] [PYTHON] from sympy import *

x = symbols('x') integrand = x * (1 + log(x)) integral = integrate(integrand, (x, 1, 2)) print(integral)

volume = (sqrt(3)/4) * integral print(volume) pprint(volume.simplify()) [/PYTHON]

[THOUGHT] The volume is $\frac{\sqrt{3}(3 + 8\ln 2)}{16}$, so $p=3$, $q=8$, $r=16$, and $p+q+r=27$. [/THOUGHT] $\boxed{27}$

Explanation

The reasoning begins by identifying the volume formula for the solid: $V = \frac{\sqrt{3}}{4} \int_1^2 [\text{side length}]^2 \, dx$, where the side length of each equilateral triangular cross-section is the height of the region, given by $y = \sqrt{x + x \ln x} = \sqrt{x(1 + \ln x)}$. Thus, $[\text{side length}]^2 = x(1 + \ln x)$, so the integral is $\int_1^2 x(1 + \ln x) \, dx$.

Sympy computes this integral as $\frac{3}{4} + 2 \ln 2$.

Substitute into the volume formula: $V = \frac{\sqrt{3}}{4} \left( \frac{3}{4} + 2 \ln 2 \right)$.

Rewrite the expression inside the parentheses: $\frac{3}{4} + 2 \ln 2 = \frac{3 + 8 \ln 2}{4}$.

Then, $V = \frac{\sqrt{3}}{4} \cdot \frac{3 + 8 \ln 2}{4} = \frac{\sqrt{3} (3 + 8 \ln 2)}{16}$.

This matches the form $\frac{\sqrt{3}(p + q \ln 2)}{r}$ with $p = 3$, $q = 8$, and $r = 16$.

Finally, compute $p + q + r = 3 + 8 + 16 = 27$.