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 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).