Problem Analysis #40

[THOUGHT] To find the volume of the solid, we need to integrate the area of the cross-sections along the x-axis from x = 1 to x = 2. The cross-sections are equilateral triangles with their bases lying on the curve ( y = \sqrt{x + x \ln x} ).

1. Determine the base length of the triangle at each x: The base of the triangle at any point ( x ) is ( 2y = 2\sqrt{x + x \ln x} ).

2. Find the area of the equilateral triangle: The formula for the area ( A ) of an equilateral triangle with side length ( s ) is ( A = \frac{\sqrt{3}}{4} s^2 ). Here, the side length ( s ) is the same as the base length, ( s = 2\sqrt{x + x \ln x} ). Therefore, the area ( A(x) ) is: [ A(x) = \frac{\sqrt{3}}{4} \left( 2\sqrt{x + x \ln x} \right)^2 = \frac{\sqrt{3}}{4} \cdot 4(x + x \ln x) = \sqrt{3}(x + x \ln x) ]

3. Set up the integral for the volume: The volume ( V ) of the solid is given by integrating the area of the cross-sections from ( x = 1 ) to ( x = 2 ): [ V = \int_{1}^{2} \sqrt{3}(x + x \ln x) , dx ] Factor out the constant ( \sqrt{3} ): [ V = \sqrt{3} \int_{1}^{2} (x + x \ln x) , dx ]

4. Evaluate the integral: We split the integral into two parts: [ \int_{1}^{2} (x + x \ln x) , dx = \int_{1}^{2} x , dx + \int_{1}^{2} x \ln x , dx ]

   • The first integral is straightforward: [ \int_{1}^{2} x , dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{2^2}{2} - \frac{1^2}{2} = 2 - \frac{1}{2} = \frac{3}{2} ]

   • For the second integral, we use integration by parts. Let ( u = \ln x ) and ( dv = x , dx ). Then, ( du = \frac{1}{x} , dx ) and ( v = \frac{x^2}{2} ).

     Using integration by parts ( \int u , dv = uv - \int v , du ): [ \int x \ln x , dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} , dx = \frac{x^2}{2} \ln x - \int \frac{x}{2} , dx ] [ = \frac{x^2}{2} \ln x - \frac{1}{2} \int x , dx = \frac{x^2}{2} \ln x - \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{2} \ln x - \frac{x^2}{4} ]

     Evaluate this from ( x = 1 ) to ( x = 2 ): [ \left[ \frac{x^2}{2} \ln x - \frac{x^2}{4} \right]_{1}^{2} = \left( \frac{2^2}{2} \ln 2 - \frac{2^2}{4} \right) - \left( \frac{1^2}{2} \ln 1 - \frac{1^2}{4} \right) ] [ = \left( 2 \ln 2 - 1 \right) - \left( 0 - \frac{1}{4} \right) = 2 \ln 2 - 1 + \frac{1}{4} = 2 \ln 2 - \frac{3}{4} ]

   Combine the results: [ \int_{1}^{2} (x + x \ln x) , dx = \frac{3}{2} + 2 \ln 2 - \frac{3}{4} = \frac{6}{4} + 2 \ln 2 - \frac{3}{4} = \frac{3}{4} + 2 \ln 2 ]

5. Multiply by (\sqrt{3}) to get the volume: [ V = \sqrt{3} \left( \frac{3}{4} + 2 \ln 2 \right) = \frac{\sqrt{3}(3 + 8 \ln 2)}{4} ]

   Therefore, ( p = 3 ), ( q = 8 ), and ( r = 4 ).

6. Compute ( p + q + r ): [ p + q + r = 3 + 8 + 4 = 15 ]

The final answer is: [ \boxed{15} ]