Problem
Let the region $\mathrm{R}$ be the area enclosed by the function $f(x)=3 x^{2}$ and $g(x)=4 x$. If the region $\mathrm{R}$ is the base of a solid such that each cross section perpendicular to the $x$ axis is an isosceles right triangle with a leg in the region $\mathrm{R}$, find tha volume of the solid. You may use a calculator and round to the nearest thousandth.
Solution
Step 1 ：Find the points of intersection between \(f(x) = 3x^2\) and \(g(x) = 4x\) by setting \(f(x) = g(x)\)
Step 2 ：Integrate the area of the cross-sectional triangles along the x-axis from the left intersection point to the right intersection point using the formula \(A(x) = \frac{1}{2}(-3x^2 + 4x)^2\)
Step 3 ：Evaluate the integral to find the volume of the solid: \(\boxed{0.632}\) cubic units
