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

From Solvely APP
Source: https://solvelyapp.com/problems/8048/

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