I am trying to determine the volume of a solid by cross-sections, and I am having trouble determining the integral I should use to calculate it.
The base is bounded from $x = -8.216$ to $x = -4.537$.
The top graph is $\frac{1}{100}x^2+0.5$, and the bottom graph is $\frac{1}{20}x^2-2.2$.
I am attempting to find the volume when the cross sections are equilateral triangles. How would I create an integral to solve this?
I'm assuming your reference to triangular cross-section means orientated like this:
If you consider one of the triangles the area of it is $\frac{\sqrt{3}}{2}b^2$ where $b$ is the base. In this situations the base is the difference between the two curves: $\frac{1}{100}x^2+\frac{1}{2}-\left(\frac{1}{20}x^2-\frac{11}{5}\right)=\frac{27}{10}-\frac{1}{25}x^2$.
To find the volume you need to integrate all the areas:
$$\int^{-4.537}_{-8.216}\frac{\sqrt{3}}{2}\left(\frac{27}{10}-\frac{1}{25}x^2\right)^2\ dx$$
$$=\frac{\sqrt{3}}{2}\int^{-4.537}_{-8.216}\frac{729}{100}-\frac{27}{125}x^2+\frac{1}{625}x^4\ dx$$
$$=\frac{\sqrt{3}}{2}\left(\frac{729}{100}x-\frac{9}{125}x^3+\frac{1}{3125}x^5\right)_{-8.216}^{-4.537}$$
$$=4.97745\cdots$$