Question

The equilateral triangular cross sections of a solid with base an arch of y = sin(x) on

Answer 1

We can define the base of each equilateral section as $b(x) = \sin(x)$ since this is the distance from the x-axis to the function $y=\sin(x)$.

An equilateral triangle's area can have a function defined solely by the base. Since all sides are equal, the height of the triangle can be found via pythagorean theorem as: $A(b)=\frac{1}{2}b\sqrt{b^2+\left (\frac{b}{2} \right)^2} = \frac{b}{2}\sqrt{\frac{5b^2}{4}}$

But we know that the base $b(x)$ is a function of x. We also can then state that Area $A(b(x))$ is also a function of x. Specifically: $A(x) = \frac{\sin(x)}{2}\sqrt{\frac{5\sin^2(x)}{4}}$.

Then if we integrate the area function A(x) from 0 to pi, we get the total volume:

$\displaystyle \int_0^{\pi} A(x) dx = \int_0^{\pi} \frac{\sin(x)}{2}\sqrt{\frac{5\sin^2(x)}{4}} dx = \int_0^{\pi} \frac{\sqrt{5} \sin^2(x)}{4}dx$