So I was given the following prompt:
"Let $R$ be the region in the first quadrant bounded by the graph of $y=x^2$, the line $x=4$, and the $x$-axis. $R$ is the base of a solid whose cross sections perpendicular to the $x$-axis are equilateral triangles. What is the volume of the solid?"
I understand how to find the area of cross sections that might be squares or rectangles, but I haven't worked with equilateral triangles before. I also understand that the formula for the area of an equilateral triangle is: $\frac {a^2\sqrt3}{4}$, but I'm a bit lost over how I'd find the area or even the sides of the triangle in this case. Any help would be appreciated!
Best Answer
Please note that the equilateral triangle cross-section is parallel to $YZ$ plane (perpendicular to $x-$axis). The base of the equilateral triangle is $x^2$ as it is between $y = 0$ and $y = x^2$.
So sides of equilateral triangle $ = x^2$ and height $ = \frac{\sqrt3}{2} x^2$
For a given value of $z$, $y$ will vary between $\frac{z}{\sqrt3}$ and $x^2-\frac{z}{\sqrt3}$.
The limits of $y$ come from the fact that at any given height ($z$) in the triangle if we drop perpendicular from two edges to the base, the distance from both vertices of the base to the foot of the perpendicular will be $ = z \cot 60^0= \frac{z}{\sqrt3} \ $.
So the integral to find volume should be
$\displaystyle \int_{0}^{4} \int_{0}^{\frac{\sqrt3}{2} x^2} \int_{\frac{z}{\sqrt3}}^{x^2- \frac{z}{\sqrt3}} dy \ dz \ dx$
If you integrate wrt $dz$ first, you will have to split the integral into two hence the choice of the order.