Integration of a rotated triangle: determining the slope of a side I'm working through MIT's OCW single-variable calculus course. The problem is taken from Problem set 7: Question 4B-5.1
Q: Find the volume of an equilateral triangle with side length a which is rotated around one of its sides.
I am actually doing the calculus side okay, I believe. My difficulty is with the trigonometry/geometry setting up the problem. Here is how I set up the integral:
$$
2 \int_0^\frac{a}{2} \pi(\frac{\sqrt{3}}{a}x)^2dx
$$
My integrand is apparently wrong. I've calculated $y$ to be a function of $x$ such that $y = \frac{\sqrt{3}}{a}x$. The solution sets it up the integral exactly the same, except that $y = \sqrt{3}x$. 
Here is how I derived the integrand:
Let the equilateral triangle sit with one side along the x-axis and its leftmost vertex at (0,0) and its rightmost vertex at (a, 0). The strategy will be to split the triangle in half, integrate the rotated solid around the x-axis, and multiply by two to get the whole triangle.
So the function to be integrated is a simple linear f(x) from 0 to $\frac{a}{2}$. Great. But what is $f(x)$? Here's where I go wrong.
I derived the slope of the line of f(x) by observing that the vertex at (0,0) is $\frac{\pi}{3}$ radians. Using sohcahtoa, I calculated the height of the triangle (i.e., $f(\frac{a}{2})$) as:$$\sin(\frac{\pi}{3})$$
To then derive the slope of f, I simply did rise-over-run, i.e. (plugging in for $\sin\frac{\pi}{3})$:
$$\frac{\sqrt{3}}{2}\div\frac{a}{2} = \frac{\sqrt{3}}{a}$$
Which is the slope of the function f(x). I will leave it here, as I don't think the calculus is the problem. Clearly, I've done something wrong in the way I derived the slope of $f(x)$. What am I missing? What would have been the correct way to calculate $f(x)$ to arrive at $y = \sqrt{3}x$?

You were close, but height needs to be $a\sin\left(\frac \pi 3\right)$. Then slope is given by $\dfrac {\frac {a\sqrt 3}2}{\frac a2} = \sqrt 3$.
