Question

The region in the first quadrant bounded by the graph of y=sec(x), x=3.14(pi)/4, and the axes is rotated about the x-axis. What is the volume of the solid generated?

Answer 1

We'll use a method called the "disk" method.
We're basically taking the sum of the volumes of very thin disks over the interval \([0,\frac{\pi}{4}]\).
These disks have a radius of \( f(x) \) and a width of \( dx \). The equation for the volume of a single disk at a given \( x \) is \( \color{red}V = \pi (f(x))^2 dx \).

So, the sum of all of these thin disks is the definite integral on the interval \([0,\frac{\pi}{4}]\):
\[~ \int_{0}^{\frac{\pi}{4}} \color{red}V = \int_{0}^{\frac{\pi}{4}} \pi (f(x))^2 dx \]
Here, \(f(x) = sec(x)\).
\[~ \int_{0}^{\frac{\pi}{4}} \pi sec^2(x) dx = \pi \int_{0}^{\frac{\pi}{4}} sec^2(x) dx \]
You can bring out the \( \pi \) since it is just a constant.