Volume of Revolution $f(x) = x^2$

Question

Suppose you are given $y = f(x)$

I want to use double integrals, instead of the traditional washers.

Suppose even better, $f(x) = x^2$

Find the volume of $f(x) = x^2$, $x = 0$, $x = 4$, $y = 0$ <--- The region spun about the $x-$axis.

The area of $f(x)$ in the region is:

$$A = \int_{0}^{4} x^2 dx$$

How do I take this into a double integral? I dont see any relation with $dy$??

Answer 1

The problem is symmetric around the $x$ axis. \begin{align} V &= \int\limits_0^{2\pi} \int\limits_0^4 \int\limits_0^{x^2} dy \, dx \, d\phi \\ &= 2\pi \int\limits_0^4 x^2 dx \\ &= 2\pi \frac{4^3}{3} \end{align}