# How do you find the surface area of the solid obtained by rotating about the y-axis the region bounded by y=x^2 on the interval 1<=x<=2 ?

Sep 5, 2014

First of all, you are missing a bound. We will assume that the other bound is $y = 0$ or the $x$-axis. The answer is $\frac{15 \pi}{2}$.

The first step is to determine whether you are rotating along an axis that is parallel to the independent axis or the axis of the parameter ($x$ in this case). And we are not, so this integration should be done with cylindrical shells. Always draw a diagram to verify what is the parameter and what is the function.

You should note that $y$ is not always a parameter of $x$. For instance, $x = {y}^{2}$, $x$ is now a parameter of $y$.

The formula for cylindrical shells is:

$V = {\int}_{a}^{b} 2 \pi r h \mathrm{dr}$
$h$ is represented by $y$, we have $y = {x}^{2}$ and $y = 0$
$r$ is represented by $x$
$V = {\int}_{1}^{2} 2 \pi x \left({x}^{2} - 0\right) \mathrm{dx}$

Now that the substitutions are done, we can solve:

$V = 2 \pi {\int}_{1}^{2} {x}^{3} \mathrm{dx}$
$= 2 \pi \frac{{x}^{4}}{4} {|}_{1}^{2}$
$= 2 \pi \frac{\left[{2}^{4} - {1}^{4}\right]}{4}$
$= \frac{15 \pi}{2}$