# How do you find the area of the surface generated by rotating the curve about the y-axis y=x^2, 0<=x<=2?

Aug 19, 2017

$\frac{\pi}{6} \left(17 \sqrt{17} - 1\right)$

#### Explanation:

Since we are rotating this solid around the $y$-axis, we are concerned with the $x$ distance from the $y$-axis to the function. This relation is given by $x = \pm \sqrt{y}$. We're only dealing with positive $x$ values, so we can reduce this to just $x = \sqrt{y}$ for our case.

The formula for the surface area of a solid generated by rotating some curve $g \left(y\right)$ around the $y$-axis on $y \in \left[c , d\right]$ is given by

$A = 2 \pi {\int}_{c}^{\mathrm{dg}} \left(y\right) \sqrt{1 + {\left(g ' \left(y\right)\right)}^{2}} \mathrm{dy}$

We go from $x = 0$ to $x = 2$, which is analogous to traveling from $y = 0$ to $y = 4$, which is what we care about.

We will use $g \left(y\right) = \sqrt{y}$. Note that $g ' \left(y\right) = \frac{1}{2 \sqrt{y}}$.

$A = 2 \pi {\int}_{0}^{4} \sqrt{y} \sqrt{1 + {\left(\frac{1}{2 \sqrt{y}}\right)}^{2}} \mathrm{dy}$

$\textcolor{w h i t e}{A} = 2 \pi {\int}_{0}^{4} \sqrt{y \left(1 + \frac{1}{4 y}\right)} \mathrm{dy}$

$\textcolor{w h i t e}{A} = 2 \pi {\int}_{0}^{4} \sqrt{y + \frac{1}{4}} \mathrm{dy}$

Let $u = y + \frac{1}{4}$. This implies that $\mathrm{du} = \mathrm{dy}$. We will also have to change the bounds.

$A = 2 \pi {\int}_{1 / 4}^{17 / 4} {u}^{\frac{1}{2}} \mathrm{du}$

$\textcolor{w h i t e}{A} = 2 \pi {\left[\frac{2}{3} {u}^{\frac{3}{2}}\right]}_{1 / 4}^{17 / 4}$

$\textcolor{w h i t e}{A} = \frac{4 \pi}{3} \left({\left(\frac{17}{4}\right)}^{\frac{3}{2}} - {\left(\frac{1}{4}\right)}^{\frac{3}{2}}\right)$

Note that ${4}^{\frac{3}{2}} = 8$:

$A = \frac{4 \pi}{3} \left(\frac{{17}^{\frac{3}{2}} - 1}{8}\right)$

$\textcolor{w h i t e}{A} = \frac{\pi}{6} \left(17 \sqrt{17} - 1\right)$