Find surface area and centroid of a parabolic surface in $\mathbb{R}^{3}$ I have come to a problem in a multivariable calculus book that I am having problems with.
The problem can be stated :
Let $a \in \mathbb{R}$ with $a > 0$. Find the surface area of the part of a paraboloid given
by $z^{2} + x^{2} = 2ay$ that is cut out by a plane given by $y =a$. Then find the centroid
of the surface.
My attempt at computing the surface area is given below :
Let the surface be denoted $S$. We see :
\begin{equation}
y = a \text{ and } z^{2} + x^{2} = 2ay \Rightarrow z^{2} + x^{2} = 2a^{2}
\end{equation}
So intersection of $S$ with $y =a$ plane is a circle of radius $\sqrt{2}a$.
Let :
\begin{equation}
\require{cancel}
E = \left\{ (r,\theta) \in \mathbb{R}^{2} \; : \; r \in [0,\sqrt{2}a] \text{ and } \theta \in [-\pi,\pi] \right\}
\end{equation}
Now let's define a map from $E$ to $S$.
\begin{equation}
\Phi : \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}
\end{equation}
Let :
\begin{align}
x & = r\cos(\theta)\\
z & = r\sin(\theta)
\end{align}
So :
\begin{align}
z^{2} + x^{2} = 2ay & \Rightarrow y = \frac{z^{2} + x^{2}}{2a} \\
   & \Rightarrow y = \frac{r^{2}\sin^{2}(\theta) + r^{2}\cos^{2}(\theta)}{2a}\\
   & \Rightarrow y = \frac{r^{2}\cancel{(\sin^{2}(\theta) + \cos^{2}(\theta))}}{2a}\\
   & \Rightarrow y = \frac{r^{2}}{2a} 
\end{align}
So :
\begin{equation}
\Phi(r,\theta) = \left( r\cos(\theta) , \frac{r^{2}}{2a} , r \sin(\theta) \right)
\end{equation}
Where :
\begin{align}
x(r,\theta)  & = r\cos(\theta)\\
y(r,\theta)  & = \frac{r^{2}}{2a} \\
z(r,\theta) & = r\sin(\theta)
\end{align}
We see :
\begin{align}
x_{r} & = \cos(\theta)\\
y_{r} & = \frac{\cancel{2}r}{\cancel{2}a} = \frac{r}{a} \\
z_{r} & = \sin(\theta)
\end{align}
and :
\begin{align}
x_{\theta} & = -r \sin(\theta)\\
y_{\theta} & = 0\\
z_{\theta} & = r\cos(\theta)
\end{align}
Let :
\begin{align}
U  & = x_{r}^{2} + y_{r}^{2} + z_{r}^{2} \\
 & = \cos^{2}(\theta) + \frac{r^{2}}{a^{2}} + \sin^{2}(\theta)\\
 & = \frac{r^{2}}{a^{2}} + \cancel{ \left( \sin^{2}(\theta) + \cos^{2}(\theta) \right)}\\
 & = \frac{r^{2}}{a^{2}} + 1
\end{align}
and :
\begin{align}
V  & = x_{\theta}^{2} + y_{\theta}^{2} + z_{\theta}^{2} \\
 & = r^{2}\sin^{2}(\theta) + 0 + r^{2}\cos^{2}(\theta)\\
 & = r^{2} \cancel{\left( \sin^{2}(\theta) + \cos^{2}(\theta) \right)}\\
 & = r^{2} 
\end{align}
and :
\begin{align}
W  & = x_{r}x_{\theta} + y_{r}y_{\theta} + z_{r}z_{\theta}\\
 & = \cos(\theta)(-r\sin(\theta)) + 0 + \sin(\theta)r\cos(\theta)\\
 & = r\sin(\theta)\cos(\theta) - r\sin(\theta)\cos(\theta)\\
 & = 0
\end{align}
so :
\begin{equation}
\sqrt{UV - W^{2}} = \sqrt{ r^{2}\left( \frac{r^{2}}{a^{2}} + 1 \right) } = r\sqrt{ \frac{r^{2}}{a^{2}} + 1 }
\end{equation}
We know :
\begin{align}
\text{Area}(S)  & = \int\int_{E} \sqrt{UV - W^{2}} d(r,\theta)\\
  & = \int\int_{E} r\left( \frac{r^{2}}{a^{2}} + 1 \right)^{\frac{1}{2}} d(r,\theta)\\
  & = \int_{-\pi}^{\pi} \int_{0}^{\sqrt{2}a} r\left( \frac{r^{2}}{a^{2}} + 1 \right)^{\frac{1}{2}} dr d\theta
\end{align}
Let :
\begin{align}
u   & = \frac{r^{2}}{a^{2}} + 1\\
\frac{du}{dr}  & = \frac{2r}{a^{2}}
\end{align}
So :
\begin{align}
\text{Area}(S)  & = \int_{-\pi}^{\pi} \int_{0}^{\sqrt{2}a} \left( \frac{a^{2}}{2} \right) \frac{du}{dr} u^{\frac{1}{2}} dr d\theta\\
  & = \frac{a^{2}}{2} \int_{-\pi}^{\pi} \left( \int_{0}^{\sqrt{2}a} u^{\frac{1}{2}} \frac{du}{dr} dr \right) d\theta\\
  & = \frac{a^{2}}{2} \int_{-\pi}^{\pi} \left( \int_{u(0)}^{u(\sqrt{2}a)} u^{\frac{1}{2}} du \right) d\theta\\
  & = \frac{a^{2}}{2} \int_{-\pi}^{\pi} \left( \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \Bigr|_{u(0)}^{u(\sqrt{2}a)} \right) d\theta\\
  & = \frac{\cancel{2}a^{2}}{\cancel{6}} \int_{-\pi}^{\pi} \left( u^{\frac{3}{2}} \Bigr|_{1}^{\frac{2\cancel{a^{2}}}{\cancel{a^{2}}} + 1} \right) d\theta\\
  & = \frac{a^{2}}{3} \int_{-\pi}^{\pi} \left( u^{\frac{3}{2}} \Bigr|_{1}^{3} \right) d\theta\\
  & = \frac{a^{2}}{3} \int_{-\pi}^{\pi} \left( 3^{\frac{3}{2}} - 1 \right) d\theta\\
  & = \frac{a^{2}}{3} \int_{-\pi}^{\pi} (3 \sqrt{3} - 1 ) d\theta\\
  & = \frac{a^{2}(3\sqrt{3} - 1)}{3} \int_{-\pi}^{\pi} d\theta\\
  & = \frac{2\pi a^{2} (3\sqrt{3} - 1)}{3} \; \checkmark 
\end{align}
Now my attempt at finding the centroid is given below :
We see that $S$ is symmetric w.r.t. the x and z coordinates. So :
\begin{align}
\overline{x} & = 0\\
\overline{z} & = 0
\end{align}
Let :
\begin{equation}
A = \text{Area}(S) = \frac{2\pi a^{2} (3 \sqrt{3} - 1) }{ 3 }
\end{equation}
We know :
\begin{equation}
\overline{y} = \frac{1}{A} \int\int_{E} y(r,\theta) \sqrt{UV - W^{2}} d(r,\theta)
\end{equation}
We know :
\begin{equation}
y(r,\theta) = \frac{r^{2}}{2a} 
\end{equation}
and :
\begin{equation}
\sqrt{UV - W^{2}} = r\sqrt{ \frac{r^{2}}{a^{2}} + 1 }
\end{equation}
So :
\begin{align}
\overline{y}  & = \frac{1}{A} \int\int_{E} \frac{r^{2}}{2a} r \sqrt{ \frac{r^{2}}{a^{2}} + 1 } d(r,\theta)\\
  & = \frac{1}{2aA} \int\int_{E} r^{3} \left( \frac{r^{2}}{a^{2}} + 1 \right)^{\frac{1}{2}} d(r,\theta)
\end{align}
I'm not sure how to compute this integral.  I suspect that maybe I did something wrong when I was finding the surface area or made a mistake in writing the expression for $\overline{y}$, but I don't see where the mistake is made. If what I have above is actually correct, then what is the best approach to evaluate the integral ?

The surface area $A$ you found is correct.
The integral with respect to $\theta$ contributes a factor of $2\pi$. The remaining integral is simple. Start by integrating by parts:
$$\int x^3\sqrt{x^2+n^2}\,\mathrm dx = uv - \int v\,\mathrm du$$
Take $u=x^2$ and $\mathrm dv = x\left(x^2+n^2\right)^{\frac12}\,\mathrm dx$, so that $\mathrm du=2x\,\mathrm dx$ and $v=\frac13\left(x^2+n^2\right)^{\frac32}$.
$$\implies \int x^3\sqrt{x^2+n^2}\,\mathrm dx=\frac13x^2\left(x^2+n^2\right)^{\frac32}-\frac23\int x\left(x^2+n^2\right)^{\frac32}\,\mathrm dx$$
Then make a substitution, etc.
