How do I compute the area of the surface where $z=x^2+y^2$ and $0\leq z\leq 1$? 
Let me define the surface where $z=x^2+y^2$ and $0\leq z\leq 1$. I want to compute the area of this surface.

My idea was to parametrize the surface with the following parametrization $$\phi(r,\theta)=(r\cos(\theta), r\sin(\theta), r^2)$$ where $\theta\in [0, 2\pi]$ and $r\in (0,1)$. Then I compute the vector $$\frac{\partial \phi}{\partial(r,\theta)}=
  \begin{bmatrix}
    \cos(\theta)&-r\sin(\theta) \\
    \sin(\theta)&r\cos(\theta) \\
    2r &0
  \end{bmatrix}
$$
Now I denote $v=
\begin{bmatrix}
\cos(\theta)\\
\sin(\theta) \\
2r
\end{bmatrix}$, and $w=\begin{bmatrix}
-r\sin(\theta)\\
r\cos(\theta) \\
0
\end{bmatrix}$. Then I compute the Gram matrix $$G(v,w)=
\begin{bmatrix}
\langle v,v\rangle & \langle v,w \rangle \\
\langle w,v \rangle & \langle w,w \rangle
\end{bmatrix}=
\begin{bmatrix}
1+4r^2 & 0 \\
0 & r^2
\end{bmatrix}$$Then the gram determinant is $$g_\phi=\det(G(v,w))=(1+4r^2)r^2$$

Then from the lecture we know that the area of a parametrization $\phi(x)$ is given by $$A(\phi)=\int_Q \sqrt{g_\phi}~~ dx$$ where $Q$ is a square on which $\phi$ is defined.

So I would say in our case $$A(\phi)=\int_0^1 \int_{0}^{2\pi}r\sqrt{1+4r^2} ~~d\phi dr=...=\frac{1}{6}(5\sqrt{5}-1)$$
Now my question is, is this correct or am I wrong, because I'm not sure if my boundaries are correct. It would be helpful if someone could explicitly tell me where I did the mistakes if there are some.
Thanks for your help.
 A: Other than leaving out a factor of $\pi$ from the integration in $\phi$, your answer looks correct. I used a slightly different approach.

Start with the parameterization
$$
(x,y,z)=\left(r\cos(\theta),r\sin(\theta),r^2\right)\tag1
$$
Next, compute
$$
\begin{align}
\frac{\partial(x,y,z)}{\partial r}\times\frac{\partial(x,y,z)}{\partial\theta}
&=(\cos(\theta),\sin(\theta),2r)\times(-r\sin(\theta),r\cos(\theta),0)\tag{2a}\\
&=\left(2r^2\cos(\theta),-2r^2\sin(\theta),r\right)\tag{2b}
\end{align}
$$
The magnitude of $(2)$ is
$$
\left|\left(2r^2\cos(\theta),-2r^2\sin(\theta),r\right)\right|=r\sqrt{4r^2+1}\tag3
$$
Therefore, the surface area is
$$
\begin{align}
\int_0^{2\pi}\int_0^1r\sqrt{4r^2+1}\,\mathrm{d}r\,\mathrm{d}\theta
&=2\pi\int_0^1r\sqrt{4r^2+1}\,\mathrm{d}r\tag{4a}\\
&=\frac\pi4\int_0^4\sqrt{u+1}\,\mathrm{d}u\tag{4b}\\[3pt]
&=\left.\frac\pi6(u+1)^{3/2}\,\right|_0^4\tag{4c}\\[6pt]
&=\frac\pi6\left(5^{3/2}-1\right)\tag{4d}
\end{align}
$$
Explanation:
$\text{(4a):}$ integrate in $\theta$
$\text{(4b):}$ substitute $u=4r^2$
$\text{(4c):}$ integrate
$\text{(4d):}$ evaluate
