Find the surface area of paraboloid $z=x^2+y^2$, for $0\leq z\leq2$.
I've done the majority of this problem and had my limits of integration to be $0$ to $\sqrt2$ and $0$ to $2\pi$, with my integrand being $\sqrt{1+4(x^2+y^2)}$. My answer ends up being $13\pi/3$. It seems a little small, are my limits of integration wrong?
On
So, we want to find the surface area of the paraboloid $z=x^2+y^2$ that lies between the planes $z=0$ and $z=2$.
Surface area is given by $S=\iint_D\sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2}dA$.
We have: $(\frac{\partial z}{\partial x})^2=4x^2$
$\hspace{17mm}$$(\frac{\partial z}{\partial y})^2=4y^2$
So, $S=\iint_D\sqrt{1+(4x^2+4y^2})dA$.
Now, converting to polar and finding our limits of integration, we know $x^2+y^2=r^2$, and since $x^2+y^2=z$ where $0\leq z \leq 2$,
$0 \leq r \leq \sqrt{2}$
Furthermore, we want to integrate around the entire paraboloid, so $0 \leq \theta \leq 2\pi$.
Therefore, we have: \begin{align} S&=\iint_D\sqrt{1+(4x^2+4y^2})dA\\ &=\int_0^{2\pi}\int_0^{\sqrt{2}}\sqrt{1+4r^2}\hspace{1mm}rdrd\theta\\ &=\int_0^{2\pi}d\theta\hspace{2mm}\cdot\int_0^{\sqrt{2}}r\sqrt{1+4r^2}dr\\ &=2\pi(\frac{13}{6})=\boxed{\frac{13\pi}{3}} \end{align}
you are correct.
$ \newcommand{\R}{\mathbb{R}} $ One of the parametrizations of the surface is $r:S \to \R^3$
$$r(u,v) = (u, v, u^2+v^2)$$
where $S = \{(x,y) \in\R^2|\rho \le \sqrt{2}\}$ and $\rho(x,y) = \sqrt{x^2+y^2}$. The derivatives of the surface are
$$r_u = (1, 0, 2u)\\ r_v = (0, 1, 2v)$$ Hence, the cross product of them is $ r_u \times r_v = -(2u, 2v, 1) $ whose Euclidean norm is $ |r_u \times r_v| = \sqrt{4u^2 + 4v^2 + 1} $.
By definition of surface area on $S$, Wikipedia
\begin{aligned} A &= \int_{S} |r_u \times r_v| du dv\\ &=\int_{0}^{2\pi} \int_{0}^{\sqrt{2}} \rho\sqrt{4\rho^2 + 1} d\rho d\theta\\ &=\dfrac{13}{3}\pi \end{aligned}
Turned out your solution is correct?