I am studying for my final and my prof gave us review questions but with no answers so I am lost with this question. If anyone can help I would really appreciate it.
Question: Find the area of the portion of the surface $z=x^2+y$, that lies over the region $0\le x \le 1$, and $0\le y\le 1$.
I know this is a surface integral of the form $$A(S) = \int\int_s dS$$
I computed dS, $dS = \sqrt{1+{\frac{\partial z}{\partial x}}^2 + {\frac{\partial z}{\partial y}}^2} = \sqrt{4x^2 +2}$
Then I integrate this function over the region... $\int_0^1 \int_0^1 \sqrt{4x^2+2} dydx$ but then this integration seems strange to me. If anyone can tell me if I am doing this right so far and perhaps what is the next step I would appreciate it.
Simplifying, using the change of variables $ x = \frac{1}{\sqrt{2}} \tan(t) $ and the identity $1+\tan^2(x)=\sec^2(x)$, the integral falls apart
$$\int_0^1 \sqrt{4x^2+2} dx = 2\int_0^1 \sqrt{x^2+\frac{1}{2}} dx$$
$$ = \int_{0}^{\tan^{-1}\sqrt{2}} \sec^3(y)dy = \frac{\sqrt {2}\sqrt {3}}{2}+\frac{1}{4}\,\ln \left( 5+2\,\sqrt {6} \right). $$
For techniques of evaluating the last integral see here.
Added: It is easier to evaluate the integral using the suggestion by Paul. We use the substitution $x= \frac{1}{\sqrt{2}}\sinh(t)$ and the identity $1+\sinh(t)^2=\cosh^2(t)$
$$ 2\int_0^1 \sqrt{x^2+\frac{1}{2}} dx = \int_{0}^{\ln(\sqrt{2}+\sqrt{3})} \cosh^2(t)dt =\dots\,. $$
You can use the identity
$$ \cosh(x) = \frac{e^{x}+e^{-x}}{2} $$
to evaluate the above integral.