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.

  • $\begingroup$ This is all correct. That integral is simply annoying, but you've gotten to it properly. $\endgroup$ – Coffee_Table Apr 25 '13 at 3:46
  • $\begingroup$ Let $x=\frac{1}{\sqrt{2}}\tan \theta$. It's still going to be nasty. $\endgroup$ – John Douma Apr 25 '13 at 3:53

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.

  • $\begingroup$ Ok. I was also looking online... Could I also use the formula for $\int \sqrt{a^2 +x^2} dx = \frac x2 \sqrt{a^2+x^2} + \frac{a^2}2 ln|x+\sqrt{a^2+x^2}|$ using $a=\sqrt{2}$ and $x=2x$ to get the final answer??? $\endgroup$ – user68203 Apr 25 '13 at 4:22
  • $\begingroup$ @user68203: I gave you you the way to calculate it, but you can use the result directly. $\endgroup$ – Mhenni Benghorbal Apr 25 '13 at 4:28
  • 2
    $\begingroup$ I've always found it more efficient to calculate these integrals using hyperbolic trig functions. You just need to know that $cosh^2(x) - sinh^2(x) = 1$, $cosh(2x) = 2cosh^2(x) - 1$, and $\frac{d}{dx}sinh(x) = cosh(x)$. $\endgroup$ – Paul Siegel Apr 25 '13 at 5:29
  • $\begingroup$ How did you go from the first line of your integral to the second line with the substitution $x=\frac{1}{\sqrt{2}} tan(t)$. I did the substitution but it brought me to the integral $\sqrt{2} \int_0^1 sec(t) dx$ $\endgroup$ – user68203 Apr 25 '13 at 5:49
  • $\begingroup$ @user68203: Do not forget that $dx=\frac{1}{\sqrt{2}}\sec^2(t) dt $. $\endgroup$ – Mhenni Benghorbal Apr 25 '13 at 6:09

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