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The prompt is to find the surface area of the surface formed by $z=x^2+y^2$ inside $x^2 + y^2 = 1$

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Here's what I did, $$f(x, y) = x^2 + y^2$$ $$\nabla f(x, y) = <2x, 2y>$$ $$Surface Area = \iint_D \sqrt{(2x)^2 + (2y)^2 + 1} dA$$ But I'm not sure how to find the domain of the region and integral limits

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    $\begingroup$ I suggest the use of cylindrical coordinates. $\endgroup$ – Emilio Novati Jun 26 '17 at 17:46
  • $\begingroup$ so do you mean $z=1$ always? $\endgroup$ – Guy Fsone Jun 26 '17 at 17:47
  • $\begingroup$ @GuyFabrice yes $\endgroup$ – Archetype2142 Jun 26 '17 at 17:51
  • $\begingroup$ I think you mean $z \leq 1$ $\endgroup$ – user456218 Jun 26 '17 at 17:52
  • $\begingroup$ I think "inside $x^2 + y^2 = 1$” means bounded by the unit circle. In other words, $(x,y)$ range over the unit disk. $\endgroup$ – Matthew Leingang Jun 26 '17 at 17:54
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You do not have to find anything, the domain $D$ is given: it is the unit disk centered at the origin of the $xy$ plane. By using polar coordinates

$$ \iint_{x^2+y^2\leq 1}\sqrt{4(x^2+y^2)+1}\,dx\,dy = \int_{0}^{2\pi}\int_{0}^{1}\rho\sqrt{4\rho^2+1}\,d\rho\,d\theta=\left[\frac{\pi}{6}(1+4\rho^2)^{3/2}\right]_{0}^{1} $$ it follows that the wanted surface area is $\color{red}{\frac{\pi}{6}\left(5\sqrt{5}-1\right)}$.

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While this question is tailor-made for cylindrical co-ordinates, I will nonetheless provide the limits since you asked. Notice that at $x^2+ y^2 =1$, you will stop. Therefore wlog you may take $x$ from $-1$ to $1$ as the outer limit. And $y$ from $-\sqrt{1-x^2}$ to $+\sqrt{1-x^2}$. By symmetry of the functions we get

$$\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \sqrt{(2x)^2 + (2y)^2 + 1}\ dydx = 4 \times \int_{0}^1 \int_{0}^{\sqrt{1-x^2}} \sqrt{(2x)^2 + (2y)^2 + 1}\ dydx $$

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  • $\begingroup$ @JackD'Aurizio, thank you for correcting me. I made a very stupid mistake. Corrected it. Thanks again $\endgroup$ – user456218 Jun 26 '17 at 18:04
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    $\begingroup$ No mistake, just a typo: it wasn't clear if you wanted to exploit symmetry or not, just that. $\endgroup$ – Jack D'Aurizio Jun 26 '17 at 18:07

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