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I have to calculate the area of {$x^2+y^2+z^2\le1$}$\cap${$\frac{1}{2}\le z \le1$}.
So i try with the follow integral:

$\int_0^{2\pi}\int_0^{\sqrt{\frac{3}{4}}}\frac{2\rho}{\sqrt{4-\rho^2}}d\rho d\theta$

I got $z$ in function of $x$ and $y$
$z=\sqrt{1-x^2-y^2}$
Then i used the formula:
$\int_a^b\int_c^d \sqrt{1+[\frac{d}{dx}f(x,y)]^2+[\frac{d}{dy}f(x,y)]^2}$
by obtainig the first integral replacing with poolar coordinates, and considering $\frac{1}{2}\le z \le1$ I got $0\le\rho\le\sqrt{\frac{3}{4}}$. But there is an error, that i can't undertand. where is?

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    If you mean the surface area of the solid, then you've forgotten the bottom of the surface, $z=1/2$. – 2017-01-13
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    @B. Goddard You are right – 2017-01-13
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    @B. Goddard and, what i can do? – 2017-01-13
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    The missing piece is a disk. The area is pretty easy to calculate. – 2017-01-13
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    @B. Goddard It is an intersection, i don't understand. – 2017-01-13
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    The two planes $z=1/2$ and $z=1$ cut a cap off the sphere. The surface of the cap has two parts. On top, it's a piece of the sphere, and you worked out that area. The bottom is a disk. You need to add the area of that disk to the area of the piece of the sphere. – 2017-01-13

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Using spherical coordinates, $$ x = \rho \cos \theta \sin \varphi \\ y = \rho \sin \theta \sin \varphi \\ z = \rho \cos \varphi\mbox{,} $$ with $\rho \in (0 , \infty)$, $\theta \in (0 , 2 \pi)$ and $\varphi \in (0 , \pi)$, you can obtain, by the inequalities in $$ A = \left\{(x , y , z) \in {\mathbb{R}}^3 : x^2 + y^2 + z^2 \leq 1, \frac{1}{2} \leq z \leq 1\right\}\mbox{,} $$ that $\rho \in \left[\frac{1}{2 \cos \varphi} , 1\right]$ and $\varphi \in \left(0 , \frac{\pi}{3}\right)$, so $$ V(A) = \int_{\theta = 0}^{2 \pi} \left(\int_{\varphi = 0}^{\frac{\pi}{3}} \left(\int_{\rho = \frac{1}{2 \cos \varphi}}^1 {\rho}^2 \sin \varphi d \rho\right) d \varphi\right) d \theta = 2 \pi \int_{\varphi = 0}^{\frac{\pi}{3}} \left(\int_{\rho = \frac{1}{2 \cos \varphi}}^1 {\rho}^2 \sin \varphi d \rho\right) d \varphi = $$ $$ = 2 \pi \int_0^{\frac{\pi}{3}} \frac{\sin \varphi}{3} \left(1 - \frac{1}{8 {\cos}^3 \varphi}\right) d \varphi = \frac{2 \pi}{3} \int_0^{\frac{\pi}{3}} \left(\sin \varphi - \frac{\sin \varphi}{8 {\cos}^3 \varphi}\right) d \varphi = \frac{5 \pi}{24}\mbox{.} $$

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    I know the result, that is $\pi$, I'm missing the method to get it. – 2017-01-14
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    The shape of your set tells you that you must use spherical coordinates, not polar, maybe you could obtain the result, that is $\frac{5 \pi}{24}$, with polar coordinates but I can't help you if you decide solve the problem with this method – 2017-01-14