Se tiene que \( M=\left \{ (x,y,z)\in \mathbb{R}^3:z=\sqrt{a^2-x^2-y^2};\ x^2+y^2\leq a^2 \right \} \) es una superficie compacta orientable con borde, entonces consideremos una carta \( h^{-1}:V\rightarrow M \) que cubre toda la variedad \( M \), definida por: \( h^{-1}(\theta,\phi)=a(cos(\theta) sen(\phi),sen(\theta) sen(\phi),cos(\phi)) \); \( a\in \mathbb{R};\ \theta \in [0,2\pi];\ \phi \in [0,\pi /2] \)
Usando los coeficientes de la métrica Riemanniana, se obtiene:
\begin{align*}
E(\theta,\phi)=g_{11}(\theta,\phi)=\left \langle \dfrac{\partial h^{-1}}{\partial \theta},\dfrac{\partial h^{-1}}{\partial \theta}  \right \rangle (\theta,\phi) &= \left \Vert \dfrac{\partial h^{-1}(\theta,\phi)}{\partial \theta} \right \Vert^2 \\
 &= \Vert a(-sen(\theta)sen(\phi),cos(\theta)sen(\phi),0)\Vert^2 \\
 &= a^2(sen^2(\theta)sen^2(\phi)+cos^2(\theta)sen^2(\phi))=a^2sen^2(\phi)\\
G(\theta,\phi)=g_{22}(\theta,\phi)=\left \langle \dfrac{\partial h^{-1}}{\partial \phi},\dfrac{\partial h^{-1}}{\partial \phi}  \right \rangle (\theta,\phi) &= \left \Vert \dfrac{\partial h^{-1}(\theta,\phi)}{\partial \phi} \right \Vert^2 \\
 &= \Vert a(cos(\theta)cos(\phi),sen(\theta)cos(\phi),-sen(\phi))\Vert^2 \\
 &= a^2(cos^2(\theta)cos^2(\phi)+sen^2(\theta)cos^2(\phi)+sen^2(\phi))=a^2\\
 F(\theta,\phi)=g_{12}(\theta,\phi) = \left \langle \dfrac{\partial h^{-1}}{\partial \theta},\dfrac{\partial h^{-1}}{\partial \phi}  \right \rangle (\theta,\phi) &= a^2(-sen(\theta)sen(\phi)cos(\theta)cos(\phi)+cos(\theta)sen(\phi)sen(\theta)cos(\phi))=0\\
\end{align*} \( \Rightarrow dM=\sqrt{g}\ d\theta \wedge d\phi=\sqrt{EG-F^2}d\theta \wedge d\phi=\sqrt{a^4sen^2(\phi)}d\theta \wedge d\phi = a^2sen(\phi)d\theta \wedge d\phi \)
Sea \( f:M\rightarrow \mathbb{R} \) una función defininda por \( f(x,y,z)=x+y+z \)
\( \Rightarrow \displaystyle \int_M fdM = \int_Mf(h^{-1})\sqrt{g}\ d\theta d\phi = \int_{0}^{\pi/2} \int_{0}^{2\pi}(acos(\theta)sen(\phi)+asen(\theta)sen(\phi)+acos(\phi))\ a^2sen(\phi)\ d\theta d\phi \)
\( = a^3 \displaystyle \int_{0}^{\pi/2}\int_{0}^{2\pi}(cos(\theta)sen^2(\phi)+sen(\theta)sen^2(\phi)+cos(\phi)sen(\phi))\ d\theta d\phi = a^3 \displaystyle \int_{0}^{\pi/2}((sen(\theta)-cos(\theta))sen^2(\phi)+\theta cos(\phi)sen(\phi))\mid _{\theta=0}^{\theta=2\pi}\ d\phi  \)
\( = 2\pi a^3 \displaystyle \int_{0}^{\pi/2}cos(\phi)sen(\phi)\ d\phi = \displaystyle 2\pi a^3 \int_{0}^{1}u\ du = 2\pi a^3 \left(\dfrac{u^2}{2} \right)_{u=0}^{u=1}=\pi a^3 \Rightarrow \displaystyle \int_M (x+y+z)\ dM=\pi a^3 \)

Sea \( S=\lbrace (x,y,z)\in \mathbb{R}^3:\sqrt{x^2+y^2}\leq z\leq 1 \rbrace \), entonces la frontera \( M=\partial S \) es descrita por:
\( \partial S=\left \lbrace (x,y,z)\in \mathbb{R}^3:x^2+y^2=z^2;\ 0\leq z\leq 1 \right \rbrace \cup \left \lbrace (x,y,z)\in \mathbb{R}^3: x^2+y^2\leq 1,\ z=1 \right \rbrace \), por lo tanto, sean \( S_1=\left \lbrace (x,y,z)\in \mathbb{R}^3:x^2+y^2=z^2;\ 0\leq z\leq 1 \right \rbrace \) y \( S_2=\left \lbrace (x,y,z)\in \mathbb{R}^3: x^2+y^2\leq 1,\ z=1 \right \rbrace \), además consideremos dos cartas \( h_1^{-1}:V\rightarrow S_1 \) y \( h_2^{-2}:V\rightarrow S_2 \) definidas por
\( h_1^{-1}(r,\theta)=(rcos(\theta),rsen(\theta),r);\ r\in [0,1];\ \theta \in [0,2\pi] \)
\( h_2^{-1}(r,\theta)=(rcos(\theta),rsen(\theta),1);\ r\in [0,1];\ \theta \in [0,2\pi] \)

Sea \( f:M\rightarrow \mathbb{R} \) una función definida por \( f(x,y,z)=x^2+y^2 \).

Para el conjunto \( S_1 \) se tiene que:
\( E(r,\theta)=g_{11}(r,\theta)=\left \langle \dfrac{\partial h_1^{-1}}{\partial r},\dfrac{\partial h_1^{-1}}{\partial r} \right \rangle (r,\theta)=\Vert (cos(\theta),sen(\theta),1)\Vert^2=cos^2(\theta)+sen^2(\theta)+1=2 \)
\( G(r,\theta)=g_{22}(r,\theta)=\left \langle \dfrac{\partial h_1^{-1}}{\partial \theta},\dfrac{\partial h_1^{-1}}{\partial \theta} \right \rangle (r,\theta)=\Vert (-rsen(\theta),rcos(\theta),0)\Vert^2=r^2sen^2(\theta)+r^2cos^2(\theta)=r^2 \)
\( F(r,\theta)=g_{12}(r,\theta)=\left \langle \dfrac{\partial h_1^{-1}}{\partial r},\dfrac{\partial h_1^{-1}}{\partial \theta} \right \rangle (r,\theta)=\langle (cos(\theta),sen(\theta),1)\ , \ (-rsen(\theta),rcos(\theta),0) \rangle=0 \)
\( \Rightarrow dS_1=\sqrt{g}\ dr\wedge d\theta=\sqrt{EG-F^2}\ dr\wedge d\theta=\sqrt{2r^2}\ dr\wedge d\theta=r\sqrt{2}\ dr\wedge d\theta \)
\( \Rightarrow \displaystyle \int_{S_1}fdS_1=\int_{S_1}f(h_1^{-1})\sqrt{g}\ drd\theta=\int_0^{2\pi}\int_{0}^1(r^2cos^2(\theta)+r^2sen^2(\theta))r\sqrt{2}\ drd\theta=\sqrt{2}\int_0^{2\pi}\int_0^1r^3\ drd\theta \)
\( =\displaystyle 2\sqrt{2}\pi \int_0^1r^3\ dr=2\sqrt{2}\pi\left(\dfrac{r^4}{4} \right)_{r=0}^{r=1}=2\sqrt{2}\pi \left(\dfrac{1}{4} \right)=\dfrac{\pi \sqrt{2}}{2} \)

Para el conjunto \( S_2 \) se tiene que:
\( E(r,\theta)=g_{11}(r,\theta)=\left \langle \dfrac{\partial h_2^{-1}}{\partial r},\dfrac{\partial h_2^{-1}}{\partial r} \right \rangle (r,\theta)=\Vert (cos(\theta),sen(\theta),0)\Vert^2=cos^2(\theta)+sen^2(\theta)=1 \)
\( G(r,\theta)=g_{22}(r,\theta)=\left \langle \dfrac{\partial h_2^{-1}}{\partial \theta},\dfrac{\partial h_2^{-1}}{\partial \theta} \right \rangle (r,\theta)=\Vert (-rsen(\theta),rcos(\theta),0)\Vert^2=r^2sen^2(\theta)+r^2cos^2(\theta)=r^2 \)
\( F(r,\theta)=g_{12}(r,\theta)=\left \langle \dfrac{\partial h_2^{-1}}{\partial r},\dfrac{\partial h_2^{-1}}{\partial \theta} \right \rangle (r,\theta)=\langle (cos(\theta),sen(\theta),0)\ , \ (-rsen(\theta),rcos(\theta),0) \rangle=0 \)
\( \Rightarrow dS_2=\sqrt{g}\ dr\wedge d\theta=\sqrt{EG-F^2}\ dr\wedge d\theta=r\ dr\wedge d\theta \)
\( \Rightarrow \displaystyle \int_{S_2}fdS_2=\int_{S_2}f(h_2^{-1})\sqrt{g}\ drd\theta=\int_0^{2\pi}\int_{0}^1(r^2cos^2(\theta)+r^2sen^2(\theta))r\ drd\theta=\int_0^{2\pi}\int_0^1r^3\ drd\theta \)
\( =2\pi \left(\dfrac{r^4}{4}\right)_{r=0}^{r=1}=2\pi \left(\dfrac{1}{4}\right)=\dfrac{\pi}{2} \)
Por lo tanto \( \displaystyle \int_MfdM=\int_{S_1}fdS_1+\int_{S_2}fdS_2=\dfrac{\pi \sqrt{2}}{2}+\dfrac{\pi}{2}=\left(\dfrac{\sqrt{2}+1}{2}\right)\pi \Rightarrow \int_M(x^2+y^2)dM=\left(\dfrac{\sqrt{2}+1}{2}\right)\pi \)