coppia relativa alle facce \(A\) e \(A’\) l’elemento di superficie è \(d\sigma =dxdy\), nella coppia \(B\) e \(B’\) è \(d\sigma =dxdz\), nella coppia \(C\) e \(C’\) è \(d\sigma =dydz\), per cui: \[\iint\limits_{S}{\left( x+y+z \right)d\sigma }=\int\limits_{0}^{1}{\left( \int\limits_{0}^{1}{\left( x+y \right)dx} \right)dy+}\int\limits_{0}^{1}{\left( \int\limits_{0}^{1}{\left( x+y+1 \right)dx} \right)dy+}\] \[+\int\limits_{0}^{1}{\left( \int\limits_{0}^{1}{\left( x+z \right)dx} \right)dz+}\int\limits_{0}^{1}{\left( \int\limits_{0}^{1}{\left( x+1+z \right)dx} \right)dz+}\] \[+\int\limits_{0}^{1}{\left( \int\limits_{0}^{1}{\left( y+z \right)dy} \right)dz+}\int\limits_{0}^{1}{\left( \int\limits_{0}^{1}{\left( 1+y+z \right)dy} \right)dz=}\] \[=\int\limits_{0}^{1}{\left[ \frac{1}{2}{{x}^{2}}+xy \right]_{0}^{1}dy+}\int\limits_{0}^{1}{\left[ \frac{1}{2}{{x}^{2}}+x\left( y+1 \right) \right]_{0}^{1}dy+}\] \[+\int\limits_{0}^{1}{\left[ \frac{1}{2}{{x}^{2}}+xz \right]_{0}^{1}dz+}\int\limits_{0}^{1}{\left[ \frac{1}{2}{{x}^{2}}+x\left( z+1 \right) \right]_{0}^{1}dz+}\] \[+\int\limits_{0}^{1}{\left[ \frac{1}{2}{{y}^{2}}+yz \right]_{0}^{1}dz+}\int\limits_{0}^{1}{\left[ \frac{1}{2}{{y}^{2}}+y\left( z+1 \right) \right]_{0}^{1}dz}=\] \[=\int\limits_{0}^{1}{\left( \frac{1}{2}+y \right)dy+}\int\limits_{0}^{1}{\left( \frac{3}{2}+y \right)dy+}2\int\limits_{0}^{1}{\left( \frac{1}{2}+z \right)dz+2}\int\limits_{0}^{1}{\left( \frac{3}{2}+z \right)dz}=\] \[=6\int\limits_{0}^{1}{\left( 1+t \right)dt}=6\left[ t+\frac{1}{2}{{t}^{2}} \right]_{0}^{1}=6\cdot \frac{3}{2}=9\quad .\]
Nel secondo caso, utilizzando coordinate sferiche, si ha: \[\iint\limits_{S}{\left( {{x}^{2}}+{{y}^{2}} \right)d\sigma }={{a}^{4}}\int\limits_{0}^{2\pi }{\left( \int\limits_{0}^{\pi }{{{\sin }^{3}}\varphi }d\varphi \right)}d\vartheta ={{a}^{4}}\int\limits_{0}^{2\pi }{\left( \int\limits_{-1}^{1}{\left( 1-{{t}^{2}} \right)dt} \right)d\vartheta =}\] \[={{a}^{4}}\int\limits_{0}^{2\pi }{\left[ t-\frac{1}{3}{{t}^{3}} \right]_{-1}^{1}d\vartheta }={{a}^{4}}\int\limits_{0}^{2\pi }{\frac{4}{3}d\vartheta =}\frac{8}{3}\pi {{a}^{4}}\quad .\]
Massimo Bergamini
