Integrali di superficie

Esercizio 10.1

Si calcoli l'area della superficie $\phi :[0,\pi ]\times [0,1]\to \mathbb {R} ^{3}$$\phi :[0,\pi ]\times [0,1]\to \mathbb {R} ^{3}$, $\phi (u,v)=(\cos u,\sin u,v)$$\phi (u,v)=(\cos u,\sin u,v)$.

 

Chiamando $K$$K$ il dominio parametrico della superficie, l'area di $\phi$$\phi$ è definita come:

$$A=\int _{K}\vert {\frac {\partial \phi }{\partial u}}\times {\frac {\partial \phi }{\partial v}}\vert \,du\,dv$$

$$A=\int _{K}\vert {\frac {\partial \phi }{\partial u}}\times {\frac {\partial \phi }{\partial v}}\vert \,du\,dv$$

$${\frac {\partial \phi }{\partial u}}=(-\sin u,\cos u,0)$$

$${\frac {\partial \phi }{\partial u}}=(-\sin u,\cos u,0)$$

$${\frac {\partial \phi }{\partial v}}=(0,0,1)$$

$${\frac {\partial \phi }{\partial v}}=(0,0,1)$$

$$\phi _{u}\times \phi _{v}=(\cos u,\sin u,0)$$

$$\phi _{u}\times \phi _{v}=(\cos u,\sin u,0)$$

$$\vert \phi _{u}\times \phi _{v}\vert ={\sqrt {\cos ^{2}u+\sin ^{2}u}}=1$$

$$\vert \phi _{u}\times \phi _{v}\vert ={\sqrt {\cos ^{2}u+\sin ^{2}u}}=1$$

$$A=\int _{0}^{\pi }[\int _{0}^{1}1\,dv]\,du$$

$$A=\int _{0}^{\pi }[\int _{0}^{1}1\,dv]\,du$$

$$A=\int _{0}^{\pi }1\,du=\pi$$

$$A=\int _{0}^{\pi }1\,du=\pi$$

Esercizio 10.2

Calcolare l'integrale di $f={\frac {1}{z^{2}}}$$f={\frac {1}{z^{2}}}$ sulla superficie

$$\Sigma =\{(x,y,z)\in \mathbb {R} ^{3}\,t.c.\,z={\frac {1}{\sqrt {x^{2}+y^{2}}}},\,1\leq z\leq 2\}.$$

$$\Sigma =\{(x,y,z)\in \mathbb {R} ^{3}\,t.c.\,z={\frac {1}{\sqrt {x^{2}+y^{2}}}},\,1\leq z\leq 2\}.$$

 

Parametrizzo la superficie come:

$$\Sigma =\{x=u,\,y=v,\,z={\frac {1}{\sqrt {u^{2}+v^{2}}}}\}$$

$$\Sigma =\{x=u,\,y=v,\,z={\frac {1}{\sqrt {u^{2}+v^{2}}}}\}$$

Il dominio di $\Sigma$$\Sigma$ è dato da:

$$E_{=}\{(u,v)\,t.c.\,1\leq {\frac {1}{\sqrt {u^{2}+v^{2}}}}\leq 2\}$$

$$E_{=}\{(u,v)\,t.c.\,1\leq {\frac {1}{\sqrt {u^{2}+v^{2}}}}\leq 2\}$$

Deve quindi essere

$$1/2\leq {\sqrt {u^{2}+v^{2}}}^{1/2}\leq 1$$

$$1/2\leq {\sqrt {u^{2}+v^{2}}}^{1/2}\leq 1$$

$$1/4\leq u^{2}+v^{2}\leq 1$$

$$1/4\leq u^{2}+v^{2}\leq 1$$

questa è una superficie di rotazione, nella quale $z=f(x,y)$$z=f(x,y)$.

$$\Sigma _{u}={\frac {\partial \Sigma }{\partial u}}=(1,0,-u*(u^{2}+v^{2})^{-3/2})$$

$$\Sigma _{u}={\frac {\partial \Sigma }{\partial u}}=(1,0,-u*(u^{2}+v^{2})^{-3/2})$$

$$\Sigma _{v}={\frac {\partial \Sigma }{\partial v}}=(0,1,-v(u^{2}+v^{2})^{-3/2})$$

$$\Sigma _{v}={\frac {\partial \Sigma }{\partial v}}=(0,1,-v(u^{2}+v^{2})^{-3/2})$$

$$\Sigma _{u}\times \Sigma _{v}=[u*(u^{2}+v^{2})^{-3/2},(u^{2}+v^{2})^{-3/2}v,1]$$

$$\Sigma _{u}\times \Sigma _{v}=[u*(u^{2}+v^{2})^{-3/2},(u^{2}+v^{2})^{-3/2}v,1]$$

$$\vert \Sigma _{u}\times \Sigma _{v}\vert ={\sqrt {(u^{2}+v^{2})^{-3}*u^{2}+(u^{2}+v^{2})^{-3}*v^{2}+1}}$$

$$\vert \Sigma _{u}\times \Sigma _{v}\vert ={\sqrt {(u^{2}+v^{2})^{-3}*u^{2}+(u^{2}+v^{2})^{-3}*v^{2}+1}}$$

$$={\sqrt {{\frac {u^{2}+v^{2}}{(u^{2}+v^{2})^{3}}}+1}}$$

$$={\sqrt {{\frac {u^{2}+v^{2}}{(u^{2}+v^{2})^{3}}}+1}}$$

$$={\sqrt {1+{\frac {1}{(u^{2}+v^{2})^{2}}}}}$$

$$={\sqrt {1+{\frac {1}{(u^{2}+v^{2})^{2}}}}}$$

$$f={\frac {1}{z^{2}}}\,\longrightarrow \,f=(u^{2}+v^{2})^{2}$$

$$f={\frac {1}{z^{2}}}\,\longrightarrow \,f=(u^{2}+v^{2})^{2}$$

Applicando la definizione, l'integrale di $f$$f$ sulla superficie è:

$$\int _{E}(u^{2}+v^{2})^{2}{\sqrt {1+{\frac {1}{(u^{2}+v^{2})^{2}}}}}\,du\,dv$$

$$\int _{E}(u^{2}+v^{2})^{2}{\sqrt {1+{\frac {1}{(u^{2}+v^{2})^{2}}}}}\,du\,dv$$

$$E=\{(u,v)\in \mathbb {R} ^{2}\,t.c.1/4\leq u^{2}+v^{2}\leq 1\}$$

$$E=\{(u,v)\in \mathbb {R} ^{2}\,t.c.1/4\leq u^{2}+v^{2}\leq 1\}$$

Per calcolare l'integrale doppio passo a coordinate polari:

$$u=\rho \cos \theta ,\,v=\rho \sin \theta$$

$$u=\rho \cos \theta ,\,v=\rho \sin \theta$$

$$T^{-1}E=\{(\rho ,\theta )\,t.c.\,1/2\leq \rho \leq 1,\,0\leq \theta \leq 2\pi \}$$

$$T^{-1}E=\{(\rho ,\theta )\,t.c.\,1/2\leq \rho \leq 1,\,0\leq \theta \leq 2\pi \}$$

$$\det J_{T}=\rho$$

$$\det J_{T}=\rho$$

$$\int _{T^{-1}E}\rho ^{4}{\sqrt {1+{\frac {1}{\rho ^{4}}}}}\rho \,d\rho \,d\theta$$

$$\int _{T^{-1}E}\rho ^{4}{\sqrt {1+{\frac {1}{\rho ^{4}}}}}\rho \,d\rho \,d\theta$$

$$=\int _{0}^{2\pi }\int _{1/2}^{1}\rho ^{4}{\sqrt {1+{\frac {1}{\rho ^{4}}}}}\rho \,d\rho \,d\theta$$

$$=\int _{0}^{2\pi }\int _{1/2}^{1}\rho ^{4}{\sqrt {1+{\frac {1}{\rho ^{4}}}}}\rho \,d\rho \,d\theta$$

$$2\pi \int _{1/2}^{1}\rho ^{5}{\sqrt {1+{\frac {1}{\rho ^{4}}}}}\,d\rho$$

$$2\pi \int _{1/2}^{1}\rho ^{5}{\sqrt {1+{\frac {1}{\rho ^{4}}}}}\,d\rho$$

$$2\pi \int _{1/2}^{1}\rho ^{5}*{\frac {1}{\rho ^{2}}}{\sqrt {\rho ^{4}+1}}\,d\rho$$

$$2\pi \int _{1/2}^{1}\rho ^{5}*{\frac {1}{\rho ^{2}}}{\sqrt {\rho ^{4}+1}}\,d\rho$$

$$2\pi \int _{1/2}^{1}\rho ^{3}{\sqrt {\rho ^{4}+1}}\,d\rho$$

$$2\pi \int _{1/2}^{1}\rho ^{3}{\sqrt {\rho ^{4}+1}}\,d\rho$$

Integro per sostituzione ponendo $\rho ^{4}+1=t$$\rho ^{4}+1=t$, $4\rho ^{3}\,d\rho =dt$$4\rho ^{3}\,d\rho =dt$,

$$\rho =1\,\longrightarrow \,t=\rho ^{4}+1=2$$

$$\rho =1\,\longrightarrow \,t=\rho ^{4}+1=2$$

$$\rho =1/2\,\longrightarrow \,t=\rho ^{4}+1=17/16$$

$$\rho =1/2\,\longrightarrow \,t=\rho ^{4}+1=17/16$$

$$2\pi \int _{17/16}^{2}1/4{\sqrt {t}}\,dt=$$

$$2\pi \int _{17/16}^{2}1/4{\sqrt {t}}\,dt=$$

$$1/2\pi [2/3t^{3/2}]_{17/16}^{2}=$$

$$1/2\pi [2/3t^{3/2}]_{17/16}^{2}=$$

$$=\pi /3[{\sqrt {8}}-{\sqrt {(17/16)^{3}}}]$$

$$=\pi /3[{\sqrt {8}}-{\sqrt {(17/16)^{3}}}]$$

Esercizio 10.3

Calcolare l'integrale:

$$\int _{\Sigma }{\frac {x^{2}+y^{2}}{z^{3}}}\,d\Sigma$$

$$\int _{\Sigma }{\frac {x^{2}+y^{2}}{z^{3}}}\,d\Sigma$$

con

$$\Sigma =x=\sin(uv),\,y=\cos(uv),\,z=u$$

$$\Sigma =x=\sin(uv),\,y=\cos(uv),\,z=u$$

il cui dominio è:

$$E=\{1/2\leq u\leq v,\quad v<1\}.$$

$$E=\{1/2\leq u\leq v,\quad v<1\}.$$

 

Calcolo:

$$\Sigma _{u}(u,v)=(\cos(uv)*v,-\sin(uv)*v,1)$$

$$\Sigma _{u}(u,v)=(\cos(uv)*v,-\sin(uv)*v,1)$$

$$\Sigma _{v}(u,v)=(\cos(uv)u,-u*\sin(uv),0)$$

$$\Sigma _{v}(u,v)=(\cos(uv)u,-u*\sin(uv),0)$$

$$\Sigma _{u}\times \Sigma _{v}=(\sin(uv)*u,\cos(uv)*u,-\cos(uv)*\sin(uv)*uv+\sin(uv)*\cos(uv)*uv)=(\sin(uv)*u,\cos(uv)*u,0)$$

$$\Sigma _{u}\times \Sigma _{v}=(\sin(uv)*u,\cos(uv)*u,-\cos(uv)*\sin(uv)*uv+\sin(uv)*\cos(uv)*uv)=(\sin(uv)*u,\cos(uv)*u,0)$$

$$\vert \Sigma _{u}\times \Sigma _{v}\vert ={\sqrt {\sin ^{2}(uv)*u^{2}+\cos ^{2}(uv)*u^{2}}}={\sqrt {u^{2}}}=|u|$$

$$\vert \Sigma _{u}\times \Sigma _{v}\vert ={\sqrt {\sin ^{2}(uv)*u^{2}+\cos ^{2}(uv)*u^{2}}}={\sqrt {u^{2}}}=|u|$$

Calcolo l'integrale:

$$\int _{E}{\frac {\cos ^{2}(uv)+\sin ^{2}(uv)}{u^{3}}}*|u|\,du\,dv$$

$$\int _{E}{\frac {\cos ^{2}(uv)+\sin ^{2}(uv)}{u^{3}}}*|u|\,du\,dv$$

$$\int _{E}{\frac {|u|}{u^{3}}}\,du\,dv$$

$$\int _{E}{\frac {|u|}{u^{3}}}\,du\,dv$$

$$E=\{1/2\leq u\leq v,\,v\leq 1\}$$

$$E=\{1/2\leq u\leq v,\,v\leq 1\}$$

In $E$$E$$u>0$$u>0$, quindi $u=|u|$$u=|u|$.

$$\int _{E}{\frac {1}{u^{2}}}\,du\,dv$$

$$\int _{E}{\frac {1}{u^{2}}}\,du\,dv$$

Il dominio, definito da:

$$1/2\leq u\leq v\leq 1$$

$$1/2\leq u\leq v\leq 1$$

è una regione $v$$v$-semplice, infatti $u\leq v\leq 1$$u\leq v\leq 1$, invece $1/2\leq u\leq 1$$1/2\leq u\leq 1$, quindi integro prima in $dv$$dv$.

$$\int _{1/2}^{1}[\int _{u}^{1}{\frac {1}{u^{2}}}\,dv]\,du$$

$$\int _{1/2}^{1}[\int _{u}^{1}{\frac {1}{u^{2}}}\,dv]\,du$$

$$\int _{1/2}^{1}{\frac {1-u}{u^{2}}}\,du$$

$$\int _{1/2}^{1}{\frac {1-u}{u^{2}}}\,du$$

$$\int _{1/2}^{1}{\frac {1}{u^{2}}}\,du-\int 1/u\,du$$

$$\int _{1/2}^{1}{\frac {1}{u^{2}}}\,du-\int 1/u\,du$$

$$[-1/u-\log u]_{1/2}^{1}=-1+2-\log 1+\log 1/2=1+\log 1/2$$

$$[-1/u-\log u]_{1/2}^{1}=-1+2-\log 1+\log 1/2=1+\log 1/2$$

Esercizio 10.4

Calcolare l'area della superficie

$$\Sigma =\{(x,y,z)\in \mathbb {R} ^{3}\,t.c.\,z=1/2(x^{2}+2y^{2})\}$$

$$\Sigma =\{(x,y,z)\in \mathbb {R} ^{3}\,t.c.\,z=1/2(x^{2}+2y^{2})\}$$

il cui dominio parametrico è dato da

$$x^{2}+4y^{2}\leq 8$$

$$x^{2}+4y^{2}\leq 8$$

 

Il dominio parametrico è un'ellisse centrata nell'origine. Inoltre

$$\Sigma (u,v)=\{x=u,\,y=v,\,z=(1/2*u^{2}+2v^{2})\}$$

$$\Sigma (u,v)=\{x=u,\,y=v,\,z=(1/2*u^{2}+2v^{2})\}$$

è una superficie cartesiana, quindi uso la formula per determinarne l'elemento d'area $d\Sigma$$d\Sigma$:

$$\vert \Sigma _{u}\times \Sigma _{v}\vert ={\sqrt {1+|\nabla z|^{2}}}$$

$$\vert \Sigma _{u}\times \Sigma _{v}\vert ={\sqrt {1+|\nabla z|^{2}}}$$

$$z_{u}=u,\quad z_{v}=2v$$

$$z_{u}=u,\quad z_{v}=2v$$

$$\vert \Sigma _{u}\times \Sigma _{v}\vert ={\sqrt {1+u^{2}+4v^{2}}}$$

$$\vert \Sigma _{u}\times \Sigma _{v}\vert ={\sqrt {1+u^{2}+4v^{2}}}$$

L'area si calcola quindi come:

$$A_{\Sigma }=\int _{\{u^{2}+4v^{2}\leq 8\}}{\sqrt {1+u^{2}+4v^{2}}}\,du\,dv$$

$$A_{\Sigma }=\int _{\{u^{2}+4v^{2}\leq 8\}}{\sqrt {1+u^{2}+4v^{2}}}\,du\,dv$$

Introduco le coordinate ellittico-polari, e riscrivo il dominio come:

$$\{(u,v)\,t.c.\,u^{2}/8+v^{2}/2\leq 1\}$$

$$\{(u,v)\,t.c.\,u^{2}/8+v^{2}/2\leq 1\}$$

Pongo:

$$u=2{\sqrt {2}}\rho \cos \theta ,\,v={\sqrt {2}}\rho \sin \theta$$

$$u=2{\sqrt {2}}\rho \cos \theta ,\,v={\sqrt {2}}\rho \sin \theta$$

Con il cambio di variabili l'equazione che definisce il dominio diventa:

$$u^{2}/8+v^{2}/2=\rho ^{2}\leq 1$$

$$u^{2}/8+v^{2}/2=\rho ^{2}\leq 1$$

quindi

$$T^{-1}(D)=\{(\rho ,\theta )\,t.c.\,0<\theta <2\pi ,\,0\leq \rho \leq 1\}$$

$$T^{-1}(D)=\{(\rho ,\theta )\,t.c.\,0<\theta <2\pi ,\,0\leq \rho \leq 1\}$$

Calcolo il determinante jacobiano della trasformazione:

$$J_{T}=\left({\begin{array}{cc}2{\sqrt {2}}\cos \theta &-2{\sqrt {2}}\rho \sin \theta \\{\sqrt {2}}\sin \theta &{\sqrt {2}}\rho \cos \theta \end{array}}\right)$$

$$J_{T}=\left({\begin{array}{cc}2{\sqrt {2}}\cos \theta &-2{\sqrt {2}}\rho \sin \theta \\{\sqrt {2}}\sin \theta &{\sqrt {2}}\rho \cos \theta \end{array}}\right)$$

$$\det J_{T}=2{\sqrt {2}}\cos \theta {\sqrt {2}}\rho \cos \theta +2{\sqrt {2}}{\sqrt {2}}\rho \sin ^{2}\theta =4\rho$$

$$\det J_{T}=2{\sqrt {2}}\cos \theta {\sqrt {2}}\rho \cos \theta +2{\sqrt {2}}{\sqrt {2}}\rho \sin ^{2}\theta =4\rho$$

$$\int _{0}^{2\pi }[\int _{0}^{1}4\rho {\sqrt {1+8\rho ^{2}\cos ^{2}\theta +8\rho ^{2}\sin ^{2}\theta }}\,d\rho ]\,d\theta$$

$$\int _{0}^{2\pi }[\int _{0}^{1}4\rho {\sqrt {1+8\rho ^{2}\cos ^{2}\theta +8\rho ^{2}\sin ^{2}\theta }}\,d\rho ]\,d\theta$$

$$\int _{0}^{2\pi }[\int _{0}^{1}4\rho {\sqrt {1+8\rho ^{2}}}\,d\rho ]\,d\theta$$

$$\int _{0}^{2\pi }[\int _{0}^{1}4\rho {\sqrt {1+8\rho ^{2}}}\,d\rho ]\,d\theta$$

Risolvo l'integrale per sostituzione:

$$t=8\rho ^{2}+1,\,\longrightarrow \,dt=16\rho d\rho$$

$$t=8\rho ^{2}+1,\,\longrightarrow \,dt=16\rho d\rho$$

$$\rho =1\,\longrightarrow \,t=9$$

$$\rho =1\,\longrightarrow \,t=9$$

$$\rho =0\,\longrightarrow \,t=1$$

$$\rho =0\,\longrightarrow \,t=1$$

$$1/4*2\pi \int _{1}^{9}{\sqrt {t}}\,dt$$

$$1/4*2\pi \int _{1}^{9}{\sqrt {t}}\,dt$$

$$\pi /22/3[t^{3/2}]_{1}^{9}=$$

$$\pi /22/3[t^{3/2}]_{1}^{9}=$$

$$\pi /3[9^{3/2}-1]=\pi /3*(27-1)=26\pi /3$$

$$\pi /3[9^{3/2}-1]=\pi /3*(27-1)=26\pi /3$$

Esercizio 10.5

Calcolare l'integrale:

$$\int _{\Sigma }{\frac {x}{\sqrt {4z+1}}}\,d\Sigma$$

$$\int _{\Sigma }{\frac {x}{\sqrt {4z+1}}}\,d\Sigma$$

dove

$$\Sigma =\{(x,y,z)\in \mathbb {R} ^{3}\,t.c.\,z=x^{2}+y^{2},\wedge x^{2}+y^{2}-y\leq 0,\,y\geq 1/2\vee x\geq 0\}.$$

$$\Sigma =\{(x,y,z)\in \mathbb {R} ^{3}\,t.c.\,z=x^{2}+y^{2},\wedge x^{2}+y^{2}-y\leq 0,\,y\geq 1/2\vee x\geq 0\}.$$

 

La disuguaglianza $x^{2}+y^{2}-y\leq 0$$x^{2}+y^{2}-y\leq 0$ definisce l'interno di un cerchio. Con il metodo del completamento dei quadrati ottengo:

$$x^{2}+y^{2}-y+1/4\leq 1/4$$

$$x^{2}+y^{2}-y+1/4\leq 1/4$$

$$x^{2}+(y-1/4)^{2}\leq 1/4$$

$$x^{2}+(y-1/4)^{2}\leq 1/4$$

cioè il dominio parametrico $A$$A$ della superficie è dato da un cerchio di raggio $1/2$$1/2$ e centro in $(0,1/2)$$(0,1/2)$, a cui tolgo lo spicchio in cui $x\leq 0$$x \le 0$ e $y\leq 1/2$$y\leq 1/2$. La superficie ha equazioni parametriche:

$${\begin{cases}x=u\\y=v\\z=u^{2}+v^{2}\end{cases}}$$

$${\begin{cases}x=u\\y=v\\z=u^{2}+v^{2}\end{cases}}$$

ed è una superficie cartesiana, con $f(x,y)=x^{2}+y^{2}$$f(x,y)=x^{2}+y^{2}$, allora vale la formula:

$$\vert \Sigma _{u}\times \Sigma _{v}\vert ={\sqrt {1+(|\nabla f|)^{2}}}={\sqrt {1+4x^{2}+4y^{2}}}$$

$$\vert \Sigma _{u}\times \Sigma _{v}\vert ={\sqrt {1+(|\nabla f|)^{2}}}={\sqrt {1+4x^{2}+4y^{2}}}$$

Allora integrando, e scrivendo ${\frac {x}{\sqrt {1+4z}}}={\frac {u}{\sqrt {1+4(u^{2}+v^{2})}}}$${\frac {x}{\sqrt {1+4z}}}={\frac {u}{\sqrt {1+4(u^{2}+v^{2})}}}$ ottengo:

$$\int _{A}{\frac {u}{\sqrt {4u^{2}+4v^{2}+1}}}*{\sqrt {4u^{2}+4v^{2}+1}}\,du\,dv$$

$$\int _{A}{\frac {u}{\sqrt {4u^{2}+4v^{2}+1}}}*{\sqrt {4u^{2}+4v^{2}+1}}\,du\,dv$$

$$=\int _{A}u\,du\,dv$$

$$=\int _{A}u\,du\,dv$$

Spezzo $A$$A$ in due parti e lo scrivo come:

$$A_{1}=\{(u,v)\,t.c.u^{2}+v^{2}-v\leq 0,\quad ,v\geq 1/2\}$$

$$A_{1}=\{(u,v)\,t.c.u^{2}+v^{2}-v\leq 0,\quad ,v\geq 1/2\}$$

$$A_{2}=\{u^{2}+v^{2}-v\leq 0,\quad u\geq 0,\,v\leq 1/2\}$$

$$A_{2}=\{u^{2}+v^{2}-v\leq 0,\quad u\geq 0,\,v\leq 1/2\}$$

$A_{1}$$A_1$ è un dominio simmetrico rispetto alla variabile $u$$u$ ed è la metà superiore del cerchio, allora l'integrale di una funzione dispari su un dominio simmetrico è nulla, e rimane da calcolare:

$$\int _{A_{2}}u\,du\,dv=$$

$$\int _{A_{2}}u\,du\,dv=$$

Scrivo le disuguaglianze che definiscono $A_{2}$$A_2$:

$$u^{2}\leq v-v^{2}$$

$$u^{2}\leq v-v^{2}$$

$$\longrightarrow -{\sqrt {v-v^{2}}}\leq u\leq {\sqrt {v-v^{2}}}$$

$$\longrightarrow -{\sqrt {v-v^{2}}}\leq u\leq {\sqrt {v-v^{2}}}$$

$$0\geq 0\longrightarrow \,0\leq u\leq {\sqrt {v-v^{2}}}$$

$$0\geq 0\longrightarrow \,0\leq u\leq {\sqrt {v-v^{2}}}$$

Inoltre si ha $0\leq v\leq 1/2$$0\leq v\leq 1/2$:

$$\int _{0}^{1/2}\,\int _{0}^{\sqrt {v-v^{2}}}u\,du\,dv=$$

$$\int _{0}^{1/2}\,\int _{0}^{\sqrt {v-v^{2}}}u\,du\,dv=$$

$$\int _{0}^{1/2}\,[u^{2}/2]_{0}^{\sqrt {v-v^{2}}}dv=$$

$$\int _{0}^{1/2}\,[u^{2}/2]_{0}^{\sqrt {v-v^{2}}}dv=$$

$$1/2\int _{0}^{1/2}v-v^{2}dv=$$

$$1/2\int _{0}^{1/2}v-v^{2}dv=$$

$$1/2[v^{2}/2-v^{3}/3]_{0}^{1/2}=$$

$$1/2[v^{2}/2-v^{3}/3]_{0}^{1/2}=$$

$$1/2[1/8-{\frac {1}{24}}]=1/24$$

$$1/2[1/8-{\frac {1}{24}}]=1/24$$

Esercizio 10.6

Si calcoli

$$\iint _{\Sigma }(z+y^{2})\,d\sigma$$

$$\iint _{\Sigma }(z+y^{2})\,d\sigma$$

dove $\Sigma$$\Sigma$ è la semisfera superiore di centro l'origine e raggio $R$$R$.

 

La semisfera si parametrizza con le coordinate:

$$x\mapsto R\sin \phi \cos \theta ,\,y\mapsto R\sin \phi \sin \theta ,\,z\mapsto R\cos \phi$$

$$x\mapsto R\sin \phi \cos \theta ,\,y\mapsto R\sin \phi \sin \theta ,\,z\mapsto R\cos \phi$$

$$f=z+y^{2}=R\cos \phi +R^{2}\sin ^{2}\phi \sin ^{2}\theta$$

$$f=z+y^{2}=R\cos \phi +R^{2}\sin ^{2}\phi \sin ^{2}\theta$$

Quindi

$$\int _{\Sigma }f\,d\sigma =\int _{K}\vert S_{\theta }\times S_{\phi }\vert f(S(\phi ,\theta ))\,d\theta \,d\phi$$

$$\int _{\Sigma }f\,d\sigma =\int _{K}\vert S_{\theta }\times S_{\phi }\vert f(S(\phi ,\theta ))\,d\theta \,d\phi$$

$$S_{\theta }=(-R\sin \phi \sin \theta ,R\sin \phi \cos \theta ,0)$$

$$S_{\theta }=(-R\sin \phi \sin \theta ,R\sin \phi \cos \theta ,0)$$

$$S_{\phi }=(R\cos \phi \cos \theta ,R\cos \phi \sin \theta ,-R\sin \phi )$$

$$S_{\phi }=(R\cos \phi \cos \theta ,R\cos \phi \sin \theta ,-R\sin \phi )$$

$$S_{\theta }\times S_{\phi }=[-R^{2}\sin ^{2}\phi \cos \theta ,\,-R^{2}\sin \theta \sin ^{2}\phi ,\,-R^{2}\sin \phi \cos \phi \sin ^{2}\theta -R^{2}\sin \phi \cos \phi \cos ^{2}\theta ]$$

$$S_{\theta }\times S_{\phi }=[-R^{2}\sin ^{2}\phi \cos \theta ,\,-R^{2}\sin \theta \sin ^{2}\phi ,\,-R^{2}\sin \phi \cos \phi \sin ^{2}\theta -R^{2}\sin \phi \cos \phi \cos ^{2}\theta ]$$

$$S_{\theta }\times S_{\phi }=[-R^{2}\sin ^{2}\phi \cos \theta ,\,-R^{2}\sin ^{2}\phi \sin \theta ,\,-R^{2}\sin \phi \cos \phi ]$$

$$S_{\theta }\times S_{\phi }=[-R^{2}\sin ^{2}\phi \cos \theta ,\,-R^{2}\sin ^{2}\phi \sin \theta ,\,-R^{2}\sin \phi \cos \phi ]$$

$$\vert S_{\theta }\times S_{\phi }\vert =$$

$$\vert S_{\theta }\times S_{\phi }\vert =$$

$${\sqrt {R^{4}\sin ^{4}\phi \cos ^{2}\theta +R^{4}\sin ^{4}\phi \sin ^{2}\theta +R^{4}\sin ^{2}\phi \cos ^{2}\phi }}$$

$${\sqrt {R^{4}\sin ^{4}\phi \cos ^{2}\theta +R^{4}\sin ^{4}\phi \sin ^{2}\theta +R^{4}\sin ^{2}\phi \cos ^{2}\phi }}$$

$${\sqrt {R^{4}\sin ^{4}\phi +R^{4}\sin ^{2}\phi \cos ^{2}\phi }}$$

$${\sqrt {R^{4}\sin ^{4}\phi +R^{4}\sin ^{2}\phi \cos ^{2}\phi }}$$

$${\sqrt {R^{4}\sin ^{2}\phi }}=R^{2}\sin \phi$$

$${\sqrt {R^{4}\sin ^{2}\phi }}=R^{2}\sin \phi$$

$$K=[0,2\pi ]\times [0,\pi /2]$$

$$K=[0,2\pi ]\times [0,\pi /2]$$

$$\int _{0}^{2\pi }[\int _{0}^{\pi /2}R^{2}\sin \phi (R\cos \phi +R^{2}\sin ^{2}\phi \sin ^{2}\theta )\,d\phi ]\,d\theta$$

$$\int _{0}^{2\pi }[\int _{0}^{\pi /2}R^{2}\sin \phi (R\cos \phi +R^{2}\sin ^{2}\phi \sin ^{2}\theta )\,d\phi ]\,d\theta$$

$$\int _{0}^{2\pi }[\int _{0}^{\pi /2}R^{2}(R\cos \phi \sin \phi +R^{2}\sin ^{3}\phi \sin ^{2}\theta )\,d\phi ]\,d\theta$$

$$\int _{0}^{2\pi }[\int _{0}^{\pi /2}R^{2}(R\cos \phi \sin \phi +R^{2}\sin ^{3}\phi \sin ^{2}\theta )\,d\phi ]\,d\theta$$

$$[\int _{0}^{2\pi }R^{3}\,d\theta ][\int _{0}^{\pi /2}1/2\sin 2\phi \,d\phi ]+[R^{4}\int _{0}^{2\pi }\sin ^{2}\theta \,d\theta ]*[\int _{0}^{\pi /2}\,\sin ^{3}\phi d\phi ]=$$

$$[\int _{0}^{2\pi }R^{3}\,d\theta ][\int _{0}^{\pi /2}1/2\sin 2\phi \,d\phi ]+[R^{4}\int _{0}^{2\pi }\sin ^{2}\theta \,d\theta ]*[\int _{0}^{\pi /2}\,\sin ^{3}\phi d\phi ]=$$

Calcolo i quattro integrali:

1. $$\int _{0}^{2\pi }R^{3}\,d\theta =2\pi R^{3}$$

   $$\int _{0}^{2\pi }R^{3}\,d\theta =2\pi R^{3}$$
2. $$\int _{0}^{\pi /2}1/2\sin(2\phi )\,d\phi =$$

   $$\int _{0}^{\pi /2}1/2\sin(2\phi )\,d\phi =$$
   $$[-1/4\cos(2\phi )]_{0}^{\pi /2}=-1/2$$

   $$[-1/4\cos(2\phi )]_{0}^{\pi /2}=-1/2$$
3. $$R^{4}\int _{0}^{2\pi }\sin ^{2}\theta \,d\theta =$$

   $$R^{4}\int _{0}^{2\pi }\sin ^{2}\theta \,d\theta =$$
   $$R^{4}[\theta /2+1/4\sin(2\theta )]_{0}^{2\pi }=2R^{4}\pi$$

   $$R^{4}[\theta /2+1/4\sin(2\theta )]_{0}^{2\pi }=2R^{4}\pi$$
4. Integro per parti $\sin ^{3}\phi$$\sin ^{3}\phi$:
   $$\int \sin ^{3}\phi \,d\phi =\int \sin ^{2}\phi \sin \phi \,d\phi =$$

   $$\int \sin ^{3}\phi \,d\phi =\int \sin ^{2}\phi \sin \phi \,d\phi =$$
   $$f(x)=\sin ^{2}\phi ,f'(x)=2\sin \phi \cos \phi ,g(x)=-\cos \phi ,\,g'(x)=\sin \phi$$

   $$f(x)=\sin ^{2}\phi ,f'(x)=2\sin \phi \cos \phi ,g(x)=-\cos \phi ,\,g'(x)=\sin \phi$$
   $$\sin ^{2}\phi \cos \phi -\int 2\sin \phi \cos \phi \cos \phi \,d\phi =$$

   $$\sin ^{2}\phi \cos \phi -\int 2\sin \phi \cos \phi \cos \phi \,d\phi =$$
   $$\sin ^{2}\phi \cos \phi -2\int \sin \phi +2\int \sin ^{3}\phi \,d\phi =$$

   $$\sin ^{2}\phi \cos \phi -2\int \sin \phi +2\int \sin ^{3}\phi \,d\phi =$$
   ed eguagliando al primo membro:
   $$\int \sin ^{3}\phi \,d\phi =-\sin ^{2}\phi \cos \phi +2\int \sin \phi \,d\phi =$$

   $$\int \sin ^{3}\phi \,d\phi =-\sin ^{2}\phi \cos \phi +2\int \sin \phi \,d\phi =$$
   Quindi si ha
   $$\int _{0}^{\pi /2}\sin ^{3}\phi \,d\phi =[-\sin ^{2}\phi \cos \phi -2\cos \phi ]_{0}^{\pi /2}=2$$

   $$\int _{0}^{\pi /2}\sin ^{3}\phi \,d\phi =[-\sin ^{2}\phi \cos \phi -2\cos \phi ]_{0}^{\pi /2}=2$$
   Quindi complessivamente si ottiene:
   $$-2\pi R^{3}*1/2+2*2\pi R^{4}=\pi R^{3}(2R-1)$$

   $$-2\pi R^{3}*1/2+2*2\pi R^{4}=\pi R^{3}(2R-1)$$

Esercizio 10.7

Si calcoli l'area della superficie regolare di sostegno

$$\Sigma =\{(x,y,z)\in \mathbb {R} ^{3}:\,x^{2}+y^{2}+z^{2}=1,\,{\sqrt {x^{2}+y^{2}}}\leq {\frac {1}{2}},\,z\geq 0\}.$$

$$\Sigma =\{(x,y,z)\in \mathbb {R} ^{3}:\,x^{2}+y^{2}+z^{2}=1,\,{\sqrt {x^{2}+y^{2}}}\leq {\frac {1}{2}},\,z\geq 0\}.$$

 

Parametrizzo la superficie come:

$$x=\sin \phi \cos \theta ,\,y=\sin \phi \sin \theta ,\,z=\cos \phi$$

$$x=\sin \phi \cos \theta ,\,y=\sin \phi \sin \theta ,\,z=\cos \phi$$

Determino il dominio su cui $\Sigma$$\Sigma$ è definita:

$${\sqrt {x^{2}+y^{2}}}\leq 1/2\,\longrightarrow \,{\sqrt {\sin ^{2}\phi \cos ^{2}\theta +\sin ^{2}\phi \sin ^{2}\theta }}\leq 1/2$$

$${\sqrt {x^{2}+y^{2}}}\leq 1/2\,\longrightarrow \,{\sqrt {\sin ^{2}\phi \cos ^{2}\theta +\sin ^{2}\phi \sin ^{2}\theta }}\leq 1/2$$

$$\longrightarrow {\sqrt {\sin ^{2}\phi }}\leq 1/2,\,\sin \phi \leq 1/2$$

$$\longrightarrow {\sqrt {\sin ^{2}\phi }}\leq 1/2,\,\sin \phi \leq 1/2$$

quindi

$$0\leq \phi \leq \pi /6\vee 5\pi /6\leq \phi \leq 2\pi$$

$$0\leq \phi \leq \pi /6\vee 5\pi /6\leq \phi \leq 2\pi$$

Inoltre:

$$z\geq 0\,\longrightarrow \,\cos \phi \geq 0$$

$$z\geq 0\,\longrightarrow \,\cos \phi \geq 0$$

$$0\leq \phi \leq \pi /2\vee 3\pi /2\leq \phi \leq 2\pi$$

$$0\leq \phi \leq \pi /2\vee 3\pi /2\leq \phi \leq 2\pi$$

Quindi in conclusione:

$$0\leq \theta \leq 2\pi ,\,0\leq \phi \leq \pi /6,\vee 3\pi /2\leq \phi \leq 2\pi$$

$$0\leq \theta \leq 2\pi ,\,0\leq \phi \leq \pi /6,\vee 3\pi /2\leq \phi \leq 2\pi$$

e quindi

$$-\pi /2\leq \phi \leq \pi /6$$

$$-\pi /2\leq \phi \leq \pi /6$$

Come già calcolato precedentemente:

$$\vert \sigma _{\phi }\times \sigma _{\theta }\vert =\sin \phi$$

$$\vert \sigma _{\phi }\times \sigma _{\theta }\vert =\sin \phi$$

Quindi l'area di questa superficie è:

$$\int _{0}^{2\pi }\,d\theta [\int _{-\pi /2}^{\pi /6}\sin \phi \,d\phi ]=$$

$$\int _{0}^{2\pi }\,d\theta [\int _{-\pi /2}^{\pi /6}\sin \phi \,d\phi ]=$$

$$=2\pi *[-\cos \phi ]_{-\pi /2}^{\pi /6}=-2\pi *{\sqrt {3}}/2={\sqrt {3}}\pi$$

$$=2\pi *[-\cos \phi ]_{-\pi /2}^{\pi /6}=-2\pi *{\sqrt {3}}/2={\sqrt {3}}\pi$$