Observe
Uma solução:
[tex3]I=\int\limits_{}^{}\int\limits_{\sigma }^{}f(x,y,z)dS[/tex3]
[tex3]I=\int\limits_{}^{}\int\limits_{\sigma }^{}f(x,y,z)\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}dxdy[/tex3]
[tex3]I=\int\limits_{}^{}\int\limits_{\sigma }^{}(x^2+y^2)\sqrt{1+\left(-\frac{x}{\sqrt{4-x^2-y^2}}\right)^2+\left(-\frac{y}{\sqrt{4-x^2-y^2}}\right)^2}dxdy[/tex3]
[tex3]I=\int\limits_{}^{}\int\limits_{\sigma }^{}(x^2+y^2)\sqrt{1+\frac{x²}{4-x^2-y^2}+\frac{y^2}{4-x^2-y^2}}dxdy[/tex3]
[tex3]\therefore [/tex3]
[tex3]I=2.\int\limits_{}^{}\int\limits_{\sigma }^{}\frac{x^2+y^2}{\sqrt{4-x^2-y^2}}dxdy[/tex3]
Passando para coordenadas polares, temos:
[tex3]I=2.\int\limits_{0}^{2π}\int\limits_{0}^{\sqrt{3}}\frac{r^2}{\sqrt{4-r^2}}rdrd\theta [/tex3]
[tex3]I=\int\limits_{0}^{2π}[\int\limits_{0}^{\sqrt{3}}\frac{r^2}{\sqrt{4-r^2}}2rdr]d\theta [/tex3]
Resolvendo a integral que está dentro dos colchetes, vem;
u = 4 - r² → du = - 2rdr → 2rdr = - du e r² = 4 - u
Como mudamos de variável devemos então mudar os limites de integração, temos que:
Para r = √3:
u = 4 - (√3)² = 4 - 3 → u = 1
Para r = 0:
u = 4 - 0² = 4 - 0 → u = 4
Assim,
[tex3]I=\int\limits_{0}^{2π}[\int\limits_{4}^{1}\frac{4-u}{\sqrt{u}}(-du)]d\theta [/tex3]
[tex3]I=\int\limits_{0}^{2π}[\int\limits_{4}^{1}\frac{u-4}{\sqrt{u}}du]d\theta [/tex3]
[tex3]I=\int\limits_{0}^{2π}[\int\limits_{4}^{1}\frac{u}{\sqrt{u}}du-4\int\limits_{4}^{1}\frac{1}{\sqrt{u}}du]d\theta [/tex3]
[tex3]I=\int\limits_{0}^{2π}[\int\limits_{4}^{1}u^{\frac{1}{2}}du-4\int\limits_{4}^{1}u^{-\frac{1}{2}}du]d\theta [/tex3]
[tex3]I=\int\limits_{0}^{2π}\{[\frac{2}{3}\sqrt{u^3}]_{4}^{1}-4.[2\sqrt{u}]_{4}^{1}\}d\theta [/tex3]
[tex3]\therefore [/tex3]
[tex3]I=\int\limits_{0}^{2π}(\frac{2}{3}-\frac{16}{3}-8+16) \ d\theta [/tex3]
[tex3]I=\int\limits_{0}^{2π}\frac{-14+24}{3} \ d\theta [/tex3]
[tex3]I=\int\limits_{0}^{2π}\frac{10}{3} \ d\theta [/tex3]
[tex3]I=\frac{10}{3}[\theta ]_{0}^{2π}[/tex3]
[tex3]I=\frac{10}{3}(2π-0)[/tex3]
Portanto,
[tex3]I=\frac{20π}{3}[/tex3]
Nota
Intersecção de x² + y² + z² = 4 com z = 1:
x² + y² + 1² = 4
x² + y² = 3 ( disco com raio √3 , ou seja , projeção da superfície [tex3]\sigma [/tex3]
sobre o plano xy )
Obs.1
x² + y² = r²
dxdy = rdrdθ
Obs.2
x² + y² + z² = 4
z² = 4 - x² - y²
z = ± √( 4 - x² - y² )
Então;
z = √( 4 - x² - y² ) ( pois z ≥ 1 )
Daí;
[tex3]\frac{\partial z}{\partial x}=\frac{(4-x^2-y^2)'}{2\sqrt{4-x^2-y^2}}[/tex3]
[tex3]\frac{\partial z}{\partial x}=\frac{0-2x-0}{2\sqrt{4-x^2-y^2}}[/tex3]
[tex3]\frac{\partial z}{\partial x}=\frac{-\cancel{2}x}{\cancel{2}\sqrt{4-x^2-y^2}}[/tex3]
[tex3]\frac{\partial z}{\partial x}=-\frac{x}{\sqrt{4-x^2-y^2}}[/tex3]
e
[tex3]\frac{\partial z}{\partial y}=\frac{(4-x^2-y^2)'}{2\sqrt{4-x^2-y^2}}[/tex3]
[tex3]\frac{\partial z}{\partial y}=\frac{0-0-2y}{2\sqrt{4-x^2-y^2}}[/tex3]
[tex3]\frac{\partial z}{\partial y}=\frac{-\cancel{2}y}{\cancel{2}\sqrt{4-x^2-y^2}}[/tex3]
[tex3]\frac{\partial z}{\partial y}=-\frac{y}{\sqrt{4-x^2-y^2}}[/tex3]
Obs.3 A figura ( gráfico ) ficará como exercício para você
Bons estudos!