Question

(CÁLCULO 2)
Determine a área da superfície:
A parte da esfera x² + y² +z² = a² que está dentro do cilindro x² +y² = ax e acima do plano xy.

Answer 1

encontrando os limites
$\Bmatrix{x^2+y^2+z^2=a^2\\\\x^2+y^2=ax\end$

substituindo o valor de x²+y² na primeira equação e isolando z

$\boxed{\boxed{z=\pm \sqrt{a^2-ax} }}$

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as derivadas parciais

$\bmatrix{ \frac{\partial z}{\partial x}= \frac{-x}{\sqrt{a^2-x^2-y^2}} \\\\ \frac{\partial z}{\partial y}= \frac{-y}{\sqrt{a^2-x^2-y^2}} \end$

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a integral que calcula a area da superficie da esfera sera

$\boxed{\boxed{A(s)=\int\int_D \left( \sqrt{1+ \left( \frac{\partial z}{\partial x \right )^2+\left( \frac{\partial z}{\partial x \right )^2} \right )}dA}}}$

.
$A(s)=\int\int_D \left( \sqrt{1+ \frac{x^2}{a^2-x^2-y^2}+ \frac{y^2}{a^2-x^2-y^2}} \right )}dA\\\\\\A(s)=\int\int_D \left( \sqrt{1+ \frac{x^2+y^2}{a^2-(x^2+y^2)}}\right)dA\\\\\\A(s)=\int\int_D \left( \sqrt{ \frac{a^2-(x^2+y^2)+x^2+y^2}{a^2-(x^2+y^2)}}\right)dA\\\\\\A(s)=\int\int_D \left( \sqrt{ \frac{a^2}{a^2-(x^2+y^2)}}\right)dA\\\\\\A(s)=\int\int_D \left( \frac{a}{\sqrt{a^2-(x^2+y^2)}}\right)dA\\\\\\\boxed{\boxed{A(s)=a*\int\int_D \left( \frac{1}{\sqrt{a^2-(x^2+y^2)}}\right)dA}}$

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reescrevendo em coordenadas polares
no plano xy o cilindro tem uma base circular 

$x^2+y^2=ax\\\\r^2=a*r*cos(\theta)\\\\\boxed{\boxed{r=a*cos(\theta)}}$

queremos a área acima do plano xy
o angulo vai variar de 0 a pi


$D\Bmatrix{0 \leq r \leq a*cos(\theta)\\\\0 \leq \theta \leq \pi \end$


temos a integral
$\boxed{\boxed{A(s)=a* \int\limits^\pi_0 \int\limits^{a*cos(\theta)}_0 \left( \frac{r}{\sqrt{a^2-r^2} \right)} drd\theta}}}$