Question

Cálculo Diferencial e Integral 2, alguém poderia ajudar?

Answer 1

1. encontremos a equação parametrica da superficie

$S=\begin{cases} x=u&u\in[0,1]\\ y=v & v\in [0,4]\\ z=\sqrt{4-u^2} \end{cases}\\ \\ \\ \text{A \'area da superficie \'e: }\\ \\ \displaystyle A=\iint_D \left\| T_u\times T_v \right\|~du~dv\\ \\ \\ T_u=\left(1,0,-\frac{u}{\sqrt{4-u^2}}\right)\\ \\ T_v=(0,1,0)\\ \\ \\ T_u\times T_v=\left(\frac{u}{\sqrt{4-u^2}},0,1\right)\\ \\ \\ \|T_u\times T_v\|=\dfrac{2}{\sqrt{4-u^2}}$

$\displaystyle A=\iint_D \dfrac{2}{\sqrt{4-u^2}} ~du~dv\\ \\ \\ A=\int_{0}^{4}\int_{0}^{1}\dfrac{2}{\sqrt{4-u^2}} ~du~dv\\ \\ \\ A=8\int_{0}^{1}\dfrac{1}{\sqrt{4-u^2}} ~du\\ \\ \\ A=8\left.\left[\arcsin\left(\frac{u}{2}\right)\right]\right|_{0}^{1}\\ \\ A=8\left[\arcsin \frac{1}{2}-\arcsin 0\right]\\ \\ \\ A=8\left(\frac{\pi}{6}-0\right)\\ \\ \\ \boxed{A=\dfrac{4\pi}{3}}$