Question

Alguém me ajuda nesta integral dupla

Answer 1

Pretendemos calcular o integral duplo:

$\displaystyle\iint\limits_R x\sqrt{1-x^2}\textrm{ d}A, \quad \textrm{com } R = \{(x,y): 0 \leq x \leq 1, 2 \leq y \leq 3\}.$

Utilizamos o teorema de Fubini para escrever o integral duplo como integrais iterados, com $\textrm{d}A = \textrm{d}x\textrm{ d}y$:

$\displaystyle\int\limits_2^3\left(\int\limits_0^1 x\sqrt{1-x^2}\textrm{ d}x\right)\textrm{d}y = \int\limits_2^3\textrm{ d}y\int\limits_0^1 x\sqrt{1-x^2}\textrm{ d}x.$

O integral em $y$ é simples:

$\displaystyle\int\limits_2^3\textrm{ d}y = y\Big\vert_2^3 = 3 - 2 = 1.$

Para o integral em $x$, consideramos a mudança de variável $u \equiv 1-x^2 \implies \textrm{d}u = -2x\textrm{ d}x$:

$\displaystyle \int\limits_0^1 x\sqrt{1-x^2}\textrm{ d}x = -\dfrac{1}{2}\int\limits_1^0 \sqrt{u}\textrm{ d}u = \dfrac{1}{2}\int\limits_0^1 u^{1/2}\textrm{ d}u = \dfrac{1}{2}\dfrac{u^{3/2}}{3/2}\Big\vert_0^1 = \dfrac{1}{3}(1-0) = \dfrac{1}{3}.$

Assim, o integral duplo pretendido é:

$\boxed{\displaystyle\iint\limits_R x\sqrt{1-x^2}\textrm{ d}A = \dfrac{1}{3}}.$