Question

Calcule a integral dupla ∫∫R (x + y) dA semicírculo limitado por y = √ 4 − x 2 e y = 0.

Answer 1

Olá, boa tarde.

Para resolvermos esta questão, devemos nos relembrar de algumas propriedades estudadas sobre integrais duplas.

Devemos resolver a seguinte integral dupla: $\displaystyle{\iint_R(x+y)\,dA$, em que $R$ é a região do semicírculo limitado por $y=\sqrt{4-x^2}$ e $y=0$.

Primeiro, parametrizamos a curva. Fazemos:

$\begin{cases}x=r\cos(\theta)\\y=r\sin(\theta)\\\end{cases}$

Calculamos os limites de integração

$r\sin(\theta)=\sqrt{4-r^2\cos^2\theta}\\\\\\r^2\sin^2(\theta)=4-r^2\cos^2(\theta)\\\\\\ r^2\sin^2(\theta)+r^2\cos^2(\theta)=4\\\\\\ r^2\cdot\underbrace{(\sin^2(\theta)+\cos^2(\theta))}_1=4\\\\\\ r^2 =4\\\\\\ r=\pm2$

Assim, os limites de integração serão $-2\leq r\leq 2$ e $0\leq\theta\leq\pi$, tendo em vista que a região é um semicírculo.

Calculamos o determinante jacobiano da transformação:

$J=\begin{bmatrix}\dfrac{\partial x}{\partial r}&\dfrac{\partial x}{\partial \theta}\\\\ \dfrac{\partial y}{\partial r}&\dfrac{\partial y}{\partial \theta}\\\end{bmatrix}\\\\\\ J=\begin{bmatrix}\cos(\theta)&-r\sin(\theta)\\\sin(\theta)&r\cos(\theta)\\\end{bmatrix}$

Calcule o determinante

$\det|J|=\begin{Vmatrix}\cos(\theta)&-r\sin(\theta)\\\sin(\theta)&r\cos(\theta)\\\end{Vmatrix}\\\\\\ \det|J|=|\cos(\theta)\cdot(r\cos(\theta))-(-r\sin(\theta))\cdot\sin(\theta)|\\\\\\ \det|J|=|r\cos^2(\theta)+r\sin^2(\theta)|\\\\\\ \det|J|=|r\cdot\underbrace{(\cos^2(\theta)+\sin^2(\theta))}_1|\\\\\\ \det|J|=|r|=r$

A integral dupla se torna: $\displaystyle{\iint_R f(x,~y)\,dA\Rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}f(r,~\theta)\cdot\det|J|\,dr\,d\theta}$. Assim, teremos:

$\displaystyle{\int_0^{\pi}\int_{-2}^2(r\cos(\theta)+r\sin(\theta))\cdot r\,dr\,d\theta$

Fatore a expressão no integrando

$\displaystyle{\int_0^{\pi}\int_{-2}^2r^2\cdot(\cos(\theta)+\sin(\theta))\,dr\,d\theta$

A integral mais interna diz respeito à variável $r$. Logo, fazemos:

$\displaystyle{\int_0^{\pi}(\cos(\theta)+\sin(\theta))\cdot\int_{-2}^2r^2\,dr\,d\theta$

Calculamos a integral utilizando a regra da potência: $\displaystyle{\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$.

$\displaystyle{\int_0^{\pi}(\cos(\theta)+\sin(\theta))\cdot\left(\dfrac{r^{2+1}}{2+1}\right)~\biggr|_{-2}^2\,d\theta$

Some os valores no expoente e denominador e aplique os limites de integração, de acordo com o Teorema fundamental do Cálculo: $\displaystyle{\int_a ^b f(x)\,dx=F(x)~\biggr|_a ^b=F(b)-F(a)$.

$\displaystyle{\int_0^{\pi}(\cos(\theta)+\sin(\theta))\cdot\left(\dfrac{r^3}{3}\right)~\biggr|_{-2}^2\,d\theta}\\\\\\ \displaystyle{\int_0^{\pi}(\cos(\theta)+\sin(\theta))\cdot\left(\dfrac{2^3}{3}-\dfrac{(-2)^3}{3}\right)\,d\theta}$

Calcule as potências, some e multiplique os valores

$\displaystyle{\int_0^{\pi}(\cos(\theta)+\sin(\theta))\cdot\left(\dfrac{8}{3}-\dfrac{-8}{3}\right)\,d\theta}\\\\\\ \displaystyle{\int_0^{\pi}(\cos(\theta)+\sin(\theta))\cdot\dfrac{16}{3}\,d\theta}$

Aplique a regra da constante

$\dfrac{16}{3}\cdot\displaystyle{\int_0^{\pi}(\cos(\theta)+\sin(\theta))}\,d\theta}$

Aplique a regra da soma: $\displaystyle{\int f(x)\pm g(x)\,dx=\int f(x)\,dx\pm \int g(x)\,dx$

$\dfrac{16}{3}\cdot\left(\displaystyle{\int_0^{\pi}\cos(\theta)\,d\theta+\int_0^{\pi}\sin(\theta)}\,d\theta}\right)$

Calcule as integrais, sabendo que $\displaystyle{\int\cos(\theta)\,d\theta=\sin(\theta)+C$ e $\displaystyle{\int\sin(\theta)\,d\theta=-\cos(\theta)+C$

$\dfrac{16}{3}\cdot\left(\sin(\theta)-\cos(\theta)\right)~\biggr|_0^{\pi}$

Aplique os limites de integração

$\dfrac{16}{3}\cdot[\sin(\pi)-\cos(\pi)-(\sin(0)-\cos(0))]$

Sabendo que $\sin(\pi)=\sin(0)=0$, $\cos(\pi)=-1$ e $\cos(0)=1$, teremos:

$\dfrac{16}{3}\cdot[0-(-1)-(0-1)]\\\\\\ \dfrac{16}{3}\cdot2\\\\\\ \dfrac{32}{3}$

Este é o resultado desta integral.