Calcule as integrais duplas na região especificada.
\left.\iint_{R}(x \operatorname{sen} y-\operatorname{sen} x)\right) \mathrm{dA}, R=\left\{(x, y) \left\lvert\, 0 \leq x \leq \frac{\pi}{2}\right., 0 \leq y \leq \frac{\pi}{3}\right\}

Passo 1

O enunciado pede a integral dupla de f(x,y)=x\,\operatorname{sen}y-\operatorname{sen}x sobre o retângulo
R=\bigl\{(x,y)\mid 0\le x\le \tfrac{\pi}{2},\;0\le y\le \tfrac{\pi}{3}\bigr\}.
O conteúdo principal aqui é a integração iterada em uma região retangular, que nos permite separar a integral dupla em integrais simples.

Passo 2

Escrevendo como integral iterada e separado em duas partes, temos
\iint_{R}(x\,\operatorname{sen}y-\operatorname{sen}x)\,dA =\int_{x=0}^{\frac{\pi}{2}}\!\int_{y=0}^{\frac{\pi}{3}}\bigl(x\,\operatorname{sen}y-\operatorname{sen}x\bigr)\,dy\,dx =I_{1}-I_{2},
onde
I_{1}=\int_{0}^{\frac{\pi}{2}}\!\int_{0}^{\frac{\pi}{3}}x\,\operatorname{sen}y\,dy\,dx, \qquad I_{2}=\int_{0}^{\frac{\pi}{2}}\!\int_{0}^{\frac{\pi}{3}}\operatorname{sen}x\,dy\,dx.

Passo 3

Calcular I_{1}. Como x é constante ao integrar em y:
I_{1} =\int_{0}^{\frac{\pi}{2}}x\Bigl[\int_{0}^{\frac{\pi}{3}}\operatorname{sen}y\,dy\Bigr]dx =\int_{0}^{\frac{\pi}{2}}x\bigl[-\cos y\bigr]_{0}^{\frac{\pi}{3}}\,dx =\int_{0}^{\frac{\pi}{2}}x\bigl(-\cos\tfrac{\pi}{3}+\cos0\bigr)\,dx.
Sabendo \cos\tfrac{\pi}{3}=\tfrac12 e \cos0=1, o fator é 1-\tfrac12=\tfrac12. Então
I_{1} =\tfrac12\int_{0}^{\frac{\pi}{2}}x\,dx =\tfrac12\Bigl[\tfrac{x^{2}}{2}\Bigr]_{0}^{\frac{\pi}{2}} =\tfrac12\cdot\tfrac{(\pi/2)^{2}}{2} =\tfrac{\pi^{2}}{16}.

Passo 4

Calcular I_{2}. Aqui \operatorname{sen}x independe de y:
I_{2} =\int_{0}^{\frac{\pi}{2}}\operatorname{sen}x\Bigl[\int_{0}^{\frac{\pi}{3}}dy\Bigr]dx =\int_{0}^{\frac{\pi}{2}}\operatorname{sen}x\cdot\tfrac{\pi}{3}\,dx =\tfrac{\pi}{3}\int_{0}^{\frac{\pi}{2}}\operatorname{sen}x\,dx =\tfrac{\pi}{3}\bigl[-\cos x\bigr]_{0}^{\frac{\pi}{2}} =\tfrac{\pi}{3}\,(1) =\tfrac{\pi}{3}.

Passo 5

Agora basta subtrair as duas partes:
\iint_{R}(x\,\operatorname{sen}y-\operatorname{sen}x)\,dA =I_{1}-I_{2} =\frac{\pi^{2}}{16}-\frac{\pi}{3}.
Opcionalmente, em denominador comum:
\frac{\pi^{2}}{16}-\frac{\pi}{3} =\frac{3\pi^{2}-16\pi}{48}.

Resposta

\displaystyle \frac{\pi^{2}}{16}-\frac{\pi}{3}.

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