Calcule \iint_{R} f(x, y) d x d y para:
f(x, y)=x y, \mathbb{R}=\left\{(x, y) \in \mathbb{R}^{2} \mid x^{2}+y^{2} \leqslant 2, y \leqslant x, x \geqslant 0\right\}.

Passo 1

O enunciado pede para calcular
\iint_{R}f(x,y)\,dx\,dy
com f(x,y)=xy e
R=\{(x,y)\in\mathbb{R}^2\mid x^2+y^2\le2,\;y\le x,\;x\ge0\}.
O conteúdo é integral dupla em coordenadas polares, pois a condição x^2+y^2\le2 sugere usar x=r\cos\theta, y=r\sin\theta.

Passo 2

Em coordenadas polares temos
x^2+y^2\le2\;\Longrightarrow\;0\le r\le\sqrt{2},
e as restrições
x\ge0\;\Longrightarrow\;\cos\theta\ge0\;\Longrightarrow\;\theta\in\bigl[-\tfrac{\pi}{2},\tfrac{\pi}{2}\bigr],
y\le x\;\Longrightarrow\;\sin\theta\le\cos\theta\;\Longrightarrow\;\tan\theta\le1\;\Longrightarrow\;\theta\le\tfrac{\pi}{4}.
Logo o ângulo varia de \theta=-\tfrac{\pi}{2} até \theta=\tfrac{\pi}{4}. A integral fica
\iint_{R}xy\,dx\,dy =\int_{\theta=-\frac{\pi}{2}}^{\frac{\pi}{4}}\int_{r=0}^{\sqrt2} \bigl(r\cos\theta\bigr)\bigl(r\sin\theta\bigr)\;r\,dr\,d\theta.

Passo 3

No integrando, xy\,dx\,dy vira
r\cos\theta\;\cdot\;r\sin\theta\;\cdot\;r\,dr\,d\theta = r^3\sin\theta\cos\theta\,dr\,d\theta.
Separemos a parte em r:
\int_{0}^{\sqrt2}r^3\,dr =\left[\frac{r^4}{4}\right]_{0}^{\sqrt2} =\frac{(\sqrt2)^4}{4} =\frac{4}{4} =1.
Logo sobra a integral em \theta:
\iint_{R}xy\,dx\,dy =\int_{-\frac{\pi}{2}}^{\frac{\pi}{4}}\sin\theta\cos\theta\,d\theta.

Passo 4

Usamos \sin\theta\cos\theta=\tfrac12\sin(2\theta), então
\int_{-\frac{\pi}{2}}^{\frac{\pi}{4}}\sin\theta\cos\theta\,d\theta =\frac12\int_{-\frac{\pi}{2}}^{\frac{\pi}{4}}\sin(2\theta)\,d\theta =\frac12\left[-\tfrac12\cos(2\theta)\right]_{-\frac{\pi}{2}}^{\frac{\pi}{4}} =-\frac14\bigl[\cos(2\theta)\bigr]_{-\frac{\pi}{2}}^{\frac{\pi}{4}}.
Calculando os cossenos:
\cos\bigl(2\cdot\tfrac{\pi}{4}\bigr)=\cos\bigl(\tfrac{\pi}{2}\bigr)=0, \quad \cos\bigl(2\cdot(-\tfrac{\pi}{2})\bigr)=\cos(-\pi)=\cos\pi=-1.
Logo
-\frac14\bigl[\,0-(-1)\bigr] =-\frac14(1) =-\frac14.

Resposta

\iint_{R}xy\,dx\,dy=-\frac14.

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