Calcule as integrais
a) \iiint_{R} y \sin (x y) z^{2} d V onde R=\left\{(x, y, z) \in \mathbb{R}^{3} \mid 1 \leq x \leq 3,0 \leq y \leq \pi / 2,-2 \leq z \leq 3\right\}

Passo 1

O enunciado pede a integral tripla
I = \iiint_{R} y\,\sin(xy)\,z^{2}\,dV
sobre o paralelepípedo
R=\{(x,y,z)\mid 1\le x\le3,\;0\le y\le \tfrac\pi2,\;-2\le z\le3\}.
O conteúdo é mudança de ordem e fatoração em integrais 1D (teorema de Fubini) e, para a parte em y, uso de integração por partes e reconhecimento de derivada de quociente.

Passo 2

Como o domínio é retangular, escrevemos
I =\int_{x=1}^{3}\int_{y=0}^{\pi/2}\int_{z=-2}^{3}y\,\sin(xy)\,z^{2}\,dz\,dy\,dx =\Bigl(\int_{z=-2}^{3}z^{2}\,dz\Bigr)\, \int_{x=1}^{3}\int_{y=0}^{\pi/2}y\,\sin(xy)\,dy\,dx.

Passo 3

Calculemos primeiro
\int_{z=-2}^{3}z^{2}\,dz =\Bigl[\tfrac{1}{3}z^{3}\Bigr]_{-2}^{3} =\tfrac{1}{3}(27-(-8)) =\tfrac{35}{3}.

Passo 4

Definamos
B(x)=\int_{0}^{\pi/2}y\,\sin(xy)\,dy.
Fazendo integração por partes com u=y, dv=\sin(xy)\,dy du=dy, v=-\tfrac{1}{x}\cos(xy), logo
B(x) =\Bigl[-\tfrac{y}{x}\cos(xy)\Bigr]_{0}^{\pi/2} +\int_{0}^{\pi/2}\tfrac{1}{x}\cos(xy)\,dy =-\tfrac{\pi}{2x}\cos\Bigl(\tfrac{\pi x}{2}\Bigr) +\tfrac{1}{x}\,\Bigl[\tfrac{1}{x}\sin(xy)\Bigr]_{0}^{\pi/2}
=-\tfrac{\pi}{2x}\cos\Bigl(\tfrac{\pi x}{2}\Bigr) +\tfrac{1}{x^{2}}\sin\Bigl(\tfrac{\pi x}{2}\Bigr).

Passo 5

Resta calcular
\int_{1}^{3}B(x)\,dx =\int_{1}^{3}\biggl( -\tfrac{\pi}{2x}\cos\bigl(\tfrac{\pi x}{2}\bigr) +\tfrac{1}{x^{2}}\sin\bigl(\tfrac{\pi x}{2}\bigr) \biggr)dx.
Note que
\frac{d}{dx}\Bigl(\frac{\sin(\tfrac{\pi x}{2})}{x}\Bigr) =\frac{\tfrac{\pi}{2}x\cos(\tfrac{\pi x}{2})-\sin(\tfrac{\pi x}{2})}{x^{2}} =\tfrac{\pi}{2x}\cos\bigl(\tfrac{\pi x}{2}\bigr) -\tfrac{1}{x^{2}}\sin\bigl(\tfrac{\pi x}{2}\bigr).
Logo o integrando é - dessa derivada, de modo que
\int_{1}^{3}B(x)\,dx =-\Bigl[\frac{\sin(\tfrac{\pi x}{2})}{x}\Bigr]_{1}^{3} =-\Bigl(\frac{\sin(\tfrac{3\pi}{2})}{3}-\frac{\sin(\tfrac{\pi}{2})}{1}\Bigr) =-\Bigl(-\tfrac{1}{3}-1\Bigr) =\tfrac{4}{3}.

Passo 6

Portanto
I =\frac{35}{3}\times\frac{4}{3} =\frac{140}{9}.

Resposta

\displaystyle \frac{140}{9}

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