Calcule as seguintes integrais no paralelepípedo retangular dado:
\iiint_{W} x y z^{2} \cos \left(x y^{2} z\right) d V, onde W=[2,3] \times[0, \sqrt{\pi} / 2] \times[1,2].

Passo 1

O enunciado pede o valor da integral
I = \iiint_{W} x\,y\,z^{2}\cos\bigl(xy^{2}z\bigr)\,dV,
onde W=[2,3]\times[0,\sqrt{\pi}/2]\times[1,2].
O conteúdo é mudança na ordem de integração em paralelepípedo retangular e uso de substituições simples para integrais de tipo \int y\cos(y^{2})dy e \int\sin(ax)dx.

Passo 2

Vamos integrar primeiro em y, mantendo x e z fixos. Escrevemos
I=\int_{x=2}^{3}\int_{z=1}^{2}\int_{y=0}^{\sqrt{\pi}/2}x\,y\,z^{2}\cos\bigl(xy^{2}z\bigr)\,dy\,dz\,dx.
Dentro, seja
u = x\,y^{2}z \quad\Longrightarrow\quad du = 2\,x\,y\,z\,dy \;\Longrightarrow\; x\,y\,dy = \frac{du}{2z}.
Logo,
\int_{y=0}^{\sqrt{\pi}/2}x\,y\,z^{2}\cos\bigl(xy^{2}z\bigr)\,dy = z^{2}\int_{u=0}^{u=x\frac{\pi}{4}z}\frac{\cos u}{2z}\,du = \frac{z}{2}\,\Bigl[\sin u\Bigr]_{0}^{x\frac{\pi}{4}z} = \frac{z}{2}\,\sin\!\Bigl(\tfrac{\pi}{4}x\,z\Bigr).

Passo 3

Substituindo de volta, agora a integral fica
I=\int_{x=2}^{3}\int_{z=1}^{2}\frac{z}{2}\,\sin\!\Bigl(\tfrac{\pi}{4}x\,z\Bigr)\,dz\,dx =\frac12\int_{z=1}^{2}z\Bigl[\int_{x=2}^{3}\sin\!\bigl(\tfrac{\pi}{4}z\,x\bigr)\,dx\Bigr]dz.
Para a integral em x, use
\int\sin(a\,x)\,dx = -\frac{\cos(a\,x)}{a}.
Aqui a=\tfrac{\pi}{4}z, então
\int_{x=2}^{3}\sin\!\bigl(\tfrac{\pi}{4}z\,x\bigr)\,dx = \Bigl[-\frac{\cos(\tfrac{\pi}{4}z\,x)}{\tfrac{\pi}{4}z}\Bigr]_{2}^{3} = \frac{4}{\pi z}\Bigl[\cos\!\bigl(\tfrac{\pi}{2}z\bigr)-\cos\!\bigl(\tfrac{3\pi}{4}z\bigr)\Bigr].

Passo 4

Substituindo isso em I, obtemos
I =\frac12\int_{1}^{2}z\;\frac{4}{\pi z}\bigl[\cos(\tfrac{\pi}{2}z)-\cos(\tfrac{3\pi}{4}z)\bigr]\,dz =\frac{2}{\pi}\int_{1}^{2}\bigl[\cos(\tfrac{\pi}{2}z)-\cos(\tfrac{3\pi}{4}z)\bigr]dz.
Agora calcule cada termo:
\int_{1}^{2}\cos(\tfrac{\pi}{2}z)\,dz =\Bigl[\frac{2}{\pi}\sin(\tfrac{\pi}{2}z)\Bigr]_{1}^{2} =\frac{2}{\pi}\bigl(\sin\pi-\sin\tfrac{\pi}{2}\bigr) =\frac{2}{\pi}(0-1) =-\frac{2}{\pi},
\int_{1}^{2}\cos(\tfrac{3\pi}{4}z)\,dz =\Bigl[\frac{4}{3\pi}\sin(\tfrac{3\pi}{4}z)\Bigr]_{1}^{2} =\frac{4}{3\pi}\bigl(\sin\tfrac{3\pi}{2}-\sin\tfrac{3\pi}{4}\bigr) =\frac{4}{3\pi}\bigl(-1-\tfrac{\sqrt2}{2}\bigr) =-\frac{2(2+\sqrt2)}{3\pi}.
Portanto,
I =\frac{2}{\pi}\Bigl(-\frac{2}{\pi} -\bigl(-\tfrac{2(2+\sqrt2)}{3\pi}\bigr)\Bigr) =\frac{2}{\pi}\,\frac{-2+\tfrac{2(2+\sqrt2)}{3}}{\pi} =\frac{4(\sqrt2-1)}{3\pi^{2}}.

Resposta

\boxed{\frac{4(\sqrt{2}-1)}{3\pi^{2}}.}

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