Resolva: Calcule V dado por \(V=\displaystyle\int_{x=0}^{3}\displaystyle\int_{y=0}^{3-x}(3-x-y)\,dy\,dx\).

Queremos calcular

$$V=\int_{0}^{3}\int_{0}^{3-x}(3-x-y)\,dy\,dx.$$

1. Integral interna (em relação a $y$):

$$\int_{0}^{3-x}(3-x-y)\,dy =\left[(3-x)y-\frac{y^2}{2}\right]_{0}^{3-x}.$$

Substituindo $y=3-x$:

$$(3-x)(3-x)-\frac{(3-x)^2}{2}= (3-x)^2\left(1-\frac12\right)=\frac{(3-x)^2}{2}.$$

Logo,

$$V=\int_{0}^{3}\frac{(3-x)^2}{2}\,dx.$$

2. Integral externa (em relação a $x$):

$$V=\frac12\int_{0}^{3}(3-x)^2\,dx.$$

Faça $u=3-x\Rightarrow du=-dx$. Quando $x=0$, $u=3$; quando $x=3$, $u=0$:

$$\int_{0}^{3}(3-x)^2\,dx=\int_{3}^{0}u^2(-du)=\int_{0}^{3}u^2\,du=\left[\frac{u^3}{3}\right]_{0}^{3}=\frac{27}{3}=9.$$

Então,

$$V=\frac12\cdot 9=\frac{9}{2}.$$