[tex3]x^2 + 4y^2 \leq 4[/tex3] e [tex3]x+y \leq z \leq x + y + 1[/tex3]
Bem, eu faço da seguinte forma:
[tex3]x^2 + 4y^2 = 4 \Rightarrow \frac{x^2}{4} + y^2 = 1[/tex3] (elipse)
[tex3]x = a \cdot r \cdot cos\theta \Rightarrow 2 \cdot r \cdot cos\theta\\ y = b \cdot r \cdot sin\theta \Rightarrow r \cdot sin\theta \\ J = a \cdot b \cdot r \Rightarrow 2r[/tex3]
Logo,
[tex3]\frac{\cancel{2^2} \cdot cos^{2}\theta}{\cancel{2^2}}+ sin^{2}\theta = 1[/tex3] com o [tex3]J = 2r[/tex3]
[tex3]V = \int_{0}^{2\pi} \int_{0}^{1} \left(x+y+1 \right) - \left( x + y \right)2r \,dr\,d\theta[/tex3]
[tex3]= \int_{0}^{2\pi}\int_{0}^{1} 1 \cdot 2r \,dr\,d\theta[/tex3]
[tex3]=\int_0^{2\pi} \left[\int_0^1 2r \,dr\,d\theta\right][/tex3]
[tex3]= \int_0^{2\pi}\left[ \cancel{2} \left.\frac{r^{2}}{\cancel{2}}\right|_0^1 \right]\,d\theta[/tex3]
[tex3]= \int_0^{2\pi}1\,d\theta = \left.\theta\right|_0^{2\pi} = 2\pi[/tex3]
Logo,[tex3]V = \int_{0}^{2\pi} \int_{0}^{1} \left(x+y+1 \right) - \left( x + y \right)2r \,dr\,d\theta = 2\pi \blacktriangleleft[/tex3]
Resposta
[tex3]2\pi[/tex3]