Question

seja o corpo Q delimitado pelo cilindro: y=x² e os planos y+z=0 e z=0.
a) Calcula o volume do corpo dito se:
0<=z<=4-y; x²<=y<=4; -2<=x<=2

Answer 1

$\\ \textrm{V} =\displaystyle\int\limits{}\displaystyle\int\limits{}\displaystyle \int\limits_E {}\textrm{dV} \\ \\ = \displaystyle \int\limits^2_{-2} {} \displaystyle \int\limits^4_{x^2}\,\displaystyle \int\limits^{4-y}_{0} \textrm{dzdydx} = \displaystyle \int\limits^2_{-2} {} \displaystyle \int\limits^4_{x^2} z \big|_{0}^{4-y} dydx \\ \\ = \displaystyle \int\limits^2_{-2} {} \displaystyle \int\limits^4_{x^2} (4-y) dydx$

$\\ = \displaystyle \int\limits^2_{-2} {} \left(4y - \frac{y^2}{2} \right) \big|_{x^2}^4 dx \\ \\ = \displaystyle \int\limits^2_{-2} {} \left(4\cdot(4) - \frac{(4)^2}{2} \right) - \left(4\cdot(x^2) - \frac{(x^2)^2}{2} \right) dx \\ \\ = \displaystyle \int\limits^2_{-2} {} \left( 8 - \left( 4x^2 - \frac{x^4}{2} \right) \right)dx =\displaystyle \int\limits^2_{-2} {} \left( \frac{x^4}{2} -4x^2 +8 \right) dx \\ \\ \textbf{Como temos uma funcao "par"}$

$\\ = 2 \cdot\displaystyle \int\limits^2_{0} {} \left( \frac{x^4}{2} -4x^2 +8 \right) dx \\ \\ = 2 \cdot \left( \frac{x^5}{10} - \frac{4x^3}{3} +8x \right) \big|_{0}^2 \\ \\ = 2 \cdot \left( \frac{2^5}{10} - \frac{4\cdot(2)^3}{3} +8\cdot(2) \right) \\ \\ = 2 \cdot \left( \frac{32}{10} - \frac{32}{3} +16\right) \\ \\ = 2 \cdot \dispstyle \frac{128}{15} = \boxed{ \frac{256}{15} u.v}$