Question

calcule integral tripla x^2+yz dydxdz , -3<y<0 , 0<x<2 , -1<z<1

Answer 1

$\displaystyle I=\int_{-3}^{0}dy\int_{0}^{2}dx\int_{-1}^{1}x^2+yz\,dz\\ \\ I=\int_{-3}^{0}dy\int_{0}^{2}\left.\left(x^2z+y\frac{z^2}{2}\right)\right|_{z=-1}^{z=1}dx\\ \\ I=\int_{-3}^{0}dy\int_{0}^{2}\left(x^2+\frac{y}{2}+x^2-\frac{y}{2}\right)dx\\ \\ I=\int_{-3}^{0}dy\int_{0}^{2}2x^2\, dx\\ \\ I=\int_{-3}^{0}\left.\left(\frac{2}{3}x^3\right)\right|_{0}^{2}dy\\ \\ I=\int_{-3}^{0}\frac{16}{3}dy$

$\displaystyle I=\frac{16}{3}(0-(-3))\\ \\ \\ \boxed{I=16}$