Question

$a\ integral\ \int\limits^{ \sqrt{2} } _0 \ \ \ \int\limits^{3y}_0\ \ \ \int\limits^{8- x^{z} - y^{z}} _{ x^{z}+3 y^{z} \, \ dzdxdy,$

vale :

a) -6
b) -4
c) -2
d) 6
e) 8

Answer 1

$\int\limits^ \frac{ \sqrt{2} }{} _0 {} \, \int\limits^ \frac{3y}{} _0 {} \, \int\limits^ \frac{8-x^2-y^2}{} _ \frac{x^2+3y^2}{} {} \, dzdxdy = \int\limits^ \frac{ \sqrt{2} }{} _0 {} \, \int\limits^ \frac{3y}{} _0 {} \,z|(x^2+3y^2, 8-x^2-y^2)dxdy \\ \\ = \int\limits^ \frac{ \sqrt{2} }{} _0 {} \, \int\limits^ \frac{3y}{} _0 {} \,(8-x^2-y^2)-(x^2+3y^2)dxdy \\ \\ = \int\limits^ \frac{ \sqrt{2} }{} _0 {} \, \int\limits^ \frac{3y}{} _0 {} \,(8-2x^2-4y^2)dxdy$

$\\= 2 \int\limits^ \frac{ \sqrt{2} }{} _0 {} \, \int\limits^ \frac{3y}{} _0 {} \,(4-x^2-2y^2)dxdy \\ \\ = 2\int\limits^ \frac{ \sqrt{2} }{} _0 {} \, (4x- \frac{x^3}{3} -2y^2x)|(0,3y)dy \\ \\ = 2\int\limits^ \frac{ \sqrt{2} }{} _0 {} \, (4.3y- \frac{(3y)^3}{3} -2y^2.3y)dy \\ \\ = 2\int\limits^ \frac{ \sqrt{2} }{} _0 {} \, (12y- 9y^3 -6y^3)dy \\ \\ = 2.3\int\limits^ \frac{ \sqrt{2} }{} _0 {} \, (4y- 3y^3 -2y^3)dy$

$\\ =6\int\limits^ \frac{ \sqrt{2} }{} _0 {} \, (4y- 3y^3 -2y^3)dy \\ \\ =6\int\limits^ \frac{ \sqrt{2} }{} _0 {} \, (4y- 5y^3)dy \\ \\ =6( \frac{4y^2}{2} - \frac{5y^4}{4} )|(0, \sqrt{2} ) \\ \\ =6(2y^2- \frac{5y^4}{4})|(0, \sqrt{2} ) \\ \\ =6(2. \sqrt{2^2} - \frac{5. \sqrt{2^4} }{4}) \\ \\ =6(2.2- \frac{5.2^2}{4} ) \\ \\ =6(4-5) \\ \\ = -6$