Question

Calcule a integral multipla

xzcosz dxdydz

limites de integração de x,y e z respectivamente:

x= de 1 a 0
y= de 1 a -1
z= de pi/2 a 0

Answer 1

$\int\limits^1_0 {} \, \int\limits^1_ \frac{-1}{} {} \, \int\limits^ \frac{ \pi }{2} _0 {} \, xzCos(z)dzdydx$

$\int\limits^1_0 {xdx} \, \int\limits^1_ \frac{-1}{} {dy} \, \int\limits^ \frac{ \pi }{2} _0 {} \, zCos(z)dz$

Calculando a integral em "Z"

Resolvendo por partes.

u = z
du = dz

dv = Coszdz
v = Senz

∫ zCos(z)dz = uv-∫vdu

= zSen(z) - ∫ Sen(z)dz

= zSen(z) - (-Cosz)

= zSen(z) + Cos(z)
--------------------------

Substituindo o limite de integração

zSen(z) + Cos(z) | (0, π/2)

π/2×Sen(π/2) + Cos(π/2) - 0 - Cos(0)

π/2 - 1
------------------------

Então ficará:

$\\= \int\limits^1_0 {xdx} \, \int\limits^1_ \frac{-1}{} {dy} \, ( \frac{ \pi }{2} -1) \\ \\ =( \frac{ \pi }{2} -1) \int\limits^1_0 {xdx} \, \int\limits^1_ \frac{-1}{} {dy} \, \\ \\ = ( \frac{ \pi }{2} -1) \int\limits^1_0 {xdx} \, [y](-1,1) \\ \\ = ( \frac{ \pi }{2} -1) \int\limits^1_0 {xdx} \, [1-(-1)] \\ \\ = ( \frac{ \pi }{2} -1) \int\limits^1_0 {2xdx} \, \\ \\ = ( \frac{ \pi }{2} -1)[ \frac{2x^2}{2} ]|(0,1) \\ \\ = ( \frac{ \pi }{2} -1)[ x^2](0,1)$

$\\ = ( \frac{ \pi }{2} -1)[ 1-0) \\ \\ = ( \frac{ \pi }{2} -1)[1] \\ \\ = \frac{ \pi }{2} -1$