Evaluate the triple integral triple integralE x6ey dV where

Evaluate the triple integral triple integral_E x^6e^y dV where E is bounded by the parabolic cylinder z = 25 - y^2 and the planes z = 0, x = 5, and x = -5.

Solution

Solution :

x⁶e^y dV

0 z 25 - y²

when z = 0, y² = 25 ==> y = ± 5 , -5 y 5

- 5 x 5

x⁶e^y dV = x⁶e^y dz dx dy  

= x⁶e^y [ z ] dx dy ( z from 0 to 25 - y²)

= x⁶e^y ( 25 - y² ) dx dy

= e^y ( 25 - y² ) [ x⁷/7 ] dy ( x from - 5 to 5 )

= e^y ( 25 - y² ) [ (1/7) - ( - 1/7) ] dy ( x from - 5 to 5 )

= 2/7e^y ( 25 - y² )] dy

Integrate by parts
u = 25 - y²
du = - 2y dy

dv = e^y dy
v = e^y

2/7e^y ( 25 - y² )] dy = 2/7 [ e^y(25 - y²) - e^y ( - 2y )] dy ]

= 2/7 [ e^y(25 - y²) + 2 e^y y dy ]

Again integrate by parts

u = y
du = dy

dv = e^y dy
v = e^y

= 2/7 [ e^y(25 - y²) + 2 { ye^y - e^y dy} ]

= 2/7 [ 25e^y - e^y y² + 2 y e^y - 2e^y ]

= 2/7 [ e^y { 2y - y² + 23} ] from - 5 to 5

= 2/7 [ e⁵(10 - 25 + 23) - 1/e⁵(- 10 - 25 + 23) ]

= 2/7 [ 8e⁵ - 1/e⁵(- 12) ]

= 2/7 [ 8e⁵ + 12/e⁵ ]

= 2( 8e¹⁰ + 12 )/7e⁵

= 339.2532