Integral over a cube 2

Calculus Level pending

Integrate the function f ( x , y , z ) = ( x y z ) 2 f(x,y,z)=(xyz)^2 over a cube of side length 6 6 centered at the origin.

If the result is a 3 a^3 , enter a a .


The answer is 18.

Charley Shi by Charley Shi , 19, New Zealand

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1 solution

Tom Engelsman
Apr 14, 2020

If we focus on the first octant of this cube (i.e. 0 ≤ x , y , z ≤ 3 0 \le x,y,z \le 3 ), then the required integral becomes:

8 ⋅ ∫ 0 3 ∫ 0 3 ∫ 0 3 ( x y z ) 2 d z d y d x = 8 ⋅ ( ∫ 0 3 x 2 d x ) 3 = 8 ⋅ 9 3 = ( 2 ⋅ 9 ) 3 = 1 8 3 . 8 \cdot \int_{0}^{3} \int_{0}^{3} \int_{0}^{3} (xyz)^{2} dzdydx = 8 \cdot (\int_{0}^{3} x^2 dx)^{3} = 8 \cdot 9^{3} = (2 \cdot 9)^3 = \boxed{18^3}.

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