We are finding  double integration over the surface of a cube bounded by coordinate planes x = y = z = a and the plane x = y = z = a

∫∫[(x³ - yz) dy dz - 2x²y dz dx + zdxdy]

surfaces are s₁ : x = 0, s₂ : x = a, s₃: y = 0, s₄ : y = a, and s₅ : z = 0, s₆ : z = a

 \(\therefore\) ∫∫[(x³ - yz) dy dz - 2x²y dz dx + z dx dy]

 = (∫∫_s1 + ∫∫_s2 + ∫∫_s3 + ∫∫_s4 + ∫∫_s5 + ∫∫_s6) [(x³ - yz) dy dz - 2x²ydz dx + zdx dy]

= \(\int\limits_{y=0}^{y=a}\int\limits_{z=0}^{z=a}(-yz)dydz\) + \(\int\limits_{y=0}^{y=a}\int\limits_{z=0}^{z=a}(a^3-yz)dydz\) + \(\int\limits_{z=0}^{z=a}\int\limits_{x=0}^{x=a}(-2x^2\times0)dzdx\) 

+ \(\int\limits_{z=0}^{z=a}\int\limits_{x=0}^{x=a}(-2x^2a)dzdx\)+ \(\int\limits_{x=0}^{x=a}\int\limits_{y=0}^{y=a}0dxdy\) + \(\int\limits_{x=0}^{x=a}\int\limits_{y=0}^{y=a}adxdy\) 

 = \(\int\limits_{y=0}^{y=a}-[\frac{yz^2}2]^{z=a}_{z=0}dy\) + \(\int\limits_{y=0}^{y=a}-[a^3z-\frac{yz^2}2]^{z=a}_{z=0}dy\) + 0 - 2a \(\int\limits_{z=0}^{z=a}-[\frac{x^3}3]^{x=a}_{x=0}dz\) + 0 + a\(\int\limits_{x=0}^{x=a}[y]^{y=a}_{y=0}dx\) 

 = \(-\frac{a^2}2\int\limits_0^aydy+\int\limits_0^a(a^4-\frac{a^2y}2)dy\) - \(\frac{2a^4}3\int\limits_0^adz+a^2\int\limits_0^adx\) 

 = \(-\frac{a^2}2[\frac{y^2}2]_0^a+[a^4y - \frac{a^2y^2}4]_0^4\) - \(\frac{2a^4}3[z]_0^a+a^2[x]_0^a\) 

 = -\(\frac{a^4}4+a^5-\frac{a^4}4-\frac{2a^5}3+a^3\) 

 = \(\frac13a^5-\frac12a^4+a^3\)