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Let E(x,y,z)=2x²î+5yĵ+3zk̂. The value of ∭_V (∇·E) dV, where V is the volume enclosed by the unit cube defined by 0≤ x≤ 1, 0≤ y≤ 1, and 0≤ z≤ 1, is

1. 3
2. 8
3. 10
4. 5

Correct answer: 10

Solution

div E = d(2x^2)/dx + d(5y)/dy + d(3z)/dz = 4x + 5 + 3 = 4x + 8. Integrating 4x+8 over the unit cube gives integral of (4x+8) dx from 0 to 1 = 2 + 8 = 10. Stored answer 8 is wrong.

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