Question

Calculate the surface integral, F.ds, where F = 2(x,y,z), and where S is the surface of the cube define by the relations 0≤x≤1, 0≤y≤1, 0≤z≤1.

Answer 1

Answer on Question#65286, Math / Differential Equations

Question

Calculate the surface integral of $\oint_{S} \vec{F} d\vec{s}$ , where $\vec{F} = 2(x, y, z)$ , and where $S$ is the surface of the cube defined by the relations $0 \leq x \leq 1, 0 \leq y \leq 1, 0 \leq z \leq 1$ .

Solution

According to divergence theorem:

$$\oint_{S} \vec{F} d\vec{s} = \iiint_{V} (\vec{\nabla} \cdot \vec{F}) dV$$ $$\vec{\nabla} \cdot \vec{F} = 2 \left( \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} \right) = 2(1 + 1 + 1) = 6$$ $$\iiint_{V} (\vec{\nabla} \cdot \vec{F}) dV = \iiint_{V} 6 dV = 6 \iiint_{V} dV = 6V$$

Volume of the unit cube is 1, hence

$$\oint_{S} \vec{F} d\vec{s} = 6$$

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