Question

Using the divergence theorem evaluate ∫∫subscript s F^vector. dS^vector where F^vector = (y^2)z×i^cap + (y^3) j^cap+ xzk^cap and S is the surface of the cube defined by - 1< = x<=1 :-1<=y<=1 : 0<=z<=2.

Answer 1

Answer on Question #70060-Physics-Other

Using the divergence theorem evaluate $\int \text{subscript} s F^{\wedge} \text{vector. } dS^{\wedge} \text{vector where } F^{\wedge} \text{vector} = (y^{\wedge}2)z \times i^{\wedge} \text{cap} + (y^{\wedge}3)j^{\wedge} \text{cap} + xzk^{\wedge} \text{cap}$ and $S$ is the surface of the cube defined by $-1 <= x <= 1: -1 <= y <= 1: 0 <= z <= 2$.

Solution

Using the divergence theorem:

$$\begin{aligned} I &= \iint F \, dS = \iiint \operatorname{div} F \, dV \\ \operatorname{div} F &= \frac{\partial}{\partial x} \left( (y^2) z \right) + \frac{\partial}{\partial y} (y^3) + \frac{\partial}{\partial z} (xz) = 0 + 3y^2 + x = x + 3y^2 \\ I &= \int_{-1}^{1} dx \int_{-1}^{1} dy \int_{0}^{2} dz \, (x + 3y^2) = \int_{-1}^{1} x \, dx \int_{-1}^{1} dy \int_{0}^{2} dz + \int_{-1}^{1} dx \int_{-1}^{1} 3y^2 \, dy \int_{0}^{2} dz \\ &= \int_{-1}^{1} x \, dx \left(1 - (-1)\right) (2 - 0) + \left(1 - (-1)\right) \int_{-1}^{1} 3y^2 \, dy \, (2 - 0) = 4\left(\frac{x^2}{2}\right)_{-1}^{1} + 4(y^3)_{-1}^{1} \\ &= 4\left(\frac{1}{2} - \frac{1}{2}\right) + 4(1 - (-1)) = 0 + 8 = 8. \end{aligned}$$

Answer: 8.

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