Question

verify gauss divergence theorem for F=x²i+zj+yzk taken over the cube bounded by x=0, x=1, y=0, y=1, z=0 and z=1.​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Given: Vector $\vec{F} = x^{2} \hat{i} + z \hat{j} + yz \hat{k}$

Dimensions of the cube: $x = 0$, $x = 1$, $y = 0$, $y = 1$, $z = 0$ and $z = 1$

To Prove: Gauss divergence theorem

Solution:

• Gauss divergence theorem

Mathematical expression for the Gauss divergence theorem is given below;

$\iint_s \vec{F} \cdot \hat{n} \ ds = \iiint_V(\vec{\nabla} \cdot \vec{F}) \ dV$

where $\vec{F}$ is the vector field; to prove the Gauss divergence theorem, we need to show that the L.H.S and the R.H.S of the theorem are equal.

• Solving L.H.S. of the theorem for the vector $\vec{F} = x^{2} \hat{i} + z \hat{j} + yz \hat{k}$

For the given vector $\vec{F} = x^{2} \hat{x} + z \hat{y} + yz \hat{k}$ and dimensions of the cube (figure attached), we can write;

$\iint_s \vec{F} \cdot \hat{n} \ ds = \iint_{ABCD} \vec{F} \cdot \hat{n} \ ds + \iint_{EFGH} \vec{F} \cdot \hat{n} \ ds + \iint_{BCFE} \vec{F} \cdot \hat{n} \ ds \\\\ + \iint_{ADGH} \vec{F} \cdot \hat{n} \ ds + \iint_{CDGF} \vec{F} \cdot \hat{n} \ ds + \iint_{ABEH} \vec{F} \cdot \hat{n} \ ds$

For each side of a cube, we can replace $\hat{n}$ with the unit vector of the respective side. Therefore,

$\Rightarrow \iint_s \vec{F} \cdot \hat{n} \ ds = \iint_{ABCD} \vec{F} \cdot \hat{k} \ ds + \iint_{EFGH} \vec{F} \cdot (-\hat{k}) \ ds + \iint_{BCFE} \vec{F} \cdot \hat{i} \ ds \\\\ + \iint_{ADGH} \vec{F} \cdot (-\hat{i}) \ ds + \iint_{CDGF} \vec{F} \cdot \hat{j} \ ds + \iint_{ABEH} \vec{F} \cdot (-\hat{j}) \ ds$

$\Rightarrow \iint_s \vec{F} \cdot \hat{n} \ ds = \int\limits^1_{y=0} \int\limits^1_{x=0} {yz} \, dx dy \ - \ \int\limits^1_{y=0} \int\limits^1_{x=0} {yz} \, dx dy \ + \ \int\limits^1_{z=0} \int\limits^1_{y=0} {x^{2} } \, dy dz \ - \ \int\limits^1_{z=0} \int\limits^1_{y=0} {x^{2} } \, dy dz \ + \ \int\limits^1_{z=0} \int\limits^1_{x=0} {z} \, dx dz \ - \ \int\limits^1_{z=0} \int\limits^1_{x=0} {z} \, dx dz$

For ABCD ⇒ z = 1, EFGH ⇒ z = 0, BCFE ⇒ x = 1, ADGH ⇒ x = 0, CDGF ⇒ y = 1, and ABEH ⇒ y = 0. Substituting these values in the above expression and solving it, you will get;

$\Rightarrow \iint_s \vec{F} \cdot \hat{n} \ ds = \frac{3}{2}$

• Solving R.H.S. of the theorem for the vector $\vec{F} = x^{2} \hat{i} + z \hat{j} + yz \hat{k}$

Considering $\iiint_V(\vec{\nabla} \cdot \vec{F}) \ dV$, we can write,

$\iiint_V(\vec{\nabla} \cdot \vec{F}) \ dV = \iiint_V(\vec{\nabla} \cdot (x^{2} \hat{i} + z \hat{j} + yz \hat{k})) \ dV = \int\limits^1_{z=0} \int\limits^1_{y=0} \int\limits^1_{x=0}{(2x+y)} \, dxdydz$

Solving the above expression further to get;

$\iiint_V(\vec{\nabla} \cdot \vec{F}) \ dV = \frac{3}{2}$ which is equal to L.H.S.

Hence, the gauss divergence theorem is proved with $\iint_s \vec{F} \cdot \hat{n} \ ds = \iiint_V(\vec{\nabla} \cdot \vec{F}) \ dV = \frac{3}{2}$