Let \( S \) be a closed surface enclosing a volume \( V \). If \( \bar{F} \) is a continuously differentiable vector point function, then
\[ \int\limits_{V} \operatorname{div} \bar{F} \, dv = \iint\limits_{S} \bar{F} \cdot \bar{n} \, ds \]
Where \( \bar{n} \) is the outward drawn normal vector at any point on \( S \).
Cartesian Form
Let
\[ \bar{F} = F_1 \hat{i} + F_2 \hat{j} + F_3 \hat{k}, \quad \bar{n} = \cos \alpha \hat{i} + \cos \beta \hat{j} + \cos \gamma \hat{k} \]
where \( \cos \alpha, \cos \beta, \cos \gamma \) are the directional cosines of \( \bar{n} \).
Then, \[ \bar{F} \cdot \bar{n} = F_1 \cos \alpha + F_2 \cos \beta + F_3 \cos \gamma \] and \[ \operatorname{div} \bar{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \]
Hence, the divergence theorem can be written as
\[ \iiint \left( \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \right) dx\,dy\,dz = \iint_S \left( F_1 \cos \alpha + F_2 \cos \beta + F_3 \cos \gamma \right) ds \]
or equivalently,
\[ \iint_S \left( F_1 \, dy\,dz + F_2 \, dz\,dx + F_3 \, dx\,dy \right) \]
Note: Gauss’s divergence theorem is applicable only for closed surfaces.
Example 1: Evaluate \[ \iint\limits_{S} \left( x^3 \, dy\,dz + x^2 y \, dz\,dx + x^2 z \, dx\,dy \right) \] where \( S \) is the closed surface of the cylinder \( x^2 + y^2 = a^2 \), bounded between the circular discs \( z = 0 \) and \( z = b \).
Solution:\[ \iiint\limits_V \operatorname{div} \bar{F} \, dv = \iint\limits_S \bar{F} \cdot \bar{n} \, ds \] where \[ \bar{F} = x^3 \hat{i} + x^2 y \hat{j} + x^2 z \hat{k} \] and \[ \operatorname{div} \bar{F} = \frac{\partial x^3}{\partial x} + \frac{\partial (x^2 y)}{\partial y} + \frac{\partial (x^2 z)}{\partial z} = 3x^2 + x^2 + x^2 = 5x^2 \]
So, \[ \iint\limits_S \left( x^3 \, dy\,dz + x^2 y \, dz\,dx + x^2 z \, dx\,dy \right) = 5 \iiint\limits_V x^2 \, dz\,dy\,dx \]
Converting to rectangular coordinates:
\[ = 5 \int_{x=-a}^{a} \int_{y=-\sqrt{a^2 - x^2}}^{\sqrt{a^2 - x^2}} \int_{z=0}^{b} x^2 \, dz\,dy\,dx \]
\[ = 5 \int_{x=-a}^{a} \int_{y=-\sqrt{a^2 - x^2}}^{\sqrt{a^2 - x^2}} x^2 (z)\big|_0^b \, dy\,dx = 5b \int_{x=-a}^{a} x^2 \left( y \big|_{-\sqrt{a^2 - x^2}}^{\sqrt{a^2 - x^2}} \right) dx \]
\[ = 10b \int_{x=-a}^{a} x^2 \sqrt{a^2 - x^2} \, dx = 20b \int_{x=0}^{a} x^2 \sqrt{a^2 - x^2} \, dx \]
Now using the substitution \( x = a \sin \theta \), \( dx = a \cos \theta \, d\theta \):
\[ x^2 = a^2 \sin^2 \theta,\quad \sqrt{a^2 - x^2} = a \cos \theta \]
So the integral becomes: \[ = 20b \int_{\theta=0}^{\pi/2} a^2 \sin^2 \theta \cdot a \cos \theta \cdot a \cos \theta \, d\theta = 20a^4 b \int_0^{\pi/2} \sin^2 \theta \cos^2 \theta \, d\theta \]
Using the identity: \[ \int_0^{\pi/2} \sin^2 \theta \cos^2 \theta \, d\theta = \frac{1}{4} \int_0^{\pi/2} \sin^2 2\theta \, d\theta = \frac{1}{8} \int_0^{\pi/2} (1 - \cos 4\theta) \, d\theta = \frac{1}{8} \cdot \frac{\pi}{2} \]
Final answer:
\[ \iint\limits_S \left( x^3 \, dy\,dz + x^2 y \, dz\,dx + x^2 z \, dx\,dy \right) = 20a^4 b \cdot \frac{1}{8} \cdot \frac{\pi}{2} = \frac{5\pi}{2} a^4 b \]
Example 2: Evaluate: \[ \iint\limits_{S} \bar{F} \cdot \bar{n} \, ds \] where \[ \bar{F} = 4x \hat{i} - 2y^2 \hat{j} + z^2 \hat{k} \] and \( S \) is the surface enclosing the cylinder \( x^2 + y^2 = 4 \), bounded between \( z = 0 \) and \( z = 3 \).
Solution: \[ \iint\limits_S \bar{F} \cdot \bar{n} \, ds = \iiint\limits_V \operatorname{div} \bar{F} \, dv \] \[ \operatorname{div} \bar{F} = \frac{\partial (4x)}{\partial x} + \frac{\partial (-2y^2)}{\partial y} + \frac{\partial (z^2)}{\partial z} = 4 - 4y + 2z \]Switch to Cylindrical Coordinates:
\[ x = r\cos\theta, \quad y = r\sin\theta, \quad dx\,dy\,dz = r\,dr\,d\theta\,dz \] \[ \Rightarrow \operatorname{div} \bar{F} = 4 - 4r\sin\theta + 2z \] \[ \iiint\limits_V (4 - 4r\sin\theta + 2z) \cdot r \, dz\,dr\,d\theta \] \[ \int_{\theta=0}^{2\pi} \int_{r=0}^{2} \int_{z=0}^{3} (4 - 4r\sin\theta + 2z) \cdot r \, dz\,dr\,d\theta \] \[ = \int_0^{2\pi} \left[ \int_0^2 r(21 - 12r\sin\theta) \, dr \right] d\theta = \int_0^{2\pi} (42 - 32\sin\theta) \, d\theta = 84\pi \]Example 3: Evaluate the Surface Integral using Gauss’s Divergence Theorem \[ \iint\limits_{S} x\,dy\,dz + y\,dz\,dx + z\,dx\,dy \] where \( S \) is the surface of the sphere \( x^2 + y^2 + z^2 = 1 \).
Solution: \[ \vec{F} = x\hat{i} + y\hat{j} + z\hat{k} \quad \Rightarrow \quad \iint\limits_S \vec{F} \cdot \vec{n} \, ds = \iiint\limits_V \nabla \cdot \vec{F} \, dv \] \[ \nabla \cdot \vec{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3 \] \[ \iiint\limits_V 3 \, dv = 3 \cdot \text{Volume of Sphere of radius 1} \] \[ = 3 \cdot \frac{4}{3} \pi (1)^3 = 4\pi \]Example 4: Evaluate the Surface Integral using Gauss’s Divergence Theorem \[ \iint\limits_{S} (x^2 - yz)\,dy\,dz + (y^2 - zx)\,dz\,dx + (z^2 - xy)\,dx\,dy \] where \( S \) is the surface of the rectangular parallelepiped: \[ 0 \le x \le a,\quad 0 \le y \le b,\quad 0 \le z \le c \].
Solution: \[ \vec{F} = (x^2 - yz)\hat{i} + (y^2 - zx)\hat{j} + (z^2 - xy)\hat{k} \] \[ \nabla \cdot \vec{F} = 2x + 2y + 2z \] \[ \iiint\limits_V (2x + 2y + 2z) \, dx\,dy\,dz = 2 \iiint_V (x + y + z) \, dx\,dy\,dz \] \[ = a^2 cb + a cb^2 + a bc^2 \]Example 5:Evaluate the Surface Integral using Gauss’s Divergence Theorem \[ \iint\limits_{S} \vec{F} \cdot \vec{n} \, ds \quad \text{where} \quad \vec{F} = (z^2 - x)\hat{i} - xy\hat{j} + 3z\hat{k} \] and \( S \) is bounded by: \[ z = 0,\quad z = 4 - y^2,\quad x = 0,\quad x = 3 \]
Solution: \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(z^2 - x) + \frac{\partial}{\partial y}(-xy) + \frac{\partial}{\partial z}(3z) = -1 + 0 + 3 = 2 - x \] \[ \iiint\limits_V (2 - x) \, dz\,dy\,dx = \int_0^3 \int_{-2}^{2} \int_0^{4 - y^2} (2 - x) \, dz\,dy\,dx \] \[ \int_0^3 \int_{-2}^{2} (2 - x)(4 - y^2) \, dy\,dx = \frac{8}{3} \int_0^3 (2 - x) \, dx = \frac{8}{3} \cdot \frac{3}{2} = 4 \]