The Divergence Theorem
Solved Problems
Example 3.
Use the Divergence Theorem to evaluate the surface integral \(\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} \) of the vector field \[\mathbf{F}\left( {x,y,z} \right) = \left( {{x^3},{y^3},{z^3}} \right),\] where \(S\) is the surface of a solid bounded by the cone \({x^2} + {y^2} - {z^2} = 0\) and the plane \(z = 1.\) (Figure \(2\)).
Solution.
The solid is sketched in Figure \(2.\)
Figure 2. Applying the Divergence Theorem, we can write:
\[I = \iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} = \iiint\limits_G {\left( {\nabla \cdot \mathbf{F}} \right)dV} = \iiint\limits_G {\left[ {\frac{\partial }{{\partial x}}\left( {{x^3}} \right) + \frac{\partial }{{\partial y}}\left( {{y^3}} \right) + \frac{\partial }{{\partial z}}\left( {{z^3}} \right)} \right]dxdydz} = 3\iiint\limits_G {\left( {{x^2} + {y^2} + {z^2}} \right)dxdydz}.\]
By changing to cylindrical coordinates, we have
\[I = 3\iiint\limits_G {\left( {{x^2} + {y^2} + {z^2}} \right)dxdydz} = 3\int\limits_0^1 {dz} \int\limits_0^{2\pi } {d\varphi } \int\limits_0^z {\left( {{r^2} + {z^2}} \right)rdr} = 6\pi \int\limits_0^1 {\left[ {\left. {\left( {\frac{{{r^4}}}{4} + \frac{{{z^2}{r^2}}}{2}} \right)} \right|_{r = 0}^z} \right]dz} = 6\pi \int\limits_0^1 {\frac{{3{z^4}}}{4}dz} = \frac{{9\pi }}{2} \left[ {\left. {\left( {\frac{{{z^5}}}{5}} \right)} \right|_0^1} \right] = \frac{{9\pi }}{{10}}.\]
Example 4.
Using the Divergence Theorem calculate the surface integral \(\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} \) of the vector field \[\mathbf{F}\left( {x,y,z} \right) = \left( {2xy, 8xz, 4yz} \right),\] where \(S\) is the surface of tetrahedron with vertices \(O\left( {0,0,0} \right),\) \(A\left( {1,0,0} \right),\) \(B\left( {0,1,0} \right),\) \(C\left( {0,0,1} \right)\) (Figure \(3\)).
Solution.
Figure 3. By Divergence Theorem,
\[I = \iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} = \iiint\limits_G {\left( {\nabla \cdot \mathbf{F}} \right)dV} = \iiint\limits_G {\Big[ {\frac{\partial }{{\partial x}}\left( {2xy} \right) + \frac{\partial }{{\partial y}}\left( {8xz} \right) + \frac{\partial }{{\partial z}}\left( {4yz} \right)} \Big]dV} = \iiint\limits_G {\left( {2y + 0 + 4y} \right)dxdydz} = 6\iiint\limits_G {ydxdydz} .\]
Find the given triple integral. The equation of the straight line \(AB\) has the form:
\[\frac{{y - {y_A}}}{{{y_B} - {y_A}}} = \frac{{x - {x_A}}}{{{x_B} - {x_A}}},\;\; \Rightarrow \frac{{y - 0}}{{1 - 0}} = \frac{{x - 1}}{{0 - 1}},\;\; \Rightarrow y = - \left( {x - 1} \right)\;\; \text{or}\;\;y = 1 - x.\]
The equation of the plane \(ABC\) is
\[\frac{x}{1} + \frac{y}{1} + \frac{z}{1} = 1,\;\; \Rightarrow x + y + z = 1\;\; \text{or}\;\;z = 1 - x - y.\]
So the integral becomes:
\[I = 6\iiint\limits_G {ydxdydz} = 6\int\limits_0^1 {dx} \int\limits_0^{1 - x} {dy} \int\limits_0^{1 - x - y} {ydz} = 6\int\limits_0^1 {dx} \int\limits_0^{1 - x} {\left( {1 - x - y} \right)ydy} = 6\int\limits_0^1 {dx} \int\limits_0^{1 - x} {\left[ {y\left( {1 - x} \right) - {y^2}} \right]dy} = 6\int\limits_0^1 {\left[ {\left. {\left( {\left( {1 - x} \right)\frac{{{y^2}}}{2} - \frac{{{y^3}}}{3}} \right)} \right|_{y = 0}^{1 - x}} \right]dx} = 6\int\limits_0^1 {\left[ {\frac{{{{\left( {1 - x} \right)}^3}}}{2} - \frac{{{{\left( {1 - x} \right)}^3}}}{3}} \right]dx} = 6 \cdot \frac{1}{6}\int\limits_0^1 {{{\left( {1 - x} \right)}^3}dx} = - \int\limits_0^1 {{{\left( {1 - x} \right)}^3}d\left( {1 - x} \right)} = - \left. {\left( {\frac{{{{\left( {1 - x} \right)}^4}}}{4}} \right)} \right|_0^1 = - \left( {0 - \frac{1}{4}} \right) = \frac{1}{4}.\]
Example 5.
Calculate the surface integral \(\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} \) of the vector field \[\mathbf{F}\left( {x,y,z} \right) = \left( {2{x^2}y,x{z^2},4yz} \right),\] where \(S\) is the surface of the rectangular box bounded by the planes \(x = 0,\) \(x = 1,\) \(y = 0,\) \(y = 2,\) \(z = 0,\) \(z = 3\) (Figure \(4\)).
Solution.
Figure 4. Using the Divergence Theorem, we can write:
\[
\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}}
= \iint\limits_S {\left( {2{x^2}y \cdot \mathbf{i} + x{z^2} \cdot \mathbf{j} + 4yz \cdot \mathbf{k}} \right) \cdot d\mathbf{S}}
= \iiint\limits_G {\left[ {\frac{\partial }{{\partial x}}\left( {2{x^2}y} \right) + \frac{\partial }{{\partial y}}\left( {x{z^2}} \right) + \frac{\partial }{{\partial z}}\left( {4yz} \right)} \right]dxdydz}
= \iiint\limits_G {\left( {4xy + 4y} \right)dxdydz}
= 4\int\limits_0^1 {dx} \int\limits_0^2 {dy} \int\limits_0^3 {\left( {x + 1} \right)ydz}
= 4\int\limits_0^1 {dx} \int\limits_0^2 {\left[ {\left. {\left( {\left( {x + 1} \right)yz} \right)} \right|_{z = 0}^3} \right]dy}
= 12\int\limits_0^1 {dx} \int\limits_0^2 {\left( {x + 1} \right)ydy}
= 12\int\limits_0^1 {\left[ {\left. {\left( {\left( {x + 1} \right)\frac{{{y^2}}}{2}} \right)} \right|_{y = 0}^2} \right]dx}
= 24\int\limits_0^1 {\left( {x + 1} \right)dx}
= 24\left[ {\left. {\left( {\frac{{{x^2}}}{2} + x} \right)} \right|_0^1} \right]
= 24 \cdot \frac{3}{2} = 36.\]
Example 6.
Find the surface integral
\[\iint\limits_S {2xdydz + \left( {3y + x} \right)dxdz + \left( {2y + 4z} \right)dxdy} ,\]
where \(S\) is the outer surface of the pyramid
\[\frac{x}{a} + \frac{y}{b} + \frac{z}{c} \le 1, x \ge 0, y \ge 0, z \ge 0\]
(see Figure \(5\)).
Solution.
Figure 5. Figure 6. Using the Divergence Theorem, we can write the initial surface integral as
\[\iint\limits_S {2xdydz + \left( {3y + x} \right)dxdz + \left( {2y + 4z} \right)dxdy} = \iiint\limits_G {\left( {2 + 3 + 4} \right)dxdydz} = 9\iiint\limits_G {dxdydz} .\]
Calculate the triple integral. The region of integration in the \(xy\)-plane is shown in Figure \(6.\) By setting \(z = 0,\) we find:
\[\frac{x}{a} + \frac{y}{b} \le 1,\;x \ge 0,\;y \ge 0,\;\; \Rightarrow \frac{y}{b} \le 1 - \frac{x}{a}\;\; \text{or}\;\;y \le b - \frac{b}{a}x.\]
Hence, the region \(D\) can be represented in the form:
\[D = \left\{ {\left( {x,y} \right) \mid 0 \le x \le a, \; 0 \le y \le b - \frac{b}{a}x} \right\}.\]
We rewrite the inequality \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} \le 1\) in terms of \(z:\)
\[\frac{x}{a} + \frac{y}{b} + \frac{z}{c} \le 1,\;\; \Rightarrow \frac{z}{c} \le 1 - \frac{x}{a} - \frac{y}{b}\;\; \text{or}\;\;z \le c\left( {1 - \frac{x}{a} - \frac{y}{b}} \right).\]
Then the triple integral becomes
\[\iiint\limits_G {dxdydz} = \int\limits_0^a {dx} \int\limits_0^{b\left( {1 - \frac{x}{a}} \right)} {dy} \int\limits_0^{c\left( {1 - \frac{x}{a} - \frac{y}{b}} \right)} {dz} = \int\limits_0^a {dx} \int\limits_0^{b\left( {1 - \frac{x}{a}} \right)} {c\left( {1 - \frac{x}{a} - \frac{y}{b}} \right)dy} = bc\int\limits_0^a {\left[ {\left( {1 - \frac{x}{a}} \right)\left( {1 - \frac{x}{a}} \right) - \frac{{{{\left( {1 - \frac{x}{a}} \right)}^2}}}{2}} \right]dx} = \frac{{bc}}{{2{a^2}}}\int\limits_0^a {{{\left( {a - x} \right)}^2}dx} = \frac{{bc}}{{2{a^2}}}\int\limits_0^a {\left( {{a^2} - 2ax + {x^2}} \right)dx} = \frac{{bc}}{{2{a^2}}}\left[ {\left. {\left( {{a^2}x - a{x^2} + \frac{{{x^3}}}{3}} \right)} \right|_0^a} \right] = \frac{{bc}}{{2{a^2}}}\left( {\cancel{a^3} - \cancel{a^3} + \frac{{{a^3}}}{3}} \right) = \frac{{abc}}{6}.\]
