Divergence Theorem

1 Divergence Theorem

Let \(F (x,y, z) \) be a space vector field, and \(p \) a point in the domain of \(F\). Let \(\Sigma \subset \mathbb {R}^3 \) be a very small sphere around \(p\), positively oriented (outwards). Then the integral

\[ \iint _{\Sigma } F \cdot ds \]

measures how much \(F\) is pushing things away from \(p\). It turns out we have another quantity that measures precisely that:

\[ \text {div}(F)(p) . \]

Overall, if we have a region \(E \subset \mathbb {R}^3\) with boundary \(\partial E\) oriented positively, the two integrals

\begin{gather*} \iint _{\partial E} F \cdot ds \\ \iiint _{E} \text {div}(F) \, dA \end{gather*}

measure how much the vector field is pointing outside of \(E\). The Divergence Theorem (aka Gauss Theorem) states that they agree.

Let \(\Sigma \subset \mathbb {R}^3\) be the unit sphere, oriented positively. Compute
\[ \iint _{\Sigma } \langle x^2 + 3 , 5xy + 2 , 2z - 4 \rangle \cdot dS \]

If \(F (x,y,z) : = \langle x^2 + 3 , 5xy + 2 , 2z - 4 \rangle \), we can compute

\[ \text {div}(F) = 2x + 5x + 2 = 7x + 2 \]

Then by the Divergence Theorem,

\[ \iint _{\Sigma } \langle x^2 + 3 , 5xy + 2 , 2z - 4 \rangle \cdot dS = \iiint _B (7x + 2) \, dV \]

where \(B \subset \mathbb {R}^3\) is the unit ball. By symmetry, the integral of \(x\) over \(B\) is zero, so we get

\[ \iiint _B (7x + 2) \, dV = 2 \cdot \text {vol} (B) = \frac {8}{3} \pi \]
Let \(\Sigma \subset \mathbb {R}^3\) be the boundary of the box \([-2,1]\times [-1,4]\times [1,6]\), oriented inwards. Compute
\[ \iint _{\Sigma } \langle 3x + y^2 - 5z, 2xy + 3x , x \cos y + e^z \rangle \cdot dS \]

If \(F (x,y,z) : = \langle 3x + y^2 - 5z, 2xy + 3x , x \cos y + e^z \rangle \), we can compute

\[ \text {div}(F) = 3 + 2x + e^z \]

Then by the Divergence Theorem,

\[ \iint _{\Sigma } \langle 3x + y^2 - 5z, 2xy + 3x , x \cos y + e^z \rangle \cdot dS = - \int _{-2}^1 \int _{-1}^4 \int _1^6 (3 + 2x + e^z)\, dzdydx \]

(the minus sign comes from the orientation). Then

\begin{align*} \int _{-2}^1 \int _{-1}^4 \int _1^6 (3 + 2x + e^z)\, dzdydx & = 225 - 25 (2^2 - 1) + 15 (e^6 - e)\\ & = 15 (10 + e^6 - e) \end{align*}

so we conclude

\[ \iint _{\Sigma } \langle 3x + y^2 - 5z, 2xy + 3x , x \cos y + e^z \rangle \cdot dS = - 15 (10 + e^6 - e ) \]
Let \(\Sigma \subset \mathbb {R}^3\) be the portion of the paraboloid \(z = 4 - x^2 - y^2\) oriented upwards. Compute
\[ \iint _{\Sigma } \langle x^2 - e^y , 5 + 2x + \sin z , 3x + 12y^2 + 3 \rangle \cdot dS \]

If \(F (x,y,z) : = \langle x^2 - e^y , 5 + 2x + \sin z , 3x + 12y^2 + 3 \rangle \), we can compute

\[ \text {div}(F) = 2x \]

Let \(E\subset \mathbb {R}^3\) be the region below \(\Sigma \) and above the \(xy\)-plane, and \(\partial E \) its boundary oriented positively. Then

\[ \iint _{\partial E} \langle x^2 - e^y , 5 + 2x + \sin z , 3x + 12y^2 + 3 \rangle \cdot dS = \iiint _E 2x \, dV \]

Note that \(E\) is symmetric with respect to \(yz\)-plane, so the integrals above are zero. Also note that \(\partial E \) consists of two parts: \(\Sigma \), and the disk \(x^2 + y^2 \leq 4\) in the \(xy\)-plane, oriented downwards, which we call \(\Sigma _1\). Then we compute

\begin{align*} \iint _{\Sigma } & \langle x^2 - e^y , 5 + 2x + \sin z , 3x + 12y^2 + 3 \rangle \cdot dS \\ & = - \iint _{\Sigma _1} \langle x^2 - e^y , 5 + 2x + \sin z , 3x + 12y^2 + 3 \rangle \cdot dS \\ & = - \iint _{\Sigma _1} \langle x^2 - e^y , 5 + 2x + \sin z , 3x + 12y^2 + 3 \rangle \cdot \langle 0,0,-1 \rangle \, dS \\ & = \iint _{\Sigma _1} ( 3x + 12y^2 + 3 ) \, dS \\ & = 0 + 48 \pi + 12 \pi \\ & = 60 \pi \end{align*}