📂Mathematical Physics

Various Formulas of Vector Integration Involving the Del Operator

Formulas¹

Let’s designate $T, U$ as a scalar function, and $\mathbf{v}$ as a vector function. Then, the following equations hold:

$$\begin{equation} \int_{\mathcal{V}} (\nabla T) d \tau = \oint_{\mathcal{S}} T d \mathbf{a} \end{equation}$$

$$\begin{equation} \int_{\mathcal{V}} (\nabla \times \mathbf{v}) d \tau = - \oint_{\mathcal{S}} \mathbf{v} \times d \mathbf{a} \end{equation}$$

$$\begin{equation} \int_{\mathcal{V}} \left[ T \nabla^{2} U + (\nabla T) \cdot (\nabla U) \right] d \tau = \oint_{\mathcal{S}} (T \nabla U) \cdot d \mathbf{a} \end{equation}$$

$$\begin{equation} \int_{\mathcal{V}} \left( T \nabla^{2} U - U \nabla^{2} T \right) d \tau = \oint_{\mathcal{S}} \left( T \nabla U - U \nabla T \right) \cdot d \mathbf{a} \end{equation}$$

$$\begin{equation} \int_{\mathcal{S}} \nabla T \times d \mathbf{a} = - \oint_{\mathcal{P}} T d \mathbf{l} \end{equation}$$

Description

Proving the above formulas is Exercise 1.61 in the 4th edition of Griffiths’ Electrodynamics.

$(3), (4)$ is also known as Green’s theorem.

Proof

In all the following proofs, $\mathbf{c}$ represents a constant vector.

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(1)

Divergence theorem

$$\int_{\mathcal{V}} (\nabla \cdot \mathbf{v}) d \tau = \oint_{\mathcal{S}} \mathbf{v} \cdot d \mathbf{a}$$

Let’s substitute $\mathbf{v} = T\mathbf{c}$ into the divergence theorem.

$$\int_{\mathcal{V}} \nabla \cdot (T\mathbf{c}) d \tau = \oint_{\mathcal{S}} (T \mathbf{c}) \cdot d \mathbf{a}$$

Then, by the product rule $\nabla \cdot (f\mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f)$, the LHS can be rewritten as follows.

$$\int_{\mathcal{V}} T(\nabla \cdot \mathbf{c}) d \tau + \int_{\mathcal{V}} \mathbf{c} \cdot (\nabla T) d \tau = \oint_{\mathcal{S}} (T \mathbf{c}) \cdot d \mathbf{a}$$

Since $\mathbf{c}$ is a constant vector in the first term on the LHS, it results in $\nabla \cdot \mathbf{c}=0$. Therefore,

$$\begin{align*} \\ && \int_{\mathcal{V}} \mathbf{c} \cdot (\nabla T) d \tau =&\ \oint_{\mathcal{S}} (T \mathbf{c}) \cdot d \mathbf{a} \\ \implies && \mathbf{c} \cdot \int_{\mathcal{V}} (\nabla T) d \tau =&\ \mathbf{c} \cdot \oint_{\mathcal{S}} T d \mathbf{a} \\ \implies && \int_{\mathcal{V}} (\nabla T) d \tau =&\ \oint_{\mathcal{S}} T d \mathbf{a} \end{align*}$$

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(2)

Divergence theorem

$$\int_{\mathcal{V}} (\nabla \cdot \mathbf{v}) d \tau = \oint_{\mathcal{S}} \mathbf{v} \cdot d \mathbf{a}$$

Let’s substitute $\mathbf{v} = \mathbf{v} \times \mathbf{c}$ into the divergence theorem.

$$\int_{\mathcal{V}} \nabla \cdot (\mathbf{v} \times \mathbf{c}) d \tau = \oint_{\mathcal{S}} (\mathbf{v} \times \mathbf{c}) \cdot d \mathbf{a}$$

Then, by the product rule $\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})$, the LHS can be rewritten as follows.

$$\int_{\mathcal{V}} \mathbf{c} \cdot (\nabla \times \mathbf{v}) d \tau - \int_{\mathcal{V}} \mathbf{v} \cdot (\nabla \times \mathbf{c}) d \tau = \oint_{\mathcal{S}} (\mathbf{v} \times \mathbf{c}) \cdot d \mathbf{a}$$

Since $\mathbf{c}$ is a constant vector in the second term on the LHS, it results in $\nabla \times \mathbf{c} = \mathbf{0}$. Therefore,

$$\int_{\mathcal{V}} \mathbf{c} \cdot (\nabla \times \mathbf{v}) d \tau = \oint_{\mathcal{S}} (\mathbf{v} \times \mathbf{c}) \cdot d \mathbf{a}$$

Also, by the scalar triple product formula and properties of the cross product, the following holds.

$$(\mathbf{v} \times \mathbf{c}) \cdot d \mathbf{a} = (d \mathbf{a} \times \mathbf{v}) \cdot \mathbf{c} = -( \mathbf{v} \times d \mathbf{a}) \cdot \mathbf{c} = - \mathbf{c} \cdot ( \mathbf{v} \times d \mathbf{a})$$

Substituting it into the RHS of the previous equation results in

$$\begin{align*} \\ && \int_{\mathcal{V}} \mathbf{c} \cdot (\nabla \times \mathbf{v}) d \tau =&\ - \oint_{\mathcal{S}} \mathbf{c} \cdot ( \mathbf{v} \times d \mathbf{a}) \\ \implies && \mathbf{c} \cdot \int_{\mathcal{V}} (\nabla \times \mathbf{v}) d \tau =&\ - \mathbf{c} \cdot \oint_{\mathcal{S}} \mathbf{v} \times d \mathbf{a} \\ \implies && \int_{\mathcal{V}} (\nabla \times \mathbf{v}) d \tau =&\ -\oint_{\mathcal{S}} \mathbf{v} \times d \mathbf{a} \end{align*}$$

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(3)

Divergence theorem

$$\int_{\mathcal{V}} (\nabla \cdot \mathbf{v}) d \tau = \oint_{\mathcal{S}} \mathbf{v} \cdot d \mathbf{a}$$

Let’s substitute $\mathbf{v} = T \nabla U$ into the divergence theorem.

$$\int_{\mathcal{V}} \nabla \cdot (T \nabla U) d \tau = \oint_{\mathcal{S}} (T \nabla U) \cdot d \mathbf{a}$$

Then, by the product rule $\nabla \cdot (f\mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f)$, the LHS can be rewritten as follows.

$$\int_{\mathcal{V}} \left[ T \nabla \cdot (\nabla U) + \nabla U \cdot (\nabla T) \right] d \tau = \oint_{\mathcal{S}} (T \nabla U) \cdot d \mathbf{a}$$

The divergence of a gradient is the Laplacian $\nabla^{2} = \nabla \cdot \nabla$, therefore

$$\int_{\mathcal{V}} \left[ T \nabla^{2} U + (\nabla T) \cdot (\nabla U) \right] d \tau = \oint_{\mathcal{S}} (T \nabla U) \cdot d \mathbf{a}$$

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(4)

The equation $(3)$ can be written as follows:

$$\int_{\mathcal{V}} (\nabla T) \cdot (\nabla U) d \tau = \oint_{\mathcal{S}} (T \nabla U) \cdot d \mathbf{a} - \int_{\mathcal{V}} T \nabla^{2} U d \tau$$

If we switch $T$ and $U$, it becomes:

$$\int_{\mathcal{V}} (\nabla U) \cdot (\nabla T) d \tau = \oint_{\mathcal{S}} (U \nabla T) \cdot d \mathbf{a} - \int_{\mathcal{V}} U \nabla^{2} T d \tau$$

Subtracting one from the other gives us:

$$0 = \left( \oint_{\mathcal{S}} (T \nabla U) \cdot d \mathbf{a} - \int_{\mathcal{V}} T \nabla^{2} U d \tau \right) - \left( \oint_{\mathcal{S}} (U \nabla T) \cdot d \mathbf{a} - \int_{\mathcal{V}} U \nabla^{2} T d \tau \right)$$

Which simplifies to:

$$\int_{\mathcal{V}} \left( T \nabla^{2} U - U \nabla^{2} T \right) d \tau = \oint_{\mathcal{S}} \left( T \nabla U - U \nabla T \right) \cdot d \mathbf{a}$$

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(5)

Stokes’ theorem

$$\int_{\mathcal{S}} (\nabla \times \mathbf{v} )\cdot d\mathbf{a} = \oint_{\mathcal{P}} \mathbf{v} \cdot d\mathbf{l}$$

Let’s substitute $\mathbf{v} = T \mathbf{c}$ into Stokes’ theorem.

$$\int_{\mathcal{S}} \left[ \nabla \times (T \mathbf{c}) \right] \cdot d\mathbf{a} = \oint_{\mathcal{P}} (T \mathbf{c}) \cdot d\mathbf{l}$$

Then, by the product rule $\nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A}) - \mathbf{A} \times (\nabla f)$, the LHS can be written as follows.

$$\int_{\mathcal{S}} T (\nabla \times \mathbf{c}) \cdot d\mathbf{a} - \int_{\mathcal{S}} \left[ \mathbf{c} \times (\nabla T) \right] \cdot d\mathbf{a} = \oint_{\mathcal{P}} (T \mathbf{c}) \cdot d\mathbf{l}$$

Since $\mathbf{c}$ is a constant vector in the first term on the LHS, it results in $\nabla \times \mathbf{c} = \mathbf{0}$. Therefore,

$$\int_{\mathcal{S}} \left[ \mathbf{c} \times (\nabla T) \right] \cdot d\mathbf{a} = - \oint_{\mathcal{P}} (T \mathbf{c}) \cdot d\mathbf{l}$$

Also, by the scalar triple product formula, the following holds.

$$\left( \mathbf{c} \times \nabla T \right) \cdot d \mathbf{a} = \left( \nabla T \times d \mathbf{a} \right) \cdot \mathbf{c}$$

Therefore,

$$\begin{align*} && \int_{\mathcal{S}} \left( \nabla T \times d \mathbf{a} \right) \cdot \mathbf{c} = - \oint_{\mathcal{P}} (T \mathbf{c}) \cdot d\mathbf{l} \\ \implies && \mathbf{c} \cdot \int_{\mathcal{S}} \left( \nabla T \times d \mathbf{a} \right) = - \mathbf{c} \cdot \oint_{\mathcal{P}} T d\mathbf{l} \\ \implies && \int_{\mathcal{S}} \left( \nabla T \times d \mathbf{a} \right) = - \oint_{\mathcal{P}} T d\mathbf{l} \end{align*}$$

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