Partial Integration of Expressions Containing the Del Operator
Formulas
The following expressions hold true for vector integration involving the del operator.
(a)
$$ \int_{\mathcal{V}}\mathbf{A} \cdot (\nabla f)d\tau = \oint_{\mathcal{S}}f\mathbf{A} \cdot d \mathbf{a}-\int_{\mathcal{V}}f(\nabla \cdot \mathbf{A})d\tau $$
(b)
$$ \int_{\mathcal{S}} f \left( \nabla \times \mathbf{A} \right)\mathbf{A} \cdot d \mathbf{a} = \int_{\mathcal{S}} \left[ \mathbf{A} \times \left( \nabla f \right) \right] \cdot d\mathbf{a} + \oint_{\mathcal{P}} f\mathbf{A} \cdot d\mathbf{l} $$
(c)
$$ \int_{\mathcal{V}} \mathbf{B} \cdot \left( \nabla \times \mathbf{A} \right) d\tau = \int_{\mathcal{V}} \mathbf{A} \cdot \left( \nabla \times \mathbf{B} \right) d\tau + \oint_{\mathcal{S}} \left( \mathbf{A} \times \mathbf{B} \right) \cdot d \mathbf{a} $$
Explanation
Partial integration is a method that simplifies the integration of the product of a function $(f\ or\ \mathbf{A}$ and the derivative of a function $(\nabla f\ or\ \nabla \cdot \mathbf{A})$.
Partial Integration $\dfrac{d}{dx}\left( fg \right) = f\dfrac{dg}{dx}+g\dfrac{df}{dx}$ Integrating both sides yields
$$ \int_{a}^b \dfrac{d}{dx} \left(fg\right) = (fg)\Big|_{a}^b=\int_{a}^b f\left(\dfrac{dg}{dx}\right)dx+\int_{a}^bg\left(\dfrac{df}{dx}\right)dx \\ \implies \int_{a}^b f\left(\dfrac{dg}{dx}\right)dx = (fg)\Big|_{a}^b-\int_{a}^bg\left(\dfrac{df}{dx}\right)dx $$
Proof
(a)
Using Product Rule 3
$$ \nabla \cdot (f\mathbf{A}) = \mathbf{A} \cdot (\nabla f) + f(\nabla \cdot \mathbf{A}) $$
Integrating both sides with respect to volume gives
$$ \int_{\mathcal{V}} \nabla \cdot (f\mathbf{A})d\tau = \int_{\mathcal{V}}\mathbf{A} \cdot (\nabla f)d\tau + \int_{\mathcal{V}}f(\nabla \cdot \mathbf{A})d\tau $$
Applying the Divergence Theorem to the left hand side gives
$$ \oint_{\mathcal{S}}f\mathbf{A} \cdot d \mathbf{a} = \int_{\mathcal{V}}\mathbf{A} \cdot (\nabla f)d\tau + \int_{\mathcal{V}}f(\nabla \cdot \mathbf{A})d\tau $$
Upon simplification, we get
$$ \int_{\mathcal{V}}f(\nabla \cdot \mathbf{A})d\tau = \oint_{\mathcal{S}}f\mathbf{A} \cdot d \mathbf{a}-\int_{\mathcal{V}}\mathbf{A} \cdot (\nabla f)d\tau $$
Or equivalently,
$$ \int_{\mathcal{V}}\mathbf{A} \cdot (\nabla f)d\tau = \oint_{\mathcal{S}}f\mathbf{A} \cdot d \mathbf{a}-\int_{\mathcal{V}}f(\nabla \cdot \mathbf{A})d\tau $$
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