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Integration By Parts 3D

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By Bowen Ho
3 min read
∭Vf(∇⋅A⃗)dV=?\iiint_V f \left(\nabla \cdot \vec A\right) dV = ?

Integration by parts is derived from the product rule for derivatives:

ddx(fg)=f(dgdx)+g(dfdx)⇔∫ddx(fg)dx=∫f(dgdx)dx+∫g(dfdx)dx⇔∫f(dgdx)dx=∫ddx(fg)dx−∫g(dfdx)dx=fg−∫g(dfdx)dx.\begin{align*} &\frac{d}{dx} (fg) = f \left(\frac{dg}{dx} \right) + g \left(\frac{df}{dx} \right) \\ &\Leftrightarrow \int \frac{d}{dx} (fg) dx = \int f \left(\frac{dg}{dx} \right) dx + \int g \left(\frac{df}{dx} \right) dx \\ &\Leftrightarrow \int f \left(\frac{dg}{dx} \right) dx = \int \frac{d}{dx} (fg) dx - \int g \left(\frac{df}{dx} \right) dx = fg - \int g \left(\frac{df}{dx} \right) dx. \end{align*}

We can do the similar to multivariables. Let ff be a scalar field, and A⃗\vec A be a vector field,

∇⋅(fA⃗)=(∂f∂xAx+f∂Ax∂x)+(∂f∂yAy+f∂Ay∂y)+(∂f∂zAz+f∂Az∂z)=∇f⋅A⃗+f(∇⋅A⃗).\begin{align*} \nabla \cdot (f \vec A) &= \left(\frac{\partial f}{\partial x} A_x + f \frac{\partial A_x}{\partial x}\right) + \left(\frac{\partial f}{\partial y} A_y + f \frac{\partial A_y}{\partial y}\right) + \left(\frac{\partial f}{\partial z} A_z + f \frac{\partial A_z}{\partial z}\right)\\ &= \nabla f \cdot \vec A + f \left(\nabla \cdot \vec A\right). \end{align*}

Integrate both sides over a volume VV and use the divergence theorem, we get

∯∂VfA⃗⋅dS⃗=∭V∇⋅(fA⃗)dV=∭V∇f⋅A⃗dV+∭Vf(∇⋅A⃗)dV⇔∭Vf(∇⋅A⃗)dV=∯∂VfA⃗⋅dS⃗−∭V∇f⋅A⃗dV.\begin{align*} &\oiint_{\partial V} f \vec A \cdot d \vec S = \iiint_V \nabla \cdot (f \vec A) dV = \iiint_V \nabla f \cdot \vec A dV + \iiint_V f \left(\nabla \cdot \vec A\right) dV\\ &\Leftrightarrow \boxed{\iiint_V f \left(\nabla \cdot \vec A\right) dV = \oiint_{\partial V} f \vec A \cdot d \vec S - \iiint_V \nabla f \cdot \vec A dV.} \end{align*}
∬Sf(∇×A⃗)dS⃗=?\iint_S f (\nabla \times \vec A) d\vec S = ?

Using the product rule (I skip the proccess here), we can show that

∇×(fA⃗)=(∇f×A⃗)+f(∇×A⃗).\nabla \times (f \vec A) = (\nabla f \times \vec A) + f(\nabla \times \vec A).

Again, integrate both sides over a surface SS and use Stokes’ Theorem to get

∮∂SfA⃗⋅dl⃗=∬S∇×(fA⃗)⋅dS⃗=∬S(∇f×A⃗)⋅dS⃗+∬Sf(∇×A⃗)⋅dS⃗⇔∬Sf(∇×A⃗)⋅dS⃗=∮∂SfA⃗⋅dl⃗−∬S(∇f×A⃗)⋅dS⃗.\begin{align*} \oint_{\partial S} f \vec A \cdot d\vec l &= \iint_S \nabla \times (f \vec A) \cdot d\vec S \\ &= \iint_S (\nabla f \times \vec A) \cdot d\vec S + \iint_S f (\nabla \times \vec A) \cdot d\vec S \\ \Leftrightarrow &\iint_S f (\nabla \times \vec A) \cdot d\vec S = \oint_{\partial S} f \vec A \cdot d\vec l - \iint_S (\nabla f \times \vec A) \cdot d\vec S. \end{align*}

Use the property of cross product, a⃗×b⃗=−b⃗×a⃗\vec a \times \vec b = - \vec b \times \vec a to cancel the minus sign:

−∬S(∇f×A⃗)⋅dS⃗=∬S(A⃗×∇f)⋅dS⃗.- \iint_S (\nabla f \times \vec A) \cdot d\vec S = \iint_S (\vec A \times \nabla f) \cdot d\vec S.

We get

∬Sf(∇×A⃗)⋅dS⃗=∮∂SfA⃗⋅dl⃗+∬S(A⃗×∇f)⋅dS⃗.\boxed{ \iint_S f (\nabla \times \vec A) \cdot d\vec S = \oint_{\partial S} f \vec A \cdot d\vec l + \iint_S (\vec A \times \nabla f) \cdot d\vec S.}
∭VB⃗⋅(∇×A⃗)dV=?\iiint_V \vec B \cdot (\nabla \times \vec A) dV = ?

Using the product rule (I skip the proccess here), we can show that

∇⋅(A⃗×B⃗)=B⃗⋅(∇×A⃗)−A⃗⋅(∇×B⃗).\nabla \cdot (\vec A \times \vec B) = \vec B \cdot (\nabla \times \vec A) - \vec A \cdot (\nabla \times \vec B).

Integrate both sides over a volume VV and use the divergence theorem to get

∯∂V(A⃗×B⃗)⋅dS⃗=∭V∇⋅(A⃗×B⃗)dV=∭VB⋅(∇×A⃗)dV−∭VA⋅(∇×B⃗)dV⇔∭VB⋅(∇×A⃗)dV=∯∂V(A⃗×B⃗)⋅dS⃗+∭VA⋅(∇×B⃗)dV.\begin{align*} \oiint_{\partial V} (\vec A \times \vec B) \cdot d \vec S &= \iiint_V \nabla \cdot (\vec A \times \vec B) dV\\ &=\iiint_V B \cdot (\nabla \times \vec A) dV - \iiint_V A \cdot (\nabla \times \vec B) dV\\ \end{align*}\\ \Leftrightarrow \boxed{\iiint_V B \cdot (\nabla \times \vec A) dV = \oiint_{\partial V} (\vec A \times \vec B) \cdot d \vec S + \iiint_V A \cdot (\nabla \times \vec B) dV.}
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