Volume Integral Change Calculation

The question asks how the volume integral \(I = \iiint_V \mathbf{A} \cdot (\nabla \times \mathbf{A}) d^3 x\) changes when the vector field \(\mathbf{A}\) is replaced by \(\mathbf{A} + \nabla \phi\). The change is denoted by \(\Delta I\). The region V is bounded by a surface \(\Sigma\), and \(d^3 x\) represents the volume element.

The new integral, let's call it \(I'\), is given by:

\(I' = \iiint_V (\mathbf{A} + \nabla \phi) \cdot (\nabla \times (\mathbf{A} + \nabla \phi)) d^3 x\)

Curl of the New Vector Field

First, let's simplify the curl term \(\nabla \times (\mathbf{A} + \nabla \phi)\). Using the linearity of the curl operator:

\(\nabla \times (\mathbf{A} + \nabla \phi) = \nabla \times \mathbf{A} + \nabla \times (\nabla \phi)\)

A fundamental identity in vector calculus states that the curl of the gradient of any scalar function \(\phi\) is always zero:

\(\nabla \times (\nabla \phi) = 0\)

Therefore, the curl of the new vector field is simply:

\(\nabla \times (\mathbf{A} + \nabla \phi) = \nabla \times \mathbf{A} + 0 = \nabla \times \mathbf{A}\)

New Integral Expression

Now, substitute this back into the expression for the new integral \(I'\):

\(I' = \iiint_V (\mathbf{A} + \nabla \phi) \cdot (\nabla \times \mathbf{A}) d^3 x\)

Expand the dot product:

\(I' = \iiint_V [\mathbf{A} \cdot (\nabla \times \mathbf{A}) + (\nabla \phi) \cdot (\nabla \times \mathbf{A})] d^3 x\)

Split the integral into two parts:

\(I' = \iiint_V \mathbf{A} \cdot (\nabla \times \mathbf{A}) d^3 x + \iiint_V (\nabla \phi) \cdot (\nabla \times \mathbf{A}) d^3 x\)

The first integral is the original integral \(I\). So, \(I' = I + \Delta I\), where \(\Delta I\) is the change in the integral.

Thus, the change is:

\(\Delta I = \iiint_V (\nabla \phi) \cdot (\nabla \times \mathbf{A}) d^3 x\)

Vector Identity Application

To simplify the integral for \(\Delta I\), we use the vector identity for the divergence of a cross product:

\(\nabla \cdot (\mathbf{F} \times \mathbf{G}) = \mathbf{G} \cdot (\nabla \times \mathbf{F}) - \mathbf{F} \cdot (\nabla \times \mathbf{G})\)

Rearranging this identity, we get:

\(\mathbf{F} \cdot (\nabla \times \mathbf{G}) = \mathbf{G} \cdot (\nabla \times \mathbf{F}) - \nabla \cdot (\mathbf{F} \times \mathbf{G})\)

Let \(\mathbf{F} = \nabla \phi\) and \(\mathbf{G} = \mathbf{A}\). Then \(\nabla \times \mathbf{F} = \nabla \times (\nabla \phi) = 0\). Substituting these into the rearranged identity:

\((\nabla \phi) \cdot (\nabla \times \mathbf{A}) = \mathbf{A} \cdot (\nabla \times (\nabla \phi)) - \nabla \cdot ((\nabla \phi) \times \mathbf{A})\)

\((\nabla \phi) \cdot (\nabla \times \mathbf{A}) = \mathbf{A} \cdot (0) - \nabla \cdot ((\nabla \phi) \times \mathbf{A})\)

\((\nabla \phi) \cdot (\nabla \times \mathbf{A}) = - \nabla \cdot ((\nabla \phi) \times \mathbf{A})\)

Integral Transformation using Divergence Theorem

Substitute this result back into the integral for \(\Delta I\):

\(\Delta I = \iiint_V [-\nabla \cdot ((\nabla \phi) \times \mathbf{A})] d^3 x\)

\(\Delta I = - \iiint_V \nabla \cdot ((\nabla \phi) \times \mathbf{A}) d^3 x\)

Now, we can use the Divergence Theorem, which states that for a vector field \(\mathbf{F}\) over a volume V bounded by a closed surface \(\Sigma\) with outward unit normal \(\hat{\mathbf{n}}\):

\(\iiint_V (\nabla \cdot \mathbf{F}) d^3 x = \iint_\Sigma \mathbf{F} \cdot \hat{\mathbf{n}} dS\)

Applying this theorem with \(\mathbf{F} = (\nabla \phi) \times \mathbf{A}\):

\(\iiint_V \nabla \cdot ((\nabla \phi) \times \mathbf{A}) d^3 x = \iint_\Sigma ((\nabla \phi) \times \mathbf{A}) \cdot \hat{\mathbf{n}} dS\)

Substitute this into the expression for \(\Delta I\):

\(\Delta I = - \iint_\Sigma ((\nabla \phi) \times \mathbf{A}) \cdot \hat{\mathbf{n}} dS\)

Comparing this result with the given options, we find that it matches option 3.

The change \(\Delta I\) in the volume integral is given by the negative of the surface integral of the cross product of the gradient of the scalar function and the original vector field, dotted with the outward unit normal vector over the bounding surface \(\Sigma\).