Question

Given the vector field \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) verify that \(\int_{S} \int \mathbf{F} \cdot \mathbf{N} d S=3 V\) where \(V\) is the volume of the solid bounded by the closed surface \(S\)

Short answer

The divergence of the given vector field \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z\mathbf{k}\) is 3. The volume integral of this divergence over the solid volume \(V\) results in \(3V\). Hence, the surface integral of \(\mathbf{F}\) over the closed surface \(S\) is equal to \(3V\), verifying the divergence theorem.

Step-by-step solution

Compute the Divergence of the Vector Field

Find the divergence of the vector field \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\). The divergence of a vector field \(\mathbf{F}=P\mathbf{i}+Q\mathbf{j}+R\mathbf{k}\) is given by \(\nabla \cdot \mathbf{F}=(\partial P/\partial x)+(\partial Q/\partial y)+(\partial R/\partial z)\). Hence, \(\nabla \cdot \mathbf{F}=\partial x/ \partial x + \partial y/ \partial y + \partial z/ \partial z =1+1+1=3\).

Compute the volume integral of the divergence of \(\mathbf{F}\) over the solid volume \(V\)

According to divergence theorem, the volume integral of the divergence of \(\mathbf{F}\) over the solid volume \(V\) is equal to the surface integral of \(\mathbf{F}\) over the closed surface S. Compute the volume integral of the divergence which is \( \int\int\int_V \nabla \cdot \mathbf{F}dV = \int\int\int_V 3 dV = 3V.\)

Verify the Equality

Verify the equality \(\int_{S} \int \mathbf{F} \cdot \mathbf{N} d S = \int\int\int_V \nabla \cdot \mathbf{F}dV\) between surface and volume integrals. Since \(\int\int\int_V \nabla \cdot \mathbf{F}dV = 3V\), the equality holds, confirming the divergence theorem.