Given \(\mathbf{V}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k},\) integrate \(\mathbf{V} \cdot \mathbf{n} d \sigma\) over the whole surface of the cube of side 1 with four of its vertices at \((0,0,0),(0,0,1),(0,1,0),(1,0,0) .\) Evaluate the same integral by means of the divergence theorem.

Given the vector field \( \mathbf{V}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k},\) and you need to integrate \( \mathbf{V} \cdot \mathbf{n} d \sigma \)overthe surface of a cube with side length 1 and vertices at \(0,0,0\), \(0,0,1\), \(0,1,0\), and \(1,0,0\). The exercise also requires evaluating the integral using the divergence theorem.

The cube has six faces with each face oriented perpendicular to one of the coordinate axes. Recall the normal vectors for each face:For faces perpendicular to the x-axis:1. \( x = 0 : \mathbf{n} = -\mathbf{i} \)2. \( x = 1 : \mathbf{n} = \mathbf{i} \)For faces perpendicular to the y-axis:3. \( y = 0 : \mathbf{n} = -\mathbf{j} \)4. \( y = 1 : \mathbf{n} = \mathbf{j} \)For faces perpendicular to the z-axis:5. \( z = 0 : \mathbf{n} = -\mathbf{k} \)6. \( z = 1 : \mathbf{n} = \mathbf{k} \)

Evaluate \( \int_{S} \mathbf{V} \cdot \mathbf{n} \, d \sigma \) for each face of the cube separately. For a face where the normal vector points in the \( \mathbf{i}, \mathbf{j}, or \mathbf{k} \)direction, the expression simplifies to the respective component of \( \mathbf{V} \)at that surface.

1. For \( x = 0 \, \mathbf{n} = -\mathbf{i} \): \( \mathbf{V} \cdot \mathbf{n} = -x^{2} \), integral from 0 to 1 for \( y \)and integral from 0 to 1 for \( z \): \( \int_{0}^{1} \int_{0}^{1} 0 \, dy \, dz = 0\)2. For \( x = 1 \, \mathbf{n} = \mathbf{i} \): \( \mathbf{V} \cdot \mathbf{n} = x^{2} = 1 \), integral from 0 to 1 for \( y \)and integral from 0 to 1 for \( z \): \( \int_{0}^{1} \int_{0}^{1} 1 \, dy \, dz = 1\)3. For \( y = 0 \, \mathbf{n} = -\mathbf{j} \): \( \mathbf{V} \cdot \mathbf{n} = -y^{2} \), integral from 0 to 1 for \( x \)and integral from 0 to 1 for \( z \): \( \int_{0}^{1} \int_{0}^{1} 0 \, dx \, dz = 0\)4. For \( y = 1 \, \mathbf{n} = \mathbf{j} \): \( \mathbf{V} \cdot \mathbf{n} = y^{2} = 1 \), integral from 0 to 1 for \( x \)and integral from 0 to 1 for \( z \): \( \int_{0}^{1} \int_{0}^{1} 1 \, dx \, dz = 1\)5. For \( z = 0 \, \mathbf{n} = -\mathbf{k} \): \( \mathbf{V} \cdot \mathbf{n} = -z^{2} \), integral from 0 to 1 for \( x \)and integral from 0 to 1 for \( y \): \( \int_{0}^{1} \int_{0}^{1} 0 \, dx \, dy = 0\)6. For \( z = 1 \, \mathbf{n} = \mathbf{k} \): \( \mathbf{V} \cdot \mathbf{n} = z^{2} = 1 \), integral from 0 to 1 for \( x \)and integral from 0 to 1 for \( y \): \( \int_{0}^{1} \int_{0}^{1} 1 \, dx \, dy = 1\)

Adding all the non-zero surface integrals:\(0 + 1 + 0 + 1 + 0 + 1 = 3 \)

The divergence theorem states:\( \int_{S} \mathbf{V} \cdot \mathbf{n} \, d \sigma = \int_{V} \mathbf{abla} \cdot \mathbf{V} \, dV\)First, calculate the divergence of \mathbf{V}:\( \mathbf{abla} \cdot \mathbf{V} = \frac{\partial}{\partial x}(x^{2}) + \frac{\partial}{\partial y}(y^{2}) + \frac{\partial}{\partial z}(z^{2}) = 2x + 2y + 2z \)Now, integrate over the volume of the cube with side length 1:\( \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (2x+2y+2z) \, dx \, dy \, dz = 2 \int_{0}^{1} x \, dx + 2 \int_{0}^{1} y \, dy + 2 \int_{0}^{1} z \, dz = 2 \left[ \frac{x^{2}}{2} \right]_{0}^{1} + 2 \left[ \frac{y^{2}}{2} \right]_{0}^{1} + 2 \left[ \frac{z^{2}}{2} \right]_{0}^{1} = 1 + 1 + 1 = 3 \)

The result from directly integrating over the surface and using the divergence theorem match. Therefore, the final result is 3.