Question

The vector field \( \vec{v} = x^3 \hat{i} + y^3 \hat{j} + z^3 \hat{k} \) is a vector field where \( \hat{i}, \hat{j}, \hat{k} \) are the base vectors of a Cartesian coordinate system.
Using the Gauss divergence theorem, the value of the outward flux of the vector field over the surface of a sphere of unit radius centered at the origin is _________ (rounded off to one decimal place).

Hint:

For problems involving Gauss's divergence theorem, compute the divergence of the vector field, then integrate over the volume enclosed by the surface.

Answer 1

Correct Answer: 7.4

We are given the vector field: \[ \vec{v} = x^3 \hat{i} + y^3 \hat{j} + z^3 \hat{k}. \] By the Gauss divergence theorem, the outward flux of the vector field over the surface of a closed surface \( S \) is given by: \[ \iint_S \vec{v} \cdot \hat{n} \, dA = \iiint_V (\nabla \cdot \vec{v}) \, dV, \] where \( \nabla \cdot \vec{v} \) is the divergence of \( \vec{v} \). The divergence of \( \vec{v} \) is: \[ \nabla \cdot \vec{v} = \frac{\partial}{\partial x} (x^3) + \frac{\partial}{\partial y} (y^3) + \frac{\partial}{\partial z} (z^3) = 3x^2 + 3y^2 + 3z^2 = 3(x^2 + y^2 + z^2). \] Since we are integrating over the surface of a unit sphere (where \( x^2 + y^2 + z^2 = 1 \)), the integral becomes: \[ \iiint_V 3 \, dV = 3 \times \text{Volume of the unit sphere}. \] The volume of a unit sphere is \( \frac{4}{3}\pi \), so the outward flux is: \[ 3 \times \frac{4}{3}\pi = 4\pi. \] Numerically, \[ 4\pi \approx 12.566. \] Thus, the value of the outward flux is approximately \( 12.6 \).