Question

$\mathbf{\iint }\mathbf{V}\mathbf{-}\mathbf{nd\sigma }$over the entire surface of a cube in the first octant with edges of length $\mathbf{2}\mathbf{}$along the coordinate axes, where$\mathbf{V}\mathbf{=}\left(x^2-y^2\right)\mathbf{i}\mathbf{+}\mathbf{3}\mathbf{yj}\mathbf{-}\mathbf{2}\mathbf{xzk}$.

Short answer

The solution derived is$\mathrm{I}=24.$

Step-by-step solution

Given Information.

The given equation is$\mathrm{V}=({\mathrm{x}}^2-{\mathrm{y}}^2)\mathrm{i}+3\mathrm{yj}-2\mathrm{xzk}$

Definition of vector.

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines its magnitude.

Apply divergence theorem.

Apply the divergence theorem and the fact that $\mathrm{\nabla }\times \mathrm{V}=2\mathrm{x}+3-2\mathrm{x}=3$then the integral becomes as mentioned below.

$\mathrm{I}=3{\iiint }_{\mathrm{V}} \mathrm{d\tau }$

$=3\times 8$

$=24$

Hence, the solution derived is $\mathrm{I}=24.$.