Question

Evaluate each of the integrals in Problems $\mathbf{3}$to $\mathbf{8}$as either a volume integral or a surface integral, whichever is easier.

$\mathbf{\iiint }\mathbf{\nabla }\mathbf{\times }\mathbf{Vd\tau }$ over the unit cube in the first octant, where $\mathbf{V}\mathbf{=}\left(x^3-x^2\right)\mathbf{yi}\mathbf{+}\left(y^3-2y^2+y\right)\mathbf{xj}\mathbf{+}\left(z^2-1\right)\mathbf{k}$

Short answer

The solution of the integrals is $\iiint \mathrm{\nabla }\times \mathrm{Vd\tau }=1$.

Step-by-step solution

Given Information.

The given integral is $\iiint \mathrm{\nabla }\times Vd\tau$.

Definition of Divergence’s Theorem.

Divergence theorem, often known as Gauss' theorem or Ostrogradsky's theorem, is a theorem that connects the flow of a vector field across a closed surface to the field's divergence in the volume enclosed. According to this theorem, the surface integral of a vector field over a closed surface, also known as the flux through the surface, equals the volume integral of the divergence over the region inside the surface.

Solve the equation.

Solve the equation as shown below.

$\mathrm{\nabla }\times \mathrm{V}=\left(3x^2-2x\right)y+\left(3y^2-4y+1\right)x+2z$

$=3x^{2}y+3x{y}^2-6xy+x+2z$

Then, the integrals become as shown below.

${\int }_0^1 {\int }_0^1 {\int }_0^1 \left(3x^{2}y+3x{y}^2-6xy+x+2z\right)dxdydz$

${\int }_0^1 {\int }_0^1 \left(y+\frac{3y^2}{2}-3y+\frac{1}{2}+2z\right)dydz$

${\int }_0^1 \left(\frac{1}{2}+\frac{1}{2}-\frac{3}{2}+\frac{1}{2}+2z\right)dz=1$

$\iiint \mathrm{\nabla }\times Vd\tau =1$

Hence, the solution of the integrals is $\iiint \mathrm{\nabla }\times Vd\tau =1$.