Question

Evaluate each of the integrals in Problems$\mathbf{3}\mathbf{}$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 volumerole="math" localid="1657334446941" ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{\le }\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{\le }\mathbf{z}\mathbf{\le }\mathbf{5}\mathbf{,}\mathbf{V}\mathbf{=}\left(\sqrt{x^2+y^2}\right)\mathbf{(}\mathbf{ix}\mathbf{+}\mathbf{jy}\mathbf{)}$

Short answer

The solution of the integrals is $\iiint \mathrm{\nabla }\times Vd\tau =80\pi$.

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.

Apply Gauss’ Theorem.

Apply Gauss' theorem and use the fact mentioned below.

$V\times \widehat{\rho }={\rho }^2,$

$\rho =2$

In cylindrical coordinates.$d\sigma =\rho d\theta dz$

Solve further as shown below.

$\iint \mathrm{V}\times \widehat{\rho }={\int }_0^5 {\int }_0^{2\pi } 2^{3}d\theta dz$

$=8\left(2\pi \right)\left(5\right)$

$=80\mathrm{\pi }$

Hence, the solution of the integrals is $\iiint \mathrm{\nabla }\times Vd\tau =80\pi$.