Question

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

$\mathbf{\iint }\mathbf{r}\mathbf{.}\mathbf{nd\sigma }$Over the whole surface of the cylinder bounded by${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{,}\mathbf{}\mathbf{z}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{z}\mathbf{=}\mathbf{3}\mathbf{}\mathbf{;}\mathbf{}\mathbf{r}\mathbf{}\mathbf{means}\mathbf{}\mathbf{ix}\mathbf{+}\mathbf{jy}\mathbf{+}\mathbf{k}z$

Short answer

The solution of the integrals is found to be as mentioned below.

$\iint r.nd\sigma =9\pi$

Step-by-step solution

Given Information.

The given integrals is $\iint r.nd\sigma$.

Definition of Divergence’s Theorem.

The 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. 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, according to this theorem.

Step 3:Apply Gauss’ Theorem.

Apply Gauss' theorem and use the fact that$\nabla .r=3$

In cylindrical coordinates$d\tau =rd\theta drdz$, then the solution is mentioned below

$$\begin{gathered} \iint r.nd\sigma ={\int }_0^3{\int }_0^{2\pi }{\int }_0^{1}3rdrd\theta dz \\ =3\left(\frac{1}{2}\right)\left(2\pi \right)\left(3\right) \\ =9\pi \end{gathered}$$

Hence, the solution of the integrals is found to be as mentioned below.

$\iint r.nd\sigma =9\pi$