Question

Use the divergence theorem to calculate the surface integral s f · ds; that is, calculate the flux of f across s. f(x, y, z) = x4i − x3z2j + 4xy2zk, s is the surface of the solid bounded by the cylinder x2 + y2 = 1 and the planes z = x + 7 and z = 0.

Answer 1

$\mathbf f(x,y,z)=x^4\,\mathbf i-x^3z^2\,\mathbf j+4xy^2z\,\mathbf k$
$\mathrm{div}(\mathbf f)=\dfrac{\partial(x^4)}{\partial x}+\dfrac{\partial(-x^3z^2)}{\partial y}+\dfrac{\partial(4xy^2z)}{\partial z}=4x^3+0+4xy^2=4x(x^2+y^2)$


Let $\mathcal D$ be the region whose boundary is $\mathcal S$. Then by the divergence theorem,

$\displaystyle\iint_{\mathcal S}\mathbf f\cdot\mathrm d\mathbf S=\iiint_{\mathcal D}4x(x^2+y^2)\,\mathrm dV$

Convert to cylindrical coordinates, setting

$x=r\cos\theta$
$y=r\sin\theta$

and keeping $z$ as is. Then the volume element becomes


$\mathrm dV=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz$

and the integral is

$\displaystyle\iiint_{\mathcal D}4x(x^2+y^2)\,\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=0}^{z=r\cos\theta+7}4r\cos\theta\cdot r^2\cdot r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle4\iiint_{\mathcal D}r^4\cos\theta\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\dfrac{2\pi}3$

Answer 2

This question is based on the divergence theorem.Therefore, the flux  is  $\bold{\int f.dS = \dfrac{2 \pi}{3}}$  by using  divergence theorem .

Given:

f(x, y, z) = $\bold{x^{4} i - x^3 z^2 j+4xy^{2}z\; k}$, s is the surface of the solid bounded by the cylinder $\bold{x^2 + y^2 = 1}$ and the planes z = x + 7 and z = 0.

We need to determined the surface integral f · ds.

According to the question,

$\bold{f(x,y,z) = x^{4} i - x^3 z^2 j+4xy^{2}z\; k}$

$div (f) = \dfrac{\partial (x^4)}{\partial x} + \dfrac{\partial (-x^3 \; z^2)}{\partial y} + \dfrac{\partial (4xy^{2}z)}{\partial z} = 4x^3 + 0 + 4 x y^2 = 4x^3 + 4 x y^2$

Let  D be the region whose boundary is . Then by the divergence theorem,

⇒ $\int\int_s f.dS = \int \int \int _D 4x ( x^2 + y^2) dV$

Convert the given coordinates into cylindrical coordinates.

We get ,

x = r cos $\Theta$

y = r sin $\Theta$

z = z

Then, the dV = r dr d$\theta$ dz.

Now the integral become,

$\int \int \int _D 4x ( x^2 + y^2) dV = \int\limits^{2\pi}_0 \int\limits^1_0 \int\limits^{ r cos\theta+7}_0 4 r cos\theta \; r^2 \; r \; dz \; dr \; d\theta\\\\=4\int\int\int_D r^4 \; cos\theta \; dz \; dr \; d\theta\\\\= \dfrac{2\pi}{3}$

Therefore, the flux  is  $\bold{\int f.dS = \dfrac{2 \pi}{3}}$  by using  divergence theorem .

For more details, prefer this link:

https://brainly.com/question/23777455