Question

Evaluate the surface integral. s y ds, s is the helicoid with vector equation r(u, v) = u cos(v), u sin(v), v , 0 ≤ u ≤ 6, 0 ≤ v ≤ π.

Answer 1

Compute the surface element:

$\mathrm dS=\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv$

$\vec r(u,v)=(u\cos v,u\sin v,v)\implies\begin{cases}\vec r_u=(\cos v,\sin v,0)\\\vec r_v=(-u\sin v,u\cos v,1)\end{cases}$

$\|\vec r_u\times\vec r_v\|=\sqrt{\sin^2v+(-\cos v)^2+u^2}=\sqrt{1+u^2}$

So the integral is

$\displaystyle\iint_Sy\,\mathrm dS=\int_0^\pi\int_0^6u\sin v\sqrt{1+u^2}\,\mathrm du\,\mathrm dv$

$=\displaystyle\left(\int_0^\pi\sin v\,\mathrm dv\right)\left(\int_0^6u\sqrt{1+u^2}\,\mathrm du\right)$

$=\dfrac23(37^{3/2}-1)$

Answer 2

The surface integral (y ds), s is the helicoid with vector equation r(u,v) = u cos(v), u sin(v), v , 0 ≤ u ≤ 6, 0 ≤ v ≤ π is $\rm \dfrac{2}{3}(37^{\frac{3}{2}}-1)$ and this can be determine by doing the integration.

Given :

The surface integral (y ds), s is the helicoid with vector equation r(u, v) = u cos(v), u sin(v), v , 0 ≤ u ≤ 6, 0 ≤ v ≤ π.

The following calculation can be used in order to evaluate the surface integral.

The surface integral is given by:

$\rm dS = ||\bar{r_u}\times \bar{r_v}||\;du\;dv$  --- (1)

where $\rm r_u$ and $\rm r_v$ is given by:

$\rm r_u = (cos \; v , sin\;v,0)$

$\rm r_v = (-u\;sin \; v , u\;cos\;v,1)$

Now, evaluate $\rm ||r_u\times r_v||$.

$\rm ||r_u\times r_v|| = \sqrt{sin^2v+(-cos\;v)^2+u^2}$

$\rm ||r_u\times r_v||=\sqrt{1+u^2}$

Now, substitute the known values in the expression (1).

$\rm \int\int_SydS = \int^{\pi}_0\int^6_0u\;(sin\;v) \sqrt{1+u^2} \;du\;dv$

Now, integrate the above expression.

$\rm \int\int_SydS = \dfrac{2}{3}(37^{\frac{3}{2}}-1)$

For more information, refer to the link given below:

https://brainly.com/question/14502499