Answer:
[tex]\bf \frac{4}{3}(53\sqrt{53}-8)[/tex]
Step-by-step explanation:
Let us recall that if f(x,y,z) is a smooth mapping defined over a surface S, and Φ(u, v) is a smooth parametrization of the surface with (u,v) varying in a certain region R, then
[tex]\bf \displaystyle\iint_Sfd\sigma=\displaystyle\iint_Rf(\phi(u,v))\left\|\displaystyle\frac{\partial \phi}{\partial u}\times\displaystyle\frac{\partial\phi}{\partial v}\right\|dudv[/tex]
Now,
[tex]\bf f(\phi(u,v))=f(2ucos(v),2usin(v),u)=2ucos(v)+2usin(v)[/tex]
[tex]\bf \displaystyle\frac{\partial \phi}{\partial u}=(2cos(v),2sin(v),1)\\\\\displaystyle\frac{\partial \phi}{\partial v}=(-2usin(v),2ucos(v),0)[/tex]
[tex]\bf \begin{vmatrix}\hat i&\hat j&\hat k\\2cos(v)&2sin(v)&1\\-2usin(v)&2ucos(v)&0\end{vmatrix}=(-2ucos(v),2usin(v),4ucos^2(v)+4usin^2(v)))=\\\\=(-2ucos(v),2usin(v),4)[/tex]
[tex]\bf \left\|(-2ucos(v),2usin(v),4)\right\|=\sqrt{4u^2cos^2(v)+4u^2sin^2(v)+16}=\\\\=\sqrt{4u^2+16}=2\sqrt{u^2+4}[/tex]
and
[tex]\bf \displaystyle\iint_Rf(\phi(u,v))\left\|\displaystyle\frac{\partial \phi}{\partial u}\times\displaystyle\frac{\partial\phi}{\partial v}\right\|dudv=\displaystyle\int_{0}^{7}\displaystyle\int_{0}^{\pi}[(2u cos(v)+2u sin(v))2\sqrt{u^2+4}]dudv=\\\\=2\displaystyle\int_{0}^{7}u\sqrt{u^2+4}du\left(\displaystyle\int_{0}^{\pi}cos(v)dv+\displaystyle\int_{0}^{\pi}sin(v)dv\right)= [/tex]
[tex]\bf =2\displaystyle\int_{0}^{7}u\sqrt{u^2+4}du\left(\displaystyle\int_{0}^{\pi}cos(v)dv+\displaystyle\int_{0}^{\pi}sin(v)dv\right)=\\\\=4\displaystyle\int_{0}^{7}u\sqrt{u^2+4}du=\boxed{\displaystyle\frac{4}{3}(53\sqrt{53}-8)}[/tex]
