Integral on a surface

Let's follow the procedure:

• Take the parametrization of the surface $S$, and compute its vectors $T_u$, $T_v$. Then compute the vector product and calculate the norm of the result.

Notice that the parametrized surface is a parabola. We calculate the vectors

$$T_r=(\cos\theta, \sin\theta, 2\cdot r)$$

$$T_{\theta}=(-r\cdot\sin\theta, r\cdot\cos\theta,0)$$

and calculate the vector product: $$T_r \times T_{\theta}=\left| \begin{matrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ \cos\theta & \sin\theta & 2r \\ -r\cdot\sin\theta & r\cdot\cos\theta & 0 \end{matrix} \right| =$$ $$= -2\cdot r^2\cdot \cos\theta\cdot \overrightarrow{i} -2 \cdot r^2\cdot \sin\theta\cdot \overrightarrow{j} + r(\sin^2\theta+\cos^2\theta)\cdot \overrightarrow{k}= $$ $$= r\cdot(-2\cdot r\cdot \cos\theta, -2r\cdot \sin\theta,1)$$

$$||T_r \times T_{\theta}=r\cdot || (-2\cdot r\cdot \cos\theta, -2r\cdot \sin\theta,1) ||=r\cdot \sqrt{4r^2+1}$$

• Substitute $x$, $y$ and $z$ by $x (u, v), y (u, v)$ and $z (u, v)$ in the function $f$, in accordance with the given parametrization. $f(x,y,z)=1$ it does not vary as it is a constant function.

• Calculate the resultant integral.

$$\int_S f \ dS=\int_0^1 \int_0^{2\pi} r\cdot\sqrt{4r^2+1}d\theta dr=\int_0^1 2\pi r\sqrt{4r^2+1}dr=$$ $$=\dfrac{\pi}{4} \int_0^1 8r\sqrt{4r^2+1}dr=\dfrac{\pi}{4}\cdot\Big[\dfrac{1}{\sqrt{4r^2+1}}\Big]_0^1=\dfrac{\pi}{4}\Big(\dfrac{1}{\sqrt{5}}-1\Big)$$