Problems from Integral on a surface

Development:

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

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

• 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.

$${\int }_{S}fdS={\int }_0^1{\int }_0^{2\pi }r\cdot \sqrt{4r^2+1}d\theta dr={\int }_0^{1}2\pi r\sqrt{4r^2+1}dr=$$ $$={\displaystyle \frac{\pi }{4}}{\int }_0^{1}8r\sqrt{4r^2+1}dr={\displaystyle \frac{\pi }{4}}\cdot [{\displaystyle \frac{1}{\sqrt{4r^2+1}}}{]}_0^1={\displaystyle \frac{\pi }{4}}({\displaystyle \frac{1}{\sqrt{5}}}-1)$$

Solution:

${\int }_{S}fdS={\displaystyle \frac{\pi }{4}}({\displaystyle \frac{1}{\sqrt{5}}}-1)$

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