# Problems from Integral on a surface

Compute the integral of $$f (x, y, z) =1$$ along the surface parametrized by $$\varphi (r, \theta)=(r \cdot \cos \theta, r \cdot \sin \theta, r^2)$$

Namely $$\left\{ \begin{array}{l} x=r \cdot \cos \theta \\ y=r \cdot \sin \theta \\ z=r^2 \end{array} \right.$$, for $$r \in [0,1]$$ and $$\theta \in [0, 2\pi]$$

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### Development:

• 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.

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

### Solution:

$$\int_S f \ dS=\dfrac{\pi}{4}\Big(\dfrac{1}{\sqrt{5}}-1\Big)$$

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