Questions: 9. Proponga una parametrización, de una de las dos curvas, que se forma por la intersección de las superficies Σ₁=(x, y, z) x²+y²=6 y Σ₂=(x, y, z) x²+y²+z²=16. No pierda de vista que en unas ecuaciones paramétricas es indispensable indicar el intervalo de variación del parámetro.

Solution Steps

We have two surfaces defined as follows:

• The cylinder \(\Sigma_{1} = \{(x, y, z) \mid x^{2} + y^{2} = 6\}\) describes a circular cylinder with radius \(\sqrt{6}\) along the z-axis.
• The sphere \(\Sigma_{2} = \{(x, y, z) \mid x^{2} + y^{2} + z^{2} = 16\}\) describes a sphere with radius 4 centered at the origin.

To find the intersection, we can parametrize the circular cross-section of the cylinder using the angle parameter \(\theta\): \[ x = \sqrt{6} \cos(\theta), \quad y = \sqrt{6} \sin(\theta) \] where \(\theta\) varies from \(0\) to \(2\pi\).

Next, we substitute the expressions for \(x\) and \(y\) into the equation of the sphere to find \(z\): \[ z = \sqrt{16 - (x^{2} + y^{2})} \] Substituting \(x^{2} + y^{2} = 6\): \[ z = \sqrt{16 - 6} = \sqrt{10} \]

The complete parametrization of the curve formed by the intersection of the two surfaces is given by: \[ \begin{align_} x(\theta) &= \sqrt{6} \cos(\theta) \\ y(\theta) &= \sqrt{6} \sin(\theta) \\ z(\theta) &= \sqrt{10} \end{align_} \] for \(\theta \in [0, 2\pi]\).

This describes a circle in the plane \(z = \sqrt{10}\) with radius \(\sqrt{6}\).

Final Answer