91影视

Using the given equation in spherical coordinates and the general formulas for conversion, we can convert the set to Cartesian coordinates: $$\rho = 2 \sec \varphi$$ Using the fact that \(\sec \varphi = \frac{1}{\cos \varphi}\), we have: $$\rho = 2 \frac{1}{\cos \varphi}$$ Now, we know that \(z = \rho \cos \varphi\), so substituting the expression for \(\rho\) gives us: $$z = 2 \frac{1}{\cos \varphi} \cos \varphi$$ $$z = 2$$ Thus, the Cartesian equation for the given set is \(z = 2\).

Now that we have converted the set to Cartesian coordinates, we can sketch it. The equation \(z = 2\) represents a horizontal plane parallel to the xy-plane at a height of 2 units above the xy-plane. To sketch this, draw the standard xyz-coordinate axes. Then, draw a horizontal plane parallel to the xy-plane at a height of 2 units above the xy-plane. Label this plane as \(z = 2\). This plane represents the given set in spherical coordinates. Since the bounds for \(\varphi\) are between \(0\) and \(\frac{\pi}{2}\), it means that only the region of the plane that lies within the upper hemisphere (i.e., above xy-plane) is relevant for this exercise. To show this, shade the region of the xy-plane below the plane \(z = 2\) and the region of the \(z = 2\) plane within the upper hemisphere to emphasize that only the latter is part of the solution set.