91Ó°ÊÓ

In spherical coordinates, a point in space is defined using the parameters - \(\rho\), \(\varphi\), and \(\theta\). Here, \(\rho\) is the distance from the origin to the point, \(\theta\) is the angle made by the projection of the point in the XY plane and the X-axis in the XY plane (also known as the azimuthal angle), and \(\varphi\) is the angle made by the point and the Z-axis (known as the polar angle). To understand the given set, we will first convert it into Cartesian coordinates using the following relations: $$x = \rho \sin \varphi \cos \theta\\ y = \rho \sin \varphi \sin \theta\\ z = \rho \cos \varphi$$

We have the following equation for the set: $$\rho = 4 \cos \varphi$$ We are also given that the range for the angle \(\varphi\) is: $$0 \leq \varphi \leq \frac{\pi}{2}$$ The range of \(\varphi\) indicates that the figure will be present only in the upper half of the space (positively oriented Z axis). We will use this equation to express \(\rho\) in terms of \(\varphi\): $$\rho = 4 \cos \varphi$$

Now, we will substitute \(\rho\) into our Cartesian coordinate relations to find expressions for x, y, and z: $$x = (4 \cos \varphi) \sin \varphi \cos \theta\\ y = (4 \cos \varphi) \sin \varphi \sin \theta\\ z = (4 \cos \varphi) \cos \varphi$$ We can simplify \(z\) as: $$z = 4 \cos^2 \varphi$$ Now, using the trigonometric identity \(\cos^2 \varphi + \sin^2 \varphi = 1\), we have: $$z = 4(1 - \sin^2 \varphi)$$ Thus, we can express \(\sin \varphi\) as: $$\sin \varphi = \sqrt{1 - \frac{z}{4}}$$ Now, substituting \(\sin \varphi\) into our \(x\) and \(y\) equations: $$x = 4 \cos \varphi (1 - \frac{z}{4}) \cos \theta\\ y = 4 \cos \varphi (1 - \frac{z}{4}) \sin \theta$$

We can see that if we eliminate \(\cos \varphi\) and \(\theta\) from the equations of \(x\) and \(y\), we get: $$\frac{x^2}{(\cos \theta)^2} + \frac{y^2}{(\sin \theta)^2} = [4\cos \varphi(1-\frac{z}{4})]^2$$ This equation represents an elliptical surface. It opens along the Z-axis from z=0 to z=4, and the cross-section parallel to the XY plane at a certain height is an ellipse. Specifically, the given set represents an elliptical cone in the upper half space for \(0 \leq \varphi \leq \frac{\pi}{2}\). To sketch the set, start with a cone shape opening along the Z-axis from z=0. As the height increases, the cross-section (ellipse) gets smaller, eventually reaching a point at z=4. The cone only exists in the upper half space as specified by the range of \(\varphi\).