Problem 37 - Calculus Early Transcendentals

Question

Identify and sketch the following sets in spherical coordinates. $$\\{(\rho, \varphi, \theta): \rho=4 \cos \varphi, 0 \leq \varphi \leq \pi / 2\\}$$

Step-by-Step Solution

Answer

Based on the given set in spherical coordinates, we determined that the shape we're sketching is a right circular cone with its apex at the origin and extending from the positive z-axis to the xy-plane. The range of z is from 0 to 2, and the azimuthal angle θ varies from 0 to 2π, creating a complete cone shape.

Step 1: Identify the given set in spherical coordinates

The given set in spherical coordinates is: $$\\{(\rho, \varphi, \theta): \rho=4 \cos \varphi, 0 \leq \varphi \leq \pi / 2\\}$$ We can observe that the equation consists of spherical coordinates, where the radius ρ depends on the angle φ. The range of φ is from 0 to π/2, while the azimuthal angle θ is not specified, which means that it can vary from 0 to 2π.

Step 2: Convert the equation into Cartesian coordinates

We know the conversion formulas for spherical to Cartesian coordinates are: $$x = \rho\sin\varphi\cos\theta$$ $$y = \rho\sin\varphi\sin\theta$$ $$z = \rho\cos\varphi$$ Replace ρ in these formulas with the given equation: $$\rho = 4\cos\varphi$$ Therefore, we have: $$x = (4\cos\varphi)\sin\varphi\cos\theta$$ $$y = (4\cos\varphi)\sin\varphi\sin\theta$$ $$z = (4\cos\varphi)\cos\varphi$$

Step 3: Identify the shape

From the given equation, we know that $$z = (4\cos\varphi)\cos\varphi = 2(\cos\varphi)(2\cos\varphi) = 2(\cos\varphi)(1 - 2\sin^2 \frac{\varphi}{2})$$ As φ varies from 0 to π/2, sin²(φ/2) varies from 0 to 1, which means z varies from 0 to 2. Now, we can rewrite the equations of x and y as follows: $$x^2 = (4\sin\varphi\cos\varphi\cos\theta)^2 = (2\sin\varphi\cos\varphi)^2 (1 - \sin^2\theta)$$ $$y^2 = (4\sin\varphi\cos\varphi\sin\theta)^2 = (2\sin\varphi\cos\varphi)^2 (\sin^2\theta)$$ Adding these two equations, we get: $$x^2 + y^2 = (2\sin\varphi\cos\varphi)^2 (1 - \sin^2\theta + \sin^2\theta)$$ Which simplifies to: $$x^2 + y^2 = 4\sin^2\varphi\cos^2\varphi$$ This implies that: $$x^2 + y^2 = (2\sin\varphi\cos\varphi) (2\sin\varphi\cos\varphi) = z^2$$ This equation represents a cone with its apex at the origin.

Step 4: Sketch the shape

Now, we will sketch the shape. Since θ varies from 0 to 2π (a complete circle), we will have a full cone shape. As φ varies from 0 to π/2, the cone extends from the positive z-axis to the xy-plane. This results in a cone where the apex is at the origin, z ranges from 0 to 2, and its bottom lies on the xy-plane. The sketch can be visualized as a right circular cone with its apex at the origin, extending from the positive z-axis to the xy-plane.