Question

Find the area of the part of the cone \(z^{2}=3\left(x^{2}+y^{2}\right)\) which is inside the sphere \(x^{2}+y^{2}+z^{2}=16\)

Short answer

The area is \[ 8\pi \].

Step-by-step solution

Identify the Intersection Curve

To find the intersection of the cone and the sphere, set the equations equal. The cone's equation is \[ z^{2} = 3 (x^{2} + y^{2}) \] and the sphere's equation is \[ x^{2} + y^{2} + z^{2} = 16 \]. Substituting the cone's equation into the sphere's equation gives \[ x^{2} + y^{2} + 3(x^{2} + y^{2}) = 16 \]. This simplifies to \[ 4(x^{2} + y^{2}) = 16 \], or \[ x^{2} + y^{2} = 4 \]. Thus, the intersection curve is a circle of radius 2 centered at the origin in the xy-plane.

Parameterize the Surface

Given the intersection curve, parameterize the cone's surface using cylindrical coordinates. Let \[ x = r\text{cos}(\theta) \], \[ y = r\text{sin}(\theta) \], and \[ z = r\text{sqrt}(3) \] for \[ 0 \le r \le 2 \] and \[ 0 \le \theta < 2\pi \]. The parameterization is thus \[ \textbf{r}(r, \theta) = (r\text{cos}(\theta), r\text{sin}(\theta), r\text{sqrt}(3)) \].

Find the Surface Element dS

Compute the partial derivatives of \[ \textbf{r}(r, \theta) \] with respect to \[ r \] and \[ \theta \]. \[ \frac{\partial \textbf{r}}{\partial r} = (\text{cos}(\theta), \text{sin}(\theta), \text{sqrt}(3)) \] and \[ \frac{\partial \textbf{r}}{\partial \theta} = (-r\text{sin}(\theta), r\text{cos}(\theta), 0) \]. The cross product \[ \frac{\partial \textbf{r}}{\partial r} \times \frac{\partial \textbf{r}}{\partial \theta} \] is \[ (-r\text{cos}(\theta)\text{sqrt}(3), -r\text{sin}(\theta)\text{sqrt}(3), r) \] with magnitude \[ \text{sqrt}((r\text{sqrt}(3)\text{cos}(\theta))^{2} + (r\text{sqrt}(3)\text{sin}(\theta))^{2} + r^{2}) = \text{sqrt}(4r^{2}) = 2r \]. Hence, the surface element \[ dS = 2r \thinspace dr \thinspace d\theta \].

Integrate the Surface Element

To find the total surface area, integrate \[ dS \] over the bounds \[ 0 \le r \le 2 \] and \[ 0 \le \theta < 2\theta \]: \[ \text{Surface Area} = \int_{0}^{2\theta} \int_{0}^{2} 2r \thinspace dr \thinspace d\theta \]. Evaluating the inner integral \[ \int_{0}^{2} 2r \thinspace dr \], compute \[ 2 \int_{0}^{2} r \thinspace dr = 2 \thinspace [\frac{r^{2}}{2}]_{0}^{2} = 2 \thinspace [2] = 4 \]. The outer integral now becomes \[ \int_{0}^{2\theta} 4 \thinspace d\theta \], which evaluates to \[ 4\theta \thinspace [\theta]_{0}^{2\theta} = 8\theta \]. Thus, the total surface area is \[ 8\pi \].

Key concepts

Intersection of Surfaces

To determine the intersection of two surfaces, you need to set their equations equal to each other. In our example, we have a cone and a sphere. The cone's equation is given by \( z^{2} = 3 (x^{2} + y^{2}) \) and the sphere’s equation is \( x^{2} + y^{2} + z^{2} = 16 \). By substituting the cone's equation into the sphere's equation, we simplify the problem to a circle in the xy-plane. This circle has a radius of 2, centered at the origin. This method allows us to visualize the intersection as a simpler shape, which is crucial for further computations.

Parameterization

Parameterizing a surface means expressing it in terms of two parameters. For the cone inside the sphere, we use cylindrical coordinates because they simplify the math and match the symmetry of the problem. Cylindrical coordinates can be defined as follows: \( x = r\text{cos}(\theta) \), \( y = r\text{sin}(\theta) \), and \( z = r\text{sqrt}(3) \). The parameters here are \( r \) (radial distance from the origin) and \( \theta \) (the angle in the xy-plane). By using these coordinates, we can express any point on the cone as \textbf{r}(r, \theta) = (r\text{cos}(\theta), r\text{sin}(\theta), r\text{sqrt}(3))\.

Cylindrical Coordinates

Cylindrical coordinates are a 3D coordinate system that extends polar coordinates by adding a height (z-coordinate). This system is particularly useful for problems involving circular symmetry, like the cone and sphere in this example. The coordinates are defined as follows: \( r \) is the radial distance from the z-axis, \( \theta \) is the angular displacement from the x-axis, and \( z \) is the height above the xy-plane. In this problem, cylindrical coordinates allow us to easily integrate over circular paths, simplifying the calculations needed to find the area of the surface.

Surface Element

The surface element, denoted as \( dS \), represents a small piece of the surface area. To compute it, we need to take partial derivatives of the parameterization with respect to \( r \) and \( \theta \). The partial derivatives yield vectors that, when crossed, give a vector perpendicular to the surface. Calculating the magnitude of this cross product then gives the area of the small surface piece. In our example, the surface element is found to be \( 2r \thinspace dr \thinspace d\theta \). This represents a tiny area on the cone's surface and helps in summing up to find the total surface area through integration.