Question

Find a vector function that represents the curve of intersection of the two surfaces

The semi ellipsoid\({x^2} + {y^2} + 4{z^2} = 4,y \ge 0\) and the cylinder\({x^2} + {z^2} = 1\)

Short answer

The required vector function that represents intersection of given two surfaces is:

\(\begin{array}{l}cos\theta i + \sqrt 3 \left| {cos\theta } \right|j + sin\theta k\\where,\;\theta \in (0,2\pi )\end{array}\)

Step-by-step solution

Solve the given equations

Here we need to find a vector function that represents the curve of intersection of the two surfaces\({x^2} + {y^2} + 4{z^2} = 4,y \ge 0\) and \({x^2} + {z^2} = 1\).

Solve both these equations as shown below, that is,

\(\begin{aligned}{x^2} + {y^2} + 4{z^2} &= 4\\{x^2} + {y^2} + {z^2} + 3{z^2} &= 4\\({x^2} + {z^2}) + {y^2} + 3{z^2} &= 4\\1 + {y^2} + 3{z^2} &= 4\\{y^2} + 3{z^2} &= 3\\\left| y \right| &= \sqrt {3 - 3{z^2}} \\y &= \sqrt {3 - 3{z^2}} .................(y \ge 0)\end{aligned}\)

Simplification and substitution

Let \(z = \sin \theta \to y = \sqrt {3 - 3{{\sin }^2}\theta } = \left| {\cos \theta } \right|\sqrt 3 \)

Also we know that \(x{}^2 + {z^2} = 1\)

Therefore\({x^2} = 1 - {\sin ^2}\theta = {\cos ^2}\theta \)

Therefore \(x = \cos \theta \)

Therefore, vector function is

\(\begin{array}{l}cos\theta i + \sqrt 3 \left| {cos\theta } \right|j + sin\theta k\\where,\;\theta \in (0,2\pi )\end{array}\)

The required vector function that represents intersection of given two surfaces is:

\(\begin{array}{l}cos\theta i + \sqrt 3 \left| {cos\theta } \right|j + sin\theta k\\where,\;\theta \in (0,2\pi )\end{array}\)