The problem asks us to identify and sketch the equations in three-space. We will solve question number 1: $4x^2 + 36y^2 = 144$.

1. Problem Description

The problem asks us to identify and sketch the equations in three-space. We will solve question number 1: $4x^2 + 36y^2 = 144$.

2. Solution Steps

First, divide the equation by 144 to get the standard form of an ellipse:

$\frac{4x^2}{144} + \frac{36y^2}{144} = 1$

$\frac{x^2}{36} + \frac{y^2}{4} = 1$

$\frac{x^2}{6^2} + \frac{y^2}{2^2} = 1$

This is an ellipse in the $xy$-plane, centered at the origin, with semi-major axis $a=6$ along the $x$-axis and semi-minor axis $b=2$ along the $y$-axis. Since there is no $z$ term, the surface extends infinitely along the $z$-axis. Therefore, the 3D shape is an elliptic cylinder.

3. Final Answer

The equation represents an elliptic cylinder.