Question

In Exercises \(27-30,\) sketch the region bounded by the graphs of the equations. $$ x^{2}+y^{2}=1, x+z=2, z=0 $$

Short answer

The bounded region formed by the three equations is a circular disk with radius 1 (from the first equation), located in the xy-plane (z=0), and cut off at z=2 by the plane represented by the second equation \(x+z=2\).

Step-by-step solution

Graph the first equation

The equation \(x^{2}+y^{2}=1\) represents a circle with radius 1 on the xy-plane (z=0). To graph it, first draw a horizontal and a vertical line representing the x and y axes respectively. On the horizontal line (x-axis), mark a point on the right side at a distance of 1 unit from the origin and similarly, mark a point on the left side at a distance of 1 unit from the origin. Repeat the same process on the vertical line (y-axis). Then draw a circle with the origin as the center and passing through the points marked.

Graph the second equation

The equation \(x+z=2\) represents a plane in 3D space. To bring this up into a 2D graph, set x=0 then z=2 and set z=0 then x=2. These points are the x and z intercepts. Draw the plane with one end at (2,0) on the x-axis and another end at (0,2) on the z-axis.

Graph the third equation

The equation z=0 represents the xy-plane itself. This is essentially the base where the circle from step 1 sits.

Visualize the 3D images

In a 3D coordinate system, the x-axis and y-axis form the horizontal plane with the z-axis extending vertically upwards and downwards. The xy-plane (z=0) is the base on which lies the circle from step 1. The plane represented by the second equation intersects this base. This 3D image visualises their interactions specifying the region bounded by the three equations.