Question

Sketch the region bounded by the surfaces \(z = \sqrt {{x^2} + {y^2}} \) and \({x^2} + {y^2} = 1\) for \(1 \le 5z \le 2\).

Short answer

The region bounded by the surfaces \(z = \sqrt {{x^2} + {y^2}} \) and \({x^2} + {y^2} = 1\) for \(1 \le z \le 2\) is sketched.

Surface equations are \(z = \sqrt {{x^2} + {y^2}} \) and \({x^2} + {y^2} = 1\).

Step-by-step solution

Standard equation of circular cone

Consider the standard equation of circular cone along the\(z\)axis.

\(\frac{{{z^2}}}{{{c^2}}} = \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}(1)\)

Consider the given surface equation.

\(z = \sqrt {{x^2} + {y^2}} \)

Take square on both sides.

\(\begin{array}{l}{z^2} = {\left( {\sqrt {{x^2} + {y^2}} } \right)^2}\\{z^2} = {x^2} + {y^2}\\\frac{{{z^2}}}{1} = \frac{{{x^2}}}{1} + \frac{{{y^2}}}{1}(2)\end{array}\)

By comparing equation (2) with (1), the given surface equation satisfies the equation of a circular cone along the \(z\)-axis.

Find the value of \({\bf{a}},{\bf{b}},{\bf{c}}\)

Find \(a,b\) and \(c\) by comparing equation (2) with equation (1).

\(\begin{array}{l}{a^2} = 1\\a = \sqrt 1 \\a = \pm 1\end{array}\)

The value of \(a\) is \( \pm 1\).

\(\begin{array}{l}{b^2} = 1\\b = \sqrt 1 \\b = \pm 1\end{array}\)

The value of \(b\) is \( \pm 1\).

\(\begin{array}{l}{c^2} = 1\\c = \sqrt 1 \\c = \pm 1\end{array}\)

The value of \(c\) is \( \pm 1\).

Put different values of \(k\)

Case i:

Let \(x = k\).

Substitute \(k\) for \(x\) in equation (2),

\(\begin{array}{l}\frac{{{z^2}}}{1} = \frac{{{k^2}}}{1} + \frac{{{y^2}}}{1}\\{z^2} - {y^2} = {k^2}(3)\end{array}\)

Modify equation (4) for \(k = 0\),

\(\begin{array}{l}{z^2} - {y^2} = {(0)^2}\\{z^2} - {y^2} = 0\\(z + y)(z - y) = 0\\z = y, - y\end{array}\)

So, the surface equation (3) has two intersecting line equations for \(k = 0\). Similarly, the surface equation (3) has a family of hyperbola traces for\(k \ne 0\).

Equation of circle

Case ii:

Let \(z = 1\).

Substitute 1 for \(z\) in equation (2),

\(\begin{array}{l}\frac{{{1^2}}}{1} = \frac{{{x^2}}}{1} + \frac{{{y^2}}}{1}\\{x^2} + {y^2} = 1\end{array}\)

This expression represents a circle with a radius of 1 along the \(z\)-axis for \(z = 1\).

Case iii:

Let \(z = 2\).

Substitute 2 for \(z\) in equation (2),

\(\begin{array}{l}\frac{{{2^2}}}{1} = \frac{{{x^2}}}{1} + \frac{{{y^2}}}{1}\\{x^2} + {y^2} = 4\end{array}\)

This expression represents a circle with a radius of 2 along the \(z\)-axis for \(z = 2\).

So, the structure of the surface equation is the circular cone along the \(z\)-axis with a circular hollow section with a radius of 1 at inside.

Consider another surface equation.

\({x^2} + {y^2} = 1\)

This surface equation represents a circle along the \(z\)-axis with the radius of 1.

The graph of the surfaces

So, the region bounded by these two surfaces \(z = \sqrt {{x^2} + {y^2}} \) and \({x^2} + {y^2} = 1\) is sketched as shown in Figure