Question

Enclosed by \(z = {x^2} + {y^2}\) and \(z = 2y\).

Short answer

We should know about computer algebra system to find the exact volume of the solid.

Step-by-step solution

Given data

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

The objective is to find the exact volume of the solid.

Maple command:

The variable \(z = {x^2} + {y^2} \ge 0\)us the following command to draw the following surfaces.

\(\begin{aligned}{l} > with(plots);\\ > implicit{\rm{ }}plot{\rm{ }}3d(\{ z = 2y,z = {x^2} + {y^2}\} ,x = - 3...3,y = - 3...3,z = 0...4);\\\end{aligned}\)

Graph representation:

Since the variable \(z\)s is positive and \(z = 2y\)the variable \(y\) is also positive.

The surfaces are intersection on

\(\begin{aligned}{l}{x^2} + {y^2} = 2y\\{x^2} + {y^2} - 2y = 1\\{x^2} + {(y - 1)^2} = {1^2}\end{aligned}\)

This circle is boundary for projection region of solid on \(xy\)-plane.

So, the projection region \(D\) is disk of radius 1 and center at (0,1).

Graph:

Maple command

The region:

\(\begin{aligned}{l}D = \{ (x,y)/0 \le {x^2} + {(y - 1)^2} \le 1\} \\ = \left\{ {(x,y)| - 1 \le x \le 1,1 - \sqrt {1 - {x^2}} \le y \le 1 + \sqrt {1 - {x^2}} } \right\}\end{aligned}\)

The lower surface is \(z = {x^2} + {y^2}\) upper surface is \(z = 2y\).

The height of solid is \(2y - {(x + y)^2}\).

The volume of the solid enclosed by the given surface is \(V = \int\limits_{x = - 1}^1 {\int\limits_{y = 1 - \sqrt {1 - {x^2}} }^{1 + \sqrt {1 - {x^2}} } {(2y - ({x^2} + {y^2})} dydx} \)

Use the following maple command to find this iterated integral:

\({\mathop{\rm int}} \left( {(2y - ({x^2} + {y^2})),y = 1 - \sqrt {1 - {x^2}} .....1 + \sqrt {1 - {x^2}} ,x = - 1....1} \right);\)

This output the volume of \(\frac{\pi }{2}\).

Hence, the volume of the solid is \(\frac{\pi }{2}\).