Question

Sketch the solid whose volume is given by the integrated integral.

\(\int\limits_0^1 {\int\limits_0^1 {\left( {2 - {x^2} - {y^2}} \right)dydx} } \)

Short answer

Given:- function- \(2 - {x^2} - {y^2}\)

To find:- Volume of solid and sketching it

Step-by-step solution

Step (1):- Defining limits

Limit is the value of function which approaches to some values.

Step (2):- Sketching solid

Consider the following integrated integral to sketch the solid

\(\int\limits_0^1 {\int\limits_0^1 {\left( {2 - {x^2} - {y^2}} \right)dydx} } \)

From the integral and find limit of variable are:

\(0 \le x \le 1,0 \le y \le 1\)

Use these limits and find limit of variable:

\(\begin{array}{l}0 \le {x^2} \le 1\\0 \le {y^2} \le 1\end{array}\)

\(\begin{array}{l}0 \le {x^2} + {y^2} \le 2\\ - 2 \le - {x^2} - {y^2} \le 2\\2 - 2 \le 2 - {x^2} - {y^2} \le 2\\0 \le 2 - {x^2} - {y^2} \le 2\end{array}\)

The solid is formed with line segment parallel to z-axis.

At any point (x,y) on rectangle \(0 \le x \le 1\& 0 \le y \le 1\)

Starting from point on plane z=0 is \(\left( {{x_0},{y_0},0} \right)\) to ending point on surface \(z = 2 - {x^2} - {y^2}\)

=\(\left( {{x_0},{y_0},2 - {x_0}^2 - {y_0}^2} \right)\)

So upper face of solid is surface,

\(z = 2 - {x^2} - {y^2}\)

And lower face is rectangle in plane z=0 is,

\(0 \le x \le 1\& 0 \le y \le 1\)

The lateral surfaces are x=1 , y=1 , y=0 , x=0. The six faces of solid is as shown figure