Chapter 1.4, Problem 64E

To determine

The domain of the function $y=\sqrt{36-x^2}$

Answer to Problem 64E

The domain of $A$ is $0<x<6$

Explanation of Solution

Given:

The rectangle bounded by the x-axis and the semicircle $y=\sqrt{36-x^2}$

Formula used:

Area of rectangle $A=lb$

Calculation:

Consider the rectangle bounded by the x-axis and the semicircle $y=\sqrt{36-x^2}$

The following is the figure of the rectangle.

  

From the above figure,

Length of rectangle $l=x+x=2x$

Breadth of rectangle $b=y$

Area of rectangle $A=lb$

  $A=2x,y\text{where}y=\sqrt{36-x^2}$

Hence,

  $A=2x\sqrt{36-x^2}\text{sq}\text{units}$

The function $2x\sqrt{36-x^2}$ is defined for $36-x^2\ge 0$ because the term under the square root cannot be negative.

  $\begin{array}{l}36-x^2\ge 0\text{or}{x}^2-36\le 0\\ Or\\ \left(x-6\right)\left(x+6\right)\le 0\end{array}$

That is, $x$ lies between -6 and 6

In the interval $\left[-6,-1\right]$ , the area $A$ is negative and if $x=0,6$ then A is zero.

A cannot be negative and zero.

That is, $x$ is non-negative and $x\ne 0,x\ne 6$

Hence, the domain of $A$ is $0<x<6$

Conclusion:

  $x$ is non-negative and $x\ne 0,x\ne 6$ . Hence, the domain of $A$ is $0<x<6$