Chapter 6.2, Problem 15E

To determine

To calculate: The integral ${\displaystyle {\int }_{-3}^3\sqrt{9-x^2}}dx$ with the help of the graph of the integrand.

Answer to Problem 15E

The value of the integral ${\displaystyle {\int }_{-3}^3\sqrt{9-x^2}}dx$ is $\frac{9\pi }{2}\text{units}$ .

Explanation of Solution

Given:

The integral is ${\displaystyle {\int }_{-3}^3\sqrt{9-x^2}}dx$ .

Concept used:

If a nonnegative function $y=f\left(x\right)$ is integrable over the interval $\left[a,b\right]$ , then the area $A$ lies below the curve $y=f\left(x\right)$ that is bounded from $a\text{ to }b$ is equal to the integral of $f$ from a to b that can be mathematically expressed as $A={\displaystyle {\int }_a^{b}f\left(x\right)dx}$ .

Formula used:

The area of the semicircle is as follows:

  $A=\frac{1}{2}\cdot \pi r^2$

Calculation:

In the given integral, the integrand function is $f\left(x\right)=\sqrt{9-x^2}$ that would be semicircle, where the values of y are positive.

The equation of circle for the semicircle $\sqrt{9-x^2}$ is $y^2=9-x^2\Rightarrow x^2+y^2=9$ ,

Comparing this with the standard equation of circle $x^2+y^2=r^2$ , we get r = 3.

Now draw the graph of $f\left(x\right)=\sqrt{9-x^2}$ and make bounds of $x=-3\text{ and }x=3$ as shown in Figure 1.

  

It can be seen that the area under a line from $x=-3$ to $x=3$ is a semicircle.

Use the formula of the semicircle to obtain the area of the semicircle, where radius is 3.

  $\begin{array}{c}A=\frac{1}{2}\pi r^2\\ =\frac{1}{2}\pi \cdot 3^2\\ =\frac{9\pi }{2}\end{array}$

Therefore, by the concept the value of the integral is equal to the area of the circle.

The value of the integral ${\displaystyle {\int }_{-3}^3\sqrt{9-x^2}}dx$ is $\frac{9\pi }{2}\text{units}$ .

Conclusion:

Thus, thevalue of the integral ${\displaystyle {\int }_{-3}^3\sqrt{9-x^2}}dx$ is $\frac{9\pi }{2}\text{units}$ .