Q. 13 - Calculus

Question

Write down a definite integral that describes the circumference of a circle of radius 5. Don’t try to solve the integral; just write it down.

Step-by-Step Solution

Answer

The definite integral that describes the circumference of a circle of radius $5 \mathrm{is} 10{\int }_{-5}^5\frac{dx}{\sqrt{25-x^2}}.$

Step 1. Given Information.

The given radius of the circle is $5.$

Step 2. Write the definite integral.

It is given that the radius of the circle is $5.$ So, the equation of the circle is $x^2+y^2=5^2.$

Now, if we express the y as a function of x,

$$\begin{gathered} y^2=25-x^2 \\ y=\pm \sqrt{25-x^2} \end{gathered}$$

As we know the positive sign shows the upper half of the circle and the negative sign shows the lower half of the circle. So, if we take the arc length of circle C as twice the arc length of the upper half of the circle.

The arc length of C for the function $y=\sqrt{25-x^2}, y\text{'}=-\frac{x}{\sqrt{25-x^2}}$ by using the definite integral is

$$\begin{gathered} C=2{\int }_{-5}^5\sqrt{1+{\left(-\frac{x}{\sqrt{25-x^2}}\right)}^2}dx \\ C=2{\int }_{-5}^5\sqrt{1+\frac{x^2}{25-x^2}}dx \\ C=2{\int }_{-5}^5\frac{5}{\sqrt{25-x^2}}dx \\ C=10{\int }_{-5}^5\frac{dx}{\sqrt{25-x^2}} \end{gathered}$$