Q. 14 - Calculus

Question

Write down a definite integral that describes the arc length around an ellipse with horizontal radius of 3 units and vertical radius of 2 units. The ellipse in this exercise extends 6 units across and 4 units vertically and has equation $\frac{x^2}{9}+\frac{y^2}{4}=1.$ Don’t try to solve the integral; just write it down.

Step-by-Step Solution

Answer

The definite integral that describes the arclength around an ellipse with a horizontal radius of 3 units and a vertical radius of 2 units is $\frac{2}{3}{\int }_{-3}^3\frac{\sqrt{81-5x^2}}{\sqrt{9-x^2}}dx.$

Step 1. Given Information.

The given equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1.$

Step 2. Write definite integral.

It is given that the equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1.$

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

$$\begin{gathered} 4x^2+9y^2=36 \\ 9y^2=36-4x^2 \\ {y}^2=\frac{1}{9}\left[36-4x^2\right] \\ y=\pm \frac{2}{3}\sqrt{9-x^2} \end{gathered}$$

As we know the positive sign shows the upper half of the ellipse and the negative sign shows the lower half of the ellipse. So, if we take the arc length of ellipse E as twice the arc length of the upper half of the ellipse, that is $E=2AB.$

The arc length of E for the function $y=\frac{2}{3}\sqrt{9-x^2}, y\text{'}=-\frac{2x}{3\sqrt{9-x^2}}$ by using the definite integral is

$$\begin{gathered} E=2{\int }_{-3}^3\sqrt{1+{(-\frac{2x}{3\sqrt{9-x^2}})}^2}dx \\ E=2{\int }_{-3}^3\sqrt{1+\frac{4x^2}{9(9-x^2)}}dx \\ E=\frac{2}{3}{\int }_{-3}^3\frac{\sqrt{81-5x^2}}{\sqrt{9-x^2}}dx \end{gathered}$$