Q. 14 Write down a definite integral t... [FREE SOLUTION]

Chapter 6: Q. 14 (page 539)

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鈥檛 try to solve the integral; just write it down.

Short Answer

Expert verified

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 $\frac{2}{3} {鈭�}_{鈭� 3}^{3} \frac{\sqrt{81 鈭� 5 x^{2}}}{\sqrt{9 鈭� x^{2}}} d x .$

Step by step solution

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 yas a function of x,

$4 x^{2} + 9 y^{2} = 36 9 y^{2} = 36 - 4 x^{2} y^{2} = \frac{1}{9} [36 - 4 x^{2}] y = 卤 \frac{2}{3} \sqrt{9 - x^{2}}$

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 $E = 2 A B .$

The arc length of Efor the function role="math" localid="1650710860569" $y = \frac{2}{3} \sqrt{9 - x^{2}}, y' = 鈭� \frac{2 x}{3 \sqrt{9 鈭� x^{2}}}$ by using the definite integral is

$E = 2 {鈭�}_{鈭� 3}^{3} \sqrt{1 + {(鈭� \frac{2 x}{3 \sqrt{9 鈭� x^{2}}})}^{2}} d x E = 2 {鈭�}_{鈭� 3}^{3} \sqrt{1 + \frac{4 x^{2}}{9 (9 鈭� x^{2})}} d x E = \frac{2}{3} {鈭�}_{鈭� 3}^{3} \frac{\sqrt{81 鈭� 5 x^{2}}}{\sqrt{9 鈭� x^{2}}} d x$