Question

Find the integration, the area of the ellipse 16x^2+25y^2=400?

Answer 1

Answer:

set the limit.

Step-by-step explanation:

Answer 2

Given,

$16x^{2} + 25y^{2} = 400$

To find ,

Area of the ellipse

Solution,

The formula to find the area of the ellipse is

$\int\limits^x_y {y} \, dx$

First write the given equation :

$16x^{2} + 25y^{2} = 400$

First we need to find the value of y :

$25y^{2} = 400 - 16x^{2}$

Now find the y value by dividing the  $400 - 16x^{2}$ with 25 and apply square root to the right hand side :

$y = \sqrt{\frac{400 - 16x^{2} }{25} }$---------(1)

Now again take the equation :

$16x^{2} + 25y^{2} = 400$ --------(2)

Divide the equation (2) with 400

$\frac{16x^{2} }{400} + \frac{25y^{2} }{400} = \frac{400}{400}$

$\frac{x^{2} }{25} + \frac{y^{2} }{16} = 1$

Area of ellipse :

$\frac{4}{5} \int\limits^5_0 {y} \, dx$  =  $\frac{4}{5} \int\limits^5_0 {\sqrt{400 - 16x^{2} } } \, dx$ --------(3)

Equation (3) is in the form of formula :

Integration of $\sqrt{a^{2} - x^{2} }$

substitute in the above formula :

$\frac{4}{5 } \int\limits^5_0 {\sqrt{16( 25 - x^{2} } } \, dx$=   $\frac{16}{5} \int\limits^5_0 {\sqrt{5^{2} - x^{2} } } \, dx$

                              = $\frac{16}{5} [ \frac{x}{2} \sqrt{25 - x^{2} } + \frac{25}{2} sin^{-1} (\frac{x}{5} )]$ 0 to 5

By substituting the limits 0 and 5 we get :

                              = $\frac{16}{5} [ \frac{25}{2} ,\frac{\pi }{2} ]$

                               = $20\pi$  

Hence the area of the ellipse = $20\pi$