# Thread: Area problem by using the Trigonometric Substitution

1. ## Area problem by using the Trigonometric Substitution

Let R be the smaller of the two regions enclosed by the elipse 169x^2 +100y^2=16900 and the line x= 5 squar root of (2).

Find the area of the region R.

thanks!~

2. Originally Posted by shannon1111
Let R be the smaller of the two regions enclosed by the elipse 169x^2 +100y^2=16900 and the line x= 5 squar root of (2).

Find the area of the region R.

thanks!~
$\displaystyle \frac{x^2}{10^2} + \frac{y^2}{13^2} = 1$

$\displaystyle x = 5\sqrt{2}$

The ellipse is centered at the origin, stretched in the y-direction, and intersects with the vertical line at:

$\displaystyle y^2 = \frac{169}{2}$
$\displaystyle y = \pm \frac{13 \sqrt{2}}{2}$

Solving for x from the ellipse's upper half, in terms of y

$\displaystyle x^2 = 100 (1 - \frac{y^2}{169})$
$\displaystyle x = \frac{10}{13} * \sqrt{169-y^2}$

The equation above represents the curve to the right. The curve to the left is simply:
$\displaystyle x= 5\sqrt{2}$

Setup the integral now as
INTEGRAL (X_right - X_left) dy where y ranges from 0 to $\displaystyle \frac{13 \sqrt{2}}{2}$

After you find the integral remember to multiply it by two to account for the other half of the area below the horizontal axis.

I hope this helps.