Let R be the region in the first quadrant bounded by the graphs of (x^2/ 9) + (y^2 /81)=1 and 3x+y=9, how do you find the area?

Mar 9, 2016

$\left(\frac{27}{4}\right) \left(\pi - 2\right)$ areal units.i

Explanation:

The ellipse and the straight line intersect at B(3, 0) and A(9, 0)
The semi-axes of the ellipse are 9 and 3. Its area is $27 \pi$
The area of the $\triangle O A B = \frac{27}{2}$.

The area bounded by the line and the ellipse in the first quadrant
= area of the ellipse in the quadrant $-$ the areal below the line in the quadrant
= $\pi \left(9\right) \left(3\right) \left(\frac{1}{4}\right) - \left(\frac{1}{2}\right) \left(3\right) \left(9\right)$