Find the area of the smaller region bounded by the ellipse x2/9 + y2/4 = 1 and the line x/3 + y/2 = 1?

Answer 1

x^2/9 + y^2/4 = 1

y^2/4 = 1 - x^2/9

y^2 = 4 - (4/9) x^2

y = +or- sqrt( 4 - (4/9)x^2)

vs.

y/2 = 1 - x/3

y = 2 - (2/3) x

THe intersection points:

2 - (2/3)x = sqrt(4 - (4/9)x^2)

[2 - (2/3)x]^2 = 4 - (4/9)x^2

4 - (8/3)x + (4/9)x^2 = 4 - (4/9)x^2

36 - 24x + 4x^2 = 36 - 4x^2

-24x + 4x^2 = -4x^2

8x^2 - 24x = 0

8x( x- 3) = 0

x = 0 or x=3

THe intersection points are (0,2) and (3,0)

Area is integral { sqrt( 4 - (4/9)x^2) - [ 2 - (2/3) x ] }, x=[0,3]

The anti-derivative integral, per the calculator, is:

x* sqrt( 1 - x^2/9) + 3* arcsin(x/3) - 2x + (1/3)x^2

for x = 3, the limit is 3 * sqrt ( 1 - 9/9) + 3 * arcsin(1) - 6 + 3 =

3 * sqrt( 0) + 3 * pi2 - 6 + 3 = pi/2

for x = 0, the limit is 0 * sqrt( 1 - o^2/9) + 3 * arcsin(0/3) - 2(0) + (1/3)*0^2

= 0 + 3 * arcsin(0) - 0 + 0

= 3 * arcsin(0)

= 0

The area is pi/2 - 0 = pi/2