The equations of the given curves are x=−3y2 and x=1−4y2.
To find the points of intersection, equate the values of x:
−3y2=1−4y2
y2=1⇒y=±1
The region is bounded between y=−1 and y=1. In this interval, 1−4y2≥−3y2.
The required area A is given by:
A=∫−11((1−4y2)−(−3y2))dy
A=∫−11(1−y2)dy
Since 1−y2 is an even function:
A=2∫01(1−y2)dy
A=2[y−3y3]01
A=2(1−31)=2(32)=34
Answer: 34