Next, we need to find where the two curves intersect, which will be the endpoints of the
region. We do this by setting the two curves equal to each other. We get y^2 = 3 − 2y^2 . The
solutions are (1, 1) and (1, −1). Therefore, in order to find the area of the region, we need to
evaluate the integral (3 − 2y^2 − y^2 ) dy = (3 − 3y^2 ) dy. We get (3 − 3y^2 ) dy = (3y − y^3 )
=(3(1) − (1)^3 )−(3(−1)−(−1)^3 ) = 4.
7.
We find the area of a region bounded by f(y) on the right and g(y) on the left at all points of the
interval [c, d] using the formula [f(y) − g(y)] dy. Here f(y) = y − 2 and g(y) = y^2 − 4y + 2.
First, let’s make a sketch of the region.
