find the area of the region between the curve y = 3-x^2 and the line y = -1 by integrating with respect to y
you first have to solve for x
x=(3-y)^(1/2)
then area
⌠ (3-y)^(1/2 dy from -1 to 3
⌡
=-2/3 (3-y)^(3/2 | from -1 to 3
= -2/3(0 - 8)
= 16/3
but this is only half the region, so the area = 32/3 units^2
Question: Why would anybody ask you to do a question the hard way, the other way with respect to x is easier.