Get the area of the full region first. You could integrate w respect to y after isolating x in the given eqn or, maybe easier, do 32 - ∫-11 16x^2dx. (The area you're looking for is the 2 x 16 rectangle - the area under the parabola.)
Take 1/2 that area and set up an integral = to that (let's call that 1/2 area B). This one probs is easier to do w respect to y; that way, the number you're looking for (call it k) becomes the upper bound of integration:
x = ±1/4√y so, to treat the ± , let's just use the + function and double it:
2· ∫0k1/4√ydy = B.
Often on the AP, they will just ask you to set this up rather than solve for k. But solving isn't hard either. You'll get an equation which reduces to k3/2 = # then just take both sides to the 2/3 power.
Happy mathing!