I'm having trouble reversing the order of integration of ∫∫dxdy from a=0 to b=2(3)^(1/2) for x and c=y^(2/6) to d=(16-y^2)^(1/2) for y.

Question

I'm having trouble reversing the order of integration of ∫∫dxdy from a=0 to  b=2(3)^(1/2) for x and c=y^(2/6) to d=(16-y^2)^(1/2) for y.

I graphed the region of integration and that still doesn't really help me.

i got approximately 7.9 for the ∫∫dxdy which im pretty sure is right.

please help

Steve · Answer

I'm having trouble interpreting your variables and limits.

c,d are the limits on dy? That's odd, since I'd expect functions of y to be limits on dx. Anyway, interpreting it as

∫[0,√12]∫[y^(1/3),sqrt(16-y^2)] dx dy I get 15.8, twice your answer

I tried to get a handle on the region using the excellent graphing tools at rechneronline dot de slash function-graphs, and it appears the region is to the right of the curve y=x^3 and inside the circle y=sqrt(16-x^2) for x in [0,√12]. Strange limits, since they don't appear to correspond to any useful portion of the picture.

??