I think it's a Type 2, but I'm not sure. I have no idea how to approach the limits of integration. Would you be able to provide an explanation on how to do so? It would be awesomeRestricting Jameson's graph of the region (including the graph of $y=-1$) appropriately gives us the region we're integrating over:
To evaluate the double integral over this region, you need to decide whether or not you should treat this as a Type I or Type II region (I'll just say that one way is much easier than the other).
Do you think you can determine the appropriate limits of integration and the double integral(s) needed to evaluate your original integral over this region?
Yes, that's correct:∫-1(lower limit) to 0 (upper limit) ∫(-1 (lower limit) to y−y^3 (upper limit)) y dxdy
∫0 (lower limit) to 1 (upper limit) ∫(√(y)-1)) (lower limit) to (y−y^3) (upper limit) y dxdy
I did mine as two different double integrals. Is that okay?
Did I get the two sets of double integrals correct this time?
My gravest apologies, the outer upper limit should be $1$...I made a silly typo there.I get a non-real result when doing from 0 to -1 for the second double integral. for the first set of double integrals i get 11/30
Glad to hear it!awesome, exactly what I got when I did it.
If you could help me with this, it would be awesome