# Thread: calc 3 question: 0=<x<=1 -x^2=<y<=x^2 for the function f(x,y) dydx

1. ## calc 3 question: 0=<x<=1 -x^2=<y<=x^2 for the function f(x,y) dydx

i couldnt get it to let me upload a picture so ill do my best to write it out

I have to rewrite a double integral

original bounds are 0=<x<=1 -x^2=<y<=x^2 for the function f(x,y) dydx

what i got when i rewrote it is -1=<y<=1 sqrt(-y)=<x<=sqrt(y) dxdy

i want to say this is right because i can't think of any other way to do this but i knew that a negative under a square root is an imaginary number so it shouldnt work. Please let me know if im right or wrong!

the follow up question to this ask what the original double integral represents

2. Originally Posted by BlitzCannon
i couldnt get it to let me upload a picture so ill do my best to write it out

I have to rewrite a double integral

original bounds are 0=<x<=1 -x^2=<y<=x^2 for the function f(x,y) dydx

what i got when i rewrote it is -1=<y<=1 sqrt(-y)=<x<=sqrt(y) dxdy

i want to say this is right because i can't think of any other way to do this but i knew that a negative under a square root is an imaginary number so it shouldnt work. Please let me know if im right or wrong!

the follow up question to this ask what the original double integral represents
Your response is not quite correct.

Did you draw an approximate sketch of the region over which you are integrating?

If not, do that first.