@TedShifrin Hi, I had to take a break but for the examples you gave: $\int_0^2\int_y^2 dxdy$ , $\int_0^4 \int_{\sqrt{y}}^{2} dxdy$, and the third one I got confused because $x^2 = x$ at $x=1$ so the inner integral doesn't make sense to me.

for the interval x=0 to x=1 I think it should be $\int_{x^2}^{x}$

so $\int_0^1\int_{x^2}^{x} dydx + \int_{1}^{2}\int_{x}^{x^2}dydx$

which I would then use the inverse functions to do $\int_{0}^{1}\int_{\sqrt{y}-y}^{1}dxdy + \int_{1}^{4}\int_{y-\sqrt{y}}^{4}dxdy$

oops disregard last one

You’re right. I messed up. I meant $\int_1^2$. But what you have is wrong.

Draw a picture.

I know the area shaded horizontally should be 1-y-sqrt{y} over the interval of y*

OH

No. We need to know what $x$ starts and ends at for each $y$ fixed.

Did you draw the picture?

yeah it's the region between y=x and y=x^2 from [0,1] and then the region b/t y=x^2 and y=x from 1,2

I switched to just $[1,2]$.

So $y$ goes from what to what?

it goes from 1,4

OK. Now fix $y$. What does $x$ do for that value of $y$?

is it right to say the area is bounded by x=2, y=x^2, y=x, from 1<x<2 ?

that's what I have drawn.

Yes.

does it make sense to have two variables in a lower limit?

No.