Sketching the region of $\int_0^2 \int^{y^2}_0dxdy$ My textbook shows the answer as being: 

Note that I've interpreted the $y^2x$ to be extraneous information for the problem of simply sketching the region the double integral is specifying; maybe I'm wrong about this? 
My question is: isn't the region negated in the answer?
It seems to me that the region should be the region above the line $y=0$ and below the curve $x=y^2$
However, the question shows the region as being the region above the curve $x=y^2$ and below the line $y=2$ 
Why is the latter and not the former correct? How can we know which is being specified by the integral?
 A: Notice that the range for the $x$-integral is from 0 to $y^2$, which means the region is right of the line $x=0$ and left of the curve $x=y^2$. 
Together with the $y$-range from 0 to 2, the region is indeed the shaded area.
A: $x <y^{2}$ iff $(x,y)$ lies above the parabola $x=y^{2}$, so the graph is correct. 
A: Notice that the integral is written in terms of $dx~dy$, not $dy~dx$.  Working from the inside out, that means the first variable of integration is $x$, not $y$.  As the other two answers note, that's consistent with how the limits of integration are treated.
A: Consider what it means to be "below" the curve $x=y^2$. The region goes from $x=0$ to $x=y^2$. So, since $x=0$ is the $y$ axis and $x=0$ bounds the region, the region must be the one that touches the y axis. This works because we're considering the region to be "below" the curve from the perspective of the $x$ axis, so "down" winds up being to the left.
A: For $\int_{x=a}^b\int_{y=f(x)}^{g(x)}F(x,y)dydx$ , the region $R$ must de drawn such that:


*

*The foot and top respectively of the infinitesimal thin strip inside $R$ must lie on curves $y=f(x)$ and $y=g(x)$ (inner limits).

*The movement of the strip must be from the end $x=a$ to $x=b$ (outer limits).
Use same procedure for $\int_{y=a}^b\int_{x=f(y)}^{g(y)}F(x,y)dxdy$ 
