Sketching the region of $\int_0^2 \int^{y^2}_0dxdy$

Question

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?

Answer 1

$x <y^{2}$ iff $(x,y)$ lies above the parabola $x=y^{2}$, so the graph is correct.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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 $\int_{y=a}^b\int_{x=f(y)}^{g(y)}F(x,y)dxdy$