Unspecific boundaries for finding area by double integration

Question

I've been given the boundaries $0≤x≤x^2+y^2≤1$. I have no set equation so it would simply be 1 integrated. Normally I have no problems when the boundaries are clearly divided, yet here I can't seem to figure out what they are for $x$ and $y$, respectively.

Answer 1

Take $0\le x \le 1$, Then you have $0 \le \sqrt{1 - x^2} $ as a conditional on $y$.

$$\int \!\! \int_{0 \le x \le x^2 + y^2 \le 1} dx dy = \int_0^1 dx \int_0^\sqrt{1 - x^2} dy$$

Answer 2

Inequality $0 \leqslant x \leqslant x^2+y^2 \leqslant 1$ implies that $$\begin{cases} x \geqslant 0,\\ \left(x-\dfrac{1}{2} \right)^2 +y^2 \geqslant \dfrac{1}{4} , \\ x^2+y^2 \leqslant 1 . \end{cases}$$

Answer 3

I suggest checking the problem statement more carefully. Does it really say $0 \leq x \leq x^2 + y^2 \leq 1$?

Notice that

• If $x$ and $y$ are real numbers, then $x^2 \geq 0$ and $y^2 \geq 0$.
• If $x^2 + y^2 \leq 1$, then $y^2 \geq 0$ implies $x^2 \leq 1$.
• $x^2 \leq 1$ implies $0 \leq x \leq 1$.
• $0 \leq x \leq 1$ implies $x^2 \leq x$.
• If $x \leq x^2$, then $x^2 \leq x$ implies $x^2 = x$.
• $x^2 = x$ implies that either $x = 0$ or $x = 1$.

In short, the only pairs of real numbers that satisfy the conditions $0 \leq x \leq x^2 + y^2 \leq 1$ are the pairs of the form $(0, y)$ for $-1 \leq y \leq 1$ and the pair $(1, 0)$. In other words, the "region" defined by $0 \leq x \leq x^2 + y^2 \leq 1$ consists of a line segment plus a single point in the $x,y$ plane. That doesn't sound to me like much of a region of integration.

If that is really what the problem statement says, I suspect it's a mistake in the problem statement, which really should have given the conditions as just $0 \leq x$ and $x^2 + y^2 \leq 1$.