Setting polar bounds for a region between circles and a parabola

Key concept: This is a polar-coordinate region problem: the main task is translating the Cartesian inequalities into bounds for $r$ and $\theta$.

Step by step

Convert the region into polar form

Use the substitutions

$x=r\cos\theta, \quad y=r\sin\theta, \quad x^2+y^2=r^2, \quad dA=r\,dr\,d\theta.$

The integrand becomes

$\sqrt{x^2+y^2}=r.$

So the integral turns into

$\iint_A \sqrt{x^2+y^2}\,dA = \iint_A r\cdot r\,dr\,d\theta = \iint_A r^2\,dr\,d\theta.$

Translate the inequalities carefully

The region is

$A=\{(x,y)\mid 2x\le x^2+y^2\le 4,\ x\ge 0,\ y\ge 0\}.$

In polar form, this becomes

$2r\cos\theta \le r^2 \le 4, \quad 0\le \theta\le \frac{\pi}{2}.$

Since we are in the first quadrant, $r\ge 0$. The right inequality gives

$r\le 2.$

For the left inequality, if $r>0$ we can divide by $r$ and obtain

$2\cos\theta \le r.$

So the radial bounds are

$2\cos\theta \le r \le 2, \qquad 0\le \theta\le \frac{\pi}{2}.$

Write the final integral

Therefore,

$\iint_A \sqrt{x^2+y^2}\,dA = \int_0^{\pi/2}\int_{2\cos\theta}^{2} r^2\,dr\,d\theta.$

That is the correct polar setup. If you want to evaluate it, integrate in $r$ first:

$\int_{2\cos\theta}^{2} r^2\,dr = \frac{1}{3}\left(8-8\cos^3\theta\right).$

Then finish with the $\theta$ integral.

Common boundary issue

The most common mistake is reversing the inequality after substitution or forgetting that $x\ge 0$ and $y\ge 0$ restrict $\theta$ to the first quadrant. Another frequent error is treating $2x\le r^2$ as $2r\cos\theta\le r^2$ and then dividing by $r$ without checking that $r=0$ is not needed for the region. Since the region lies away from the origin except at the boundary, the bound $r=2\cos\theta$ is valid for the interior description.

Pitfall alert

Do not write the bounds as $2\cos\theta\le r\le 2$ without also stating the angle range. The first-quadrant condition is essential; otherwise the same radial inequality would describe points in other quadrants as well. Another subtle mistake is forgetting the Jacobian factor $r$ in $dA$. Since the integrand already contains $\sqrt{x^2+y^2}=r$, the final integrand is $r^2$, not just $r$.

Try different conditions

If the region were instead $A=\{(x,y)\mid x^2+y^2\le 4,\ x\ge 0,\ y\ge 0\}$, then the polar bounds would simplify to $0\le r\le 2$ and $0\le \theta\le \pi/2$. If the inner inequality were $r\ge 2\cos\theta$ with the same outer circle, the setup would be unchanged in form but the lower radial boundary would be the curve $r=2\cos\theta$, which traces a circle shifted to the right. The method stays the same; only the boundary functions change.

Further reading

polar coordinates, Jacobian factor, region bounds