How to Integrate This Double Integral in Polar Coordinates

Interpreting the integral as

$$\int_1^2\int_0^{\sqrt{2x-x^2}}\frac{1}{(x^2+y^2)^2}\,dy\,dx,$$

the key is to describe its region in polar coordinates before changing the integrand. The upper boundary is part of a circle, while the lower bound $x\ge1$ cuts away the portion near the origin. Those two boundaries determine both the radial limits and the angular range.

Identify the region

The condition on $y$ implies

$$y^2\le 2x-x^2,$$

so

$$x^2+y^2\le2x \quad\Longleftrightarrow\quad (x-1)^2+y^2\le1.$$

This is the disk centered at $(1,0)$ with radius $1$. Since $y$ runs from zero to the upper semicircle, only the upper half is included. The outer integral further restricts the region to $x\ge1$. The bound $x\le2$ is already satisfied throughout this disk.

In polar coordinates, $x=r\cos\theta$ and $y=r\sin\theta$. The circle boundary becomes

$$r^2\le2r\cos\theta,$$

or $r\le2\cos\theta$ for points in this region. The vertical boundary $x=1$ becomes $r\cos\theta=1$, so points to its right satisfy $r\ge\sec\theta$.

Determine the angular limits

The ray at angle $\theta$ intersects the region only if its lower radial bound does not exceed its upper radial bound:

$$\sec\theta\le2\cos\theta.$$

For angles in the upper half-plane and with $x\ge1$, $\cos\theta>0$. Multiplying by $\cos\theta$ gives $1\le2\cos^2\theta$, hence $\cos\theta\ge1/\sqrt2$. The region is above the $x$-axis, so $\theta\ge0$. Therefore,

$$0\le\theta\le\frac{\pi}{4}, \qquad \sec\theta\le r\le2\cos\theta.$$

At $\theta=\pi/4$, the two radial boundaries meet. This gives the corner point where the line $x=1$ meets the circle.

Transform and evaluate the integral

Since $x^2+y^2=r^2$, the integrand is $r^{-4}$. The polar area element is $dA=r\,dr\,d\theta$, so the transformed integrand is $r^{-3}$—not $r^{-4}$.

Thus the integral is

$$\begin{aligned} I&=\int_0^{\pi/4}\int_{\sec\theta}^{2\cos\theta}r^{-3}\,dr\,d\theta\\ &=\int_0^{\pi/4}\left[-\frac{1}{2r^2}\right]_{\sec\theta}^{2\cos\theta}\,d\theta\\ &=\frac12\int_0^{\pi/4}\left(\cos^2\theta-\frac{1}{4\cos^2\theta}\right)\,d\theta. \end{aligned}$$

Evaluate the two terms using $\int\cos^2\theta\,d\theta=\theta/2+\sin(2\theta)/4$ and $\int\sec^2\theta\,d\theta=\tan\theta$:

$$\begin{aligned} I&=\frac12\left(\frac{\pi}{8}+\frac14\right)-\frac18(1)\\ &=\boxed{\frac{\pi}{16}}. \end{aligned}$$

The constants cancel because the contribution $1/8$ from the cosine-squared integral is exactly offset by the secant-squared term.

Common mistakes

• Using the disk’s full radial range. The circle gives $r\le2\cos\theta$, but the original condition $x\ge1$ also requires $r\ge\sec\theta$.
• Choosing the wrong angle range. The upper-half condition gives $\theta\ge0$; requiring the two radial bounds to overlap gives $\theta\le\pi/4$.
• Forgetting the Jacobian. The area element contributes a factor of $r$, changing $r^{-4}$ into $r^{-3}$.
• Treating $x\le2$ as another active boundary. The circle already has rightmost point $(2,0)$, so this limit does not further restrict the region.

If the missing upper limit in the written integral was intended to be something other than $\sqrt{2x-x^2}$, the region and answer may differ. For the displayed interpretation, the value is $\pi/16$.