Finding $\lim_{c \rightarrow 0} \iint_{R} \frac{1}{(x^2+y^2)^{3/4}}\,dA$ where $R$ is unit disk with square removed

Question

For each $0 \leq c \leq \frac{1}{\sqrt{2}}$ define the following region

$$R=\{(x,y): x^2+y^2 \leq 1\} \setminus ([-c,c] \times [-c,c])$$

Compute

$$\lim_{c \rightarrow 0} \iint_{R} \frac{1}{(x^2+y^2)^{3/4}}\,dA$$

Now, if we remove a circle instead of a square then using polar coordinates, we can easily find the answer which is $4\pi$. If we consider R as a type one or two region, then I do not know how to calculate the integral because it seems that we cannot make use of the polar coordinate anymore. Any hint would be appreciated.

Answer 1

Let $I_c$ be the integral of the function on the unit disk with $[-c,c]\times [-c,c]$ square removed. Let $J_c$ be the integral on the unit disk with the radius $c$ smaller disk centered on the origin removed.

As you’ve noticed using polar coordinates, $J_c=\int_c^1 2\pi r/r^{3/2}dr=[4\pi\sqrt{r}]^1_c=4\pi (1-\sqrt{c})$.

Since the function is positive, increasing the covered area will only increase the integral. Since the square in $I_c$ covers a circle of radius $c$ and lies within a circle of radius $2c$, we have that the region of $J_{2c}$ is within that of $I_c$ which is within that of $J_{c}$.

Thus,

$$ 4\pi(1-\sqrt{2c})=J_{2c}\leq I_c\leq J_c = 4\pi(1-\sqrt{c})$$

Applying the Squeeze Theorem gives that $\lim_{c \rightarrow 0^+} I_c=4\pi$ as desired.

Answer 2

If you wish to do this directly (which to be fair is proper, since taking the limit along different shapes is akin to taking limits along different paths for multivariable limits, and you don't necessarily know ahead of time that the limit exists and therefore the limits along different paths equal each other), then consider the symmetries of the integral to rewrite it as

$$I_c = 8 \iint\limits_{R\:\cap\:x>y\:\cap\:y>0}\frac{dxdy}{(x^2+y^2)^{\frac{3}{4}}}$$

Transforming the equation of the line to polar coordinates, we obtain

$$c = x =r\cos\theta \implies r = c\sec\theta$$

giving us the bounds

$$I_c = 8 \int_0^{\frac{\pi}{4}}\int_{c\sec\theta}^1 r^{-\frac{1}{2}}\:dr\:d\theta = 4\pi-16\sqrt{c}\int_0^{\frac{\pi}{4}}\sqrt{\sec\theta}\:d\theta$$

From here, taking the limit requires not much further justification to say $$\lim_{c\to0}I_c = \boxed{4\pi}$$