Verify the relation: 1.96 < double integral over
Verify the relation: $$ 1.96 < \iint_{|x|+|y|\leqslant10} \frac{dx\,dy}{100+\cos^2x+\cos^2y} < 2. $$
Explanation of Method
The region is a diamond of area $200$. Because $0\leqslant\cos^2x,\cos^2y\leqslant1$, the denominator stays in the very narrow range $[100,102]$, so the integrand itself stays in a narrow range. Multiplying by the area gives tight two-sided bounds.
- Compute the area of the diamond $|x|+|y|\leqslant10$.
- Bound the denominator using $0\leqslant\cos^2\leqslant1$.
- Apply the estimation theorem and compare to $1.96$ and $2$.
Step 1 · Identify the region
$D=\{(x,y):|x|+|y|\leqslant10\}$ is a diamond (a square rotated $45^\circ$) centered at the origin with vertices $(\pm10,0)$ and $(0,\pm10)$.
Step 2 · Calculate the area of $D$
By symmetry, the area of one quadrant's triangle (legs of length $10$) is $\frac12\cdot10\cdot10=50$, so
$$ A = 4\cdot50 = 200. $$
Step 3 · Analyze the function
$$ f(x,y)=\frac{1}{100+\cos^2x+\cos^2y}. $$
Step 4 · Bound the denominator
Since $0\leqslant\cos^2x\leqslant1$ and $0\leqslant\cos^2y\leqslant1$:
$$ 100 \leqslant 100+\cos^2x+\cos^2y \leqslant 102. $$
Step 5 · Bound $f$
$$ \frac{1}{102} \leqslant f(x,y) \leqslant \frac{1}{100}. $$
Step 6 · Apply the estimation theorem
$$ \frac{1}{102}\cdot200 \leqslant \iint_D \frac{dx\,dy}{100+\cos^2x+\cos^2y} \leqslant \frac{1}{100}\cdot200 \quad\Longrightarrow\quad \frac{100}{51}\leqslant I\leqslant2. $$
Numerically, $\frac{100}{51}\approx1.9608$.
Step 7 · Sharpen to strict inequalities
The extreme values $\frac1{100}$ and $\frac1{102}$ are attained only where $\cos^2x=0,\cos^2y=0$ or $\cos^2x=1,\cos^2y=1$ respectively — sets of measure zero within $D$. Hence the bounds are strict:
$$ \frac{100}{51} < I < 2. $$
Step 8 · Conclude the verification
Since $\frac{100}{51}\approx1.9608 > 1.96$:
$$ 1.96 < \frac{100}{51} < I < 2. \quad\checkmark $$
Step 9 · Symmetry observation
$f(x,y)=\frac{1}{100+\cos^2x+\cos^2y}$ is even in both $x$ and $y$, and $D$ is symmetric about both axes and both diagonals — this symmetry is not needed for the bound but is a useful check on more detailed calculations.
Summary
Key steps: the diamond $D$ has area $200$; the denominator $100+\cos^2x+\cos^2y$ stays in $[100,102]$, so $f\in\left[\frac1{102},\frac1{100}\right]$; multiplying by the area gives $\frac{200}{102}<I<\frac{200}{100}$, i.e. $\frac{100}{51}\approx1.9608<I<2$, which confirms $1.96<I<2$.
Additional Notes
- The bounds are unusually tight here because the cosine terms can add at most $2$ to a denominator already near $100$.
- This is a good illustration of the estimation theorem giving a useful result without ever evaluating the integral exactly.
Result
The verified inequality is: