Find the average value of the function: z(x,y) = sin²x
Find the average value of the function: $$ z(x,y) = \sin^2x\sin^2y $$ over the square $0\leqslant x\leqslant\pi$, $0\leqslant y\leqslant\pi$.
Explanation of Method
Since the integrand factors as a product of a function of $x$ alone and a function of $y$ alone, the double integral over a rectangle splits into a product of two single integrals, each evaluated with the identity $\sin^2\theta=\frac{1-\cos2\theta}{2}$.
- Split the double integral into a product of two identical single integrals.
- Evaluate $\int_0^\pi\sin^2x\,dx$.
- Divide by the area of the square.
Step 1 · Area of the region
$$ \operatorname{Area}(D) = \pi\cdot\pi = \pi^2. $$
Step 2 · Separate the integral
$$ \iint_D \sin^2x\sin^2y\,dx\,dy = \int_0^\pi \sin^2x\,dx \int_0^\pi \sin^2y\,dy = \left(\int_0^\pi\sin^2x\,dx\right)^2. $$
Step 3 · Evaluate the single integral
$$ \int_0^\pi \sin^2x\,dx = \int_0^\pi\frac{1-\cos2x}{2}\,dx = \frac12\left[x-\frac{\sin2x}{2}\right]_0^\pi = \frac{\pi}{2}. $$
Step 4 · Combine the results
$$ \iint_D \sin^2x\sin^2y\,dx\,dy = \left(\frac\pi2\right)^2 = \frac{\pi^2}{4}. $$
Step 5 · Compute the average value
$$ \bar z = \frac{1}{\pi^2}\cdot\frac{\pi^2}{4} = \frac14. $$
Additional Notes
- Since $\sin^2\theta$ averages to $\frac12$ over any full half-period like $[0,\pi]$, the product average is $\frac12\cdot\frac12=\frac14$, matching directly.
Result
The solution is: