Question

If f(x, y) = esin(x + y) and D = [−π, π] × [−π, π], show the following. 1 e ≤ 1 4π2 D f(x, y) dA ≤ e The area of the region D is given by . Then since −1 ≤ sin(x + y) ≤ 1 for all (x, y) is in D, we know Correct: Your answer is correct. ≤ esin(x + y) ≤ Correct: Your answer is correct. for all (x, y) is in D. Now integrating over D we have 4π2 Incorrect: Your answer is incorrect. ≤ D f(x, y) dA ≤ , and finally dividing through by the area we obtain 1 e ≤ 1 4π2 D f(x, y) dA ≤ e.

Answer 1

I'm guessing the purpose of this exercise is to find the average value of $f(x,y)=e^{\sin(x+y)}$ over the region $D$,

$D=\left\{(x,y)\mid (x,y)\in[-\pi,\pi]^2\right\}$

which is

$\displaystyle\frac{\displaystyle\iint_Df(x,y)\,\mathrm dx\,\mathrm dy}{\displaystyle\iint_D\mathrm dx\,\mathrm dy}$

The denominator is the measure (area) of $D$, which is easy to compute:

$\displaystyle\iint_D\mathrm dx\,\mathrm dy=\int_{-\pi}^\pi\int_{-\pi}^\pi\mathrm dx\,\mathrm dy=(2\pi)^2=4\pi^2$

Not to be confused with the integral in the numerator:

$\displaystyle\iint_Df(x,y)\,\mathrm dx\,\mathrm dy=\int_{-\pi}^\pi\int_{-\pi}^\pi e^{\sin(x+y)}\,\mathrm dx\,\mathrm dy$

but this integral is difficult to compute, hence the inequalities. We have

$-1\le\sin(x+y)\le1$

$\implies\dfrac1e\le e^{\sin(x+y)}\le e$

$\implies\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}e\le\iint_De^{\sin(x+y)}\,\mathrm dx\,\mathrm dy\le\iint_De\,\mathrm dx\,\mathrm dy$

$\implies\displaystyle\frac{4\pi^2}e\le\iint_De^{\sin(x+y)}\,\mathrm dx\,\mathrm dy\le4\pi^2e$