Question

Find the average value of f(x+y)=sin(x+y) over

Answer 1

For each given region $D$, the average value of $f$ over $D$ is the integral of $f$ over $D$ divided by the area of $D$. In both cases, $D$ is a rectangle so the area is trivial.

a. The area of $D$ is $\dfrac\pi3\cdot\dfrac{5\pi}3=\dfrac{5\pi^2}9$. The integral of $f$ over $D$ is

$\displaystyle\int_0^{5\pi/3}\int_0^{\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$\displaystyle=\int_0^{5\pi/3}\left(\cos y-\cos\left(y+\dfrac\pi3\right)\right)\,\mathrm dy$

$=\sin\dfrac{5\pi}3-\sin\left(\dfrac{5\pi}3+\dfrac\pi3\right) - \sin0+\sin\dfrac\pi3=0$

So the average value of $f$ over this region is 0.

b. The area of $D$ is $\dfrac{2\pi}3\cdot\dfrac{7\pi}6=\dfrac{7\pi^2}9$. The integral of $f$ is

$\displaystyle\int_0^{7\pi/6}\int_0^{2\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^{7\pi/6}\left(\cos y-\cos\left(y+\frac{2\pi}3\right)\right)\,\mathrm dy$

$=\sin\dfrac{7\pi}6-\sin\left(\dfrac{7\pi}6+\dfrac{2\pi}3\right) - \sin0+\sin\dfrac{2\pi}3=\dfrac{\sqrt3}2$

The average vale of $f$ over this region is then $\dfrac{\frac{\sqrt3}2}{\frac{7\pi^2}9}=\dfrac{9\sqrt3}{14\pi^2}$.