Find the average value of the function: z(x,y) =
Find the average value of the function: $$ z(x,y) = \sqrt{a^2-x^2-y^2} $$ over the region $x^2+y^2\leqslant a^2$.
Explanation of Method
The average value of $z$ over a region $D$ is $\bar z=\dfrac{1}{\operatorname{Area}(D)}\iint_D z\,dA$. The numerator is the familiar hemisphere-volume integral, computed via polar coordinates, and the denominator is just the area of the disk.
- Compute the area of the disk.
- Compute $\iint_D\sqrt{a^2-x^2-y^2}\,dA$ using polar coordinates.
- Divide to get the average value.
Step 1 · Area of the region
$$ \operatorname{Area}(D) = \pi a^2. $$
Step 2 · Set up the integral in polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$, $0\leqslant r\leqslant a$, $0\leqslant\theta\leqslant2\pi$:
$$ \iint_D \sqrt{a^2-x^2-y^2}\,dx\,dy = \int_0^{2\pi}\int_0^a \sqrt{a^2-r^2}\,r\,dr\,d\theta. $$
Step 3 · Integrate with respect to $r$
With $u=a^2-r^2$:
$$ \int_0^a r\sqrt{a^2-r^2}\,dr = \frac12\int_0^{a^2}\sqrt u\,du = \frac{a^3}{3}. $$
Step 4 · Integrate with respect to $\theta$
$$ \iint_D \sqrt{a^2-x^2-y^2}\,dx\,dy = 2\pi\cdot\frac{a^3}{3} = \frac{2\pi a^3}{3}. $$
Step 5 · Compute the average value
$$ \bar z = \frac{1}{\pi a^2}\cdot\frac{2\pi a^3}{3} = \frac{2a}{3}. $$
Additional Notes
- Geometrically, $z=\sqrt{a^2-x^2-y^2}$ is the upper hemisphere of radius $a$; its average height over the base disk works out to $\frac23$ of the radius, consistent with the hemisphere's volume ($\frac{2\pi a^3}{3}$) divided by its base area ($\pi a^2$).
Result
The solution is: