# Find the average value of this function

Find the average value of $e^{-z}$ over the ball $x^2+y^2+z^2 \leq 1$.

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What have you tried so far? What exactly is the problem you're having? –  smackcrane Nov 8 '11 at 18:03

The average will just be the integral over the sphere, divided by the volume of the sphere.

Integrating over the sphere: Suppose we start by integrating over $z$. For a fixed $z_0$, the function is invariant, so we need only know the area of the cross-section of the sphere perpendicular to the $z$-axis at that height $z_0$. By Pythagorus, the radius of that circle will be $\sqrt{1-z^2}$ so that its area is $\pi(1-z^2)$. Hence our integral becomes $$\pi \int_{-1}^1 e^{-z} (1-z^2)dz.$$

Computing this we see that this is $$\frac{4\pi}{e}.$$ As the volume of the sphere is $\frac{4}{3}\pi$, we conclude that the average will be $$\frac{3}{e}.$$

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Let us reason by analogy. Suppose that the problem is 'Find the average value of $x$ over the line segment defined by $S \equiv (x:0 \le x \le 1)$'. We know that the answer is 0.5 but how would we compute it?

$$\bar{x} = \frac{\int_{x \in S} x dx}{\int_{x \in S} dx}$$

Doing the integrals on the right, we get:

$$\bar{x} = \frac{(1^2-0^2)/2}{(1-0)} = 0.5$$

A quick observation that may or may not be useful: by Stokes' theorem you can transform the integral of $e^{-z}$ on the ball to the integral of $ze^z$ on the sphere. –  user7530 Nov 8 '11 at 18:36