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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

2 Answers 2

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:

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

Extending the above idea to your problem should be helpful.

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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
    
@user19021 It looks what you wrote in the comment is on the right track but hard to tell for sure. Please texify your equations. –  tards Nov 8 '11 at 18:36
    
@tards: Can you explain how this analogy is helpful? I honestly can't see it since we are dealing with an exponential. –  Eric Naslund Nov 8 '11 at 18:37
    
@eric It provides a generic approach to extending the concept of finding 'averages' in higher dimensions. Your answer essentially extends the above idea except for the small difference that you omitted the denominator as it is a sphere of volume 1. –  tards Nov 8 '11 at 18:39
    
@tards: Sorry, I misread your answer. I though you were suggesting a clever trick to solve (that I didn't think would work) to solve the 3 dimensional integral. –  Eric Naslund Nov 8 '11 at 18:48

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