Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Evaluate the integral $\int x\,dV$ inside domain $V$, where $V$ is bounded by the planes $x=0$, $y=x$, $z=0$, and the surface $x^2+y^2+z^2=1$.

Answer given: $\dfrac{1}{8} - \dfrac{\sqrt{2}}{16}$

Uh, so I did it in spherical coordinates, which equals

$$\iiint p^2 \sin φ \;dp dφ dθ$$

$∫dp$ runs from $0$ to $1$

$∫dφ$ runs from $0$ to $\frac{\pi}{2}$ (right??)

$∫dθ$ runs from $-\frac{\pi}{2}$ to $\frac{\pi}{4}$ (because of the line $y = x$ in the $xy$ plane)

I do not get the given answer though.

share|improve this question
1  
Hmm, on second thought, the question seems ambiguous. There are two possible interpretations of the region $V$: the one I was thinking of and the one you were thinking of (the other six are all equal in volume, though their $x$-values might be different in sign). Does the question specify which one is intended? –  Zev Chonoles Apr 12 '12 at 3:19
2  
Don't forget the integrand. We have $x = r \cos \theta = p \sin \varphi \cos\theta$ so the integrand should be $(p\sin \varphi \cos \theta)(p^2\sin\varphi)$. –  nullUser Apr 12 '12 at 3:32
    
The question doesn't specify. In either case, the integral of dφ would run from zero to pi/2, right? I worked it your way and almost got the answer, but then the denominators changed to 32.. –  Anon Apr 12 '12 at 3:45
    
Yes, $\varphi$ should be going from $0$ to $\pi/2$. –  Zev Chonoles Apr 12 '12 at 3:47
    
@ZevChonoles I think you mean phi goes from 0 to pi/2. Theta should be between 0 and pi/4 correct? –  gsingh2011 Apr 12 '12 at 3:51

2 Answers 2

up vote 0 down vote accepted

As Zev pointed out, the question is ill-posed since there's more than one region bounded by these surfaces. From the answer, it seems that the region $0\le p\le1$, $0\le\varphi\le\pi/2$, $\pi/4\le\theta\le\pi/2$ was intended, but even then the given answer is missing a factor of $\pi/2$. So I think the main conclusion from this exercise should be not to put too much stock in its source :-)

As has been pointed out in comments, your integrand is just the Jacobian and you forgot to include the original integrand $x=p\sin\varphi\cos\theta$. The required integral is

$$\int_0^1\int_0^{\pi/2}\int_{\pi/4}^{\pi/2}p^3\sin^2\varphi\cos\theta\,\mathrm d\theta\,\mathrm d\varphi\,\mathrm dp=\frac14\cdot\frac\pi4\left(1-\frac1{\sqrt2}\right)=\frac\pi2\left(\frac18-\frac{\sqrt2}{16}\right)\;.$$

share|improve this answer

I may have made a mistake, but it seems easier to use rectilinear coordinates here. I get:

$$I = \int_{z=0}^1 \int_{y=0}^{\sqrt{1-z^2}} \int_{x=0}^{\min(y,\sqrt{1-y^2-z^2})} x \; dx dy dz$$

The min can be removed by splitting the integral into:

$$I = \int_{z=0}^1 \int_{y=0}^{\frac{1}{\sqrt{2}}\sqrt{1-z^2}} \int_{x=0}^{y} x \; dx dy dz + \int_{z=0}^1 \int_{y=\frac{1}{\sqrt{2}}\sqrt{1-z^2}}^{\sqrt{1-z^2}} \int_{x=0}^{\sqrt{1-y^2-z^2}} x \; dx dy dz$$

However, when I integrate this (a little tedious, but not particularly difficult), I get $I = \frac{(2-\sqrt{2}) \pi}{32}$.

share|improve this answer
    
You can use double dollar signs to get displayed equations, which look nicer and are easier to read. –  joriki Apr 12 '12 at 8:18
    
@joriki: I added the double dollars; I think it just indents more? –  copper.hat Apr 12 '12 at 8:23
    
Which browser are you using? It might make sense to report this as a bug at meta.math.stackexchange.com. The displayed equations should look quite noticably different, less condensed, with much larger integral signs. (Also they're centred rather than indented.) –  joriki Apr 12 '12 at 8:25
    
The splitting point is $\sqrt{(1-z^2)/2}$, not $\sqrt{1-z^2}/2$. With that fixed, this gives the right result (Wolfram|Alpha computations for the first and second integral). –  joriki Apr 12 '12 at 10:35
    
@joriki: good catch! –  copper.hat Apr 12 '12 at 17:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.