I want to compute the following triple integral:

$$\iiint_{x^2+y^2+z^2\le 2x} x + z \,dV $$

What I've done so far is that I decided I will use spherical coordinates to solve the integral, as the region over which it should be integrated is a sphere centered at the point: $(1,0,0)$. I know this by bringing $2x$ to the other side of the equation and completing the square. Next I set up the triple integral. In spherical we first integrate with respect to $r$, to find the bounds of the integral for r, I know that $r$ is always positive so the lower bound is $0$, and the upper bound can by found by substituting $r^2$ into $x^2+y^2+z^2$ we then get $r^2\le 2x $, $x = rsin\phi cos\theta $ therefore $r^2\le 2rsin\phi cos\theta $ and $r\le 2sin\phi cos\theta$ so the bounds for $r$ are $0\le r\le 2sin\phi cos\theta$ I'm wondering if this is correct or that $r$ is simply between 0 and 1 because the radius of the sphere is $1$. Then I found the bounds for $\phi$ as we integrate with respect to $\phi$ next. I set $x=1$ then we get $y^2+z^2 \le 1 $ which is a circle of radius $1$ on the $yz$-plane since $\phi$ is never greater than $\pi$ I set the boundary for $\phi$ as $0\le \phi \le \pi $. Next I find the bounds for $\theta$, I set $z=0$ and get $(x-1)^2 + y^2 \le 1$ this is a circle of radius one in the $xy$-plane centered at $(1,0) $ so I set the boundary $\frac{-\pi}2 \le \theta \le \frac{\pi}2 $. Then I changed the function $f(x,y,z)$ to $f(r,\phi,\theta)$ $$x + z = rsin\phi cos\theta + rcos\phi$$ Finally I get the integral( with the Jacobian ) :

$$\int_{\frac{-\pi}2}^{\frac{\pi}2} \int_0^{\pi} \int_0^{2sin\phi cos\theta} \, rsin\phi cos\theta + rcos\phi\, (r^2sin\phi) \,drd\phi d\theta $$

When I started integrating I started getting very nasty integrals with sin to the power of 6 and 5 and even if r=1 I would still have quite complicated integrals. I feel like I'm doing something wrong. Could someone please help me develop my work and point out any possible mistakes?.

  • $\begingroup$ notice that you can rearrange $x^2+y^2+z^2\le 2x$ into $(x-1)^2+y^2+z^2\le 1$ so this has a domain which is a sphere of radius $1$ and centre $(1,0,0)$ $\endgroup$
    – Henry Lee
    Apr 17 '19 at 18:02
  • $\begingroup$ I do notice that I have stated the same thing in my explanation, I'm wondering am I solving the integral correctly even though it is a very tedious way to do it? $\endgroup$ Apr 17 '19 at 18:08
  • $\begingroup$ I think you have complicated the limits somehow, I think they should be: $0\le r\le 1$ and $0\le\theta\le 2\pi$ and $0\le\phi\le\pi$ $\endgroup$
    – Henry Lee
    Apr 17 '19 at 18:11
  • $\begingroup$ So r depends on the radius of the sphere? and not on where the sphere is centered? I explained how I found the limits, I used the methods in my textbook. Could you please comment where I went wrong in my reasoning. I believe theta is not between $0 ≤ θ ≤ 2π$ for example because the circle on the xy plane is only on the right side of the y axis. $\endgroup$ Apr 17 '19 at 18:14
  • $\begingroup$ as shown in @J.G.'s answer we can redefine x as its actual position is arbitrary, and make the centre $(0,0,0)$ which makes it much easier to define the bounds $\endgroup$
    – Henry Lee
    Apr 17 '19 at 18:29

With $X:=x-1$ your problem is $\iiint_{X^2+y^2+z^2\le 1}(X+z+1)dV$, an integral over a unit sphere. The simplest solution uses symmetry to note neither $X$ nor $z$ contributes to the final result, so we just get the volume $\frac{4\pi}{3}$.

  • $\begingroup$ I don't understand how X or z do not contribute to the final result. I'm completely new to this topic so please elaborate. $\endgroup$ Apr 17 '19 at 18:11
  • 1
    $\begingroup$ For $X$ and $z$, the portions of the volume of integration where each variable is positive is exactly cancelled by the portions of the volume of integration where each variable is negative. That's what J.G. means by "uses symmetry." $\endgroup$ Apr 17 '19 at 18:18
  • $\begingroup$ Ok It's starting to make more sense, I get the concept that when they cancel themselves out we are left with 1 and when you do the integral of that you just get the volume of the unit sphere however I don't see that they cancel themselves out just by looking at the function. How do you think of this? $\endgroup$ Apr 17 '19 at 18:42
  • $\begingroup$ @marky Experience. If you're integrating over a symmetrical shape, there's a good chance that symmetry is informative. $\endgroup$
    – J.G.
    Apr 17 '19 at 18:51

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