The original problem:
Find the Moment of Inertia $I$ of a solid sphere of uniform density $\rho$.


I thought of doing it using Triple Integrals since I'm not very fond of the usual Single Integral involving manipulations with differentials.

Moment of Inertia for Continuous Distribution of particles is $\int r^2 dm$ with suitable limits ($r$ represents the radius while $dm$ is an elementary mass). $$dm=\rho dV$$ $$$$ Thus, the moment of Inertia can be represented as $$\int\int\int_V r^2\rho dV$$

From the Jacobian, $dV=r dr d\theta dz$ (converting into cylindrical coordinates).

$$$$ Hence, $$I=\int\int\int r^3 \rho drd\theta dz$$ $$$$ However, I cannot understand how to set limits for this Triple Integral. I know that $\theta$ will vary from $0$ to $2\pi$. However, I can't set limits on $r$ and $z$.

In general, how do we set limits on Integrals in Cylindrical or Spherical Co-ordinates? This is something which has perpetually confused me.

I would be truly grateful if somebody could kindly help me. Many thanks in anticipation!


Another word for a solid sphere is "ball".

The problem statement does not say what the radius of the ball is. Since the moment of inertia will be a function of the radius of the ball, let's use $R$ to denote that radius.

One way to find the limits of integration is via Cartesian coordinates. The ball of radius $R$ centered at the origin consists of all points $(x,y,z)$ such that $$x^2 + y^2 + z^2 \leq R^2. \tag 1$$ Having set up the order of integration as you did, $I=\iiint r^3 \rho\;dr\;d\theta \;dz$, in general (for any region over which you want to integrate) the limits of $z$ are independent of $r$ and $\theta$, the limits of $\theta$ depend only on $z$, and the limits of $r$ may depend on both $\theta$ and $z$. Hence $z$ is limited only by Inequality $(1)$. From that inequality it follows that $z^2 \leq R^2 - (x^2 + y^2)$. Since $x^2 + y^2$ can be zero but cannot be less than zero, this tells us that $z^2 \leq R^2$. Moreover, $(0,0,R)$ and $(0,0,-R)$ both satisfy Inequality $(1)$, so the limits of $z$ are $-R$ and $R$.

As you already observed, when $-R < z < R$ the ball has points at every possible angle $\theta$, so we can integrate $\theta$ from $0$ to $2\pi$.

For $r$, we have $r^2 = x^2 + y^2$, so from Inequality $(1)$ we have $r^2 + z^2 \leq R^2$ and $r^2 \leq R^2 - z^2$. Any $x$ and $y$ that satisfy this inequality give a point in the ball, regardless of $\theta$. We also want $r \geq 0$. Therefore $0 \leq r \leq \sqrt{R^2 - z^2}$. The limits of integration are

$$\int_{-R}^R \int_0^{2\pi} \int_0^{\sqrt{R^2 - z^2}} r^3\rho \;dr\;d\theta\;dz.$$

A second way to look at this is directly in cylindrical coordinates. The surface of the ball intersects the $z$-axis at $z=-R$ and $z=R$, and the ball intersects all points of the $z$-axis between those limits. But any point with $|z|>R$ is outside the ball, so we just have to integrate $z$ from $-R$ to $R$. Again, we integrate over $\theta$ from $0$ to $2\pi$ regardless of $z$, by symmetry, but taking the origin of coordinates as the center of the ball, the distance from the center of the ball to the point at cylindrical coordinates $(r,\theta,z)$ is $\sqrt{r^2 + z^2}$. This distance must not be greater than $R$, that is, $\sqrt{r^2 + z^2} \leq R$, from which we get $r^2 + z^2 \leq R^2$ and (as before) $0 \leq r \leq \sqrt{R^2 - z^2}$.

A third approach is to consider that this order of integration is an application of the "disc" method of integration. Viewing the ball as a stack of "discs", the disc on the "bottom" is at $z = -R$ and the disk on "top" is at $z = R$, so $-R \leq z \leq R$. The "discs" are in fact actually circular discs, so we can integrate over $\theta$ from $0$ to $2\pi$. The radius of the disc at $z$-coordinate $z$ is a leg of a right triangle whose other leg is a segment of length $|z|$ along the axis and whose hypotenuse is $R$, so the radius of the disc is $\sqrt{R^2 - z^2}$, and we want to integrate $r$ over the interval $0 \leq r \leq \sqrt{R^2 - z^2}$.

There are a number of closely-related questions that may also be of interest, such as Using triple integral to find the volume of a sphere with cylindrical coordinates.

  • $\begingroup$ Thanks very much Sir. Sir, I'm going through your answer. In general, are the limits of $z$ dependent on $r$ and $\theta$? $\endgroup$ – Ishan Sep 11 '15 at 13:41
  • 1
    $\begingroup$ When you choose the order of integration, that determines which variable is dependent on which other variable. For $\iiint dr\;d\theta\;dz$, no matter which way you determine the limits of integration, $z$ cannot depend on $r$ or $\theta$. In general, whichever variable is "outermost" will be independent of all other variables of integration, since those variables have no meaning outside of their own integrals within the multiple integral. $\endgroup$ – David K Sep 11 '15 at 13:50
  • $\begingroup$ Thank you Sir. Sir, I have a slight confusion about setting the limits for $r$ in method 2 (involving cylindrical co-ordinates). How would we use the same method for determining the range of $r$ for a right cone of height $h$ and base radius $R$? $\theta$ would vary from 0 to 2$\pi$ while $z$ would vary from 0 to $h$. However, what would be the limits of $r$? $\endgroup$ – Ishan Sep 11 '15 at 13:55
  • $\begingroup$ In a general case, how do we use method 2 (with cylindrical coordinates) to determine the range of $r$? $\endgroup$ – Ishan Sep 11 '15 at 13:55

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