Consider all of the different ways that you can find the total mass of $1\over8$ of the unit sphere, lying in the first octant, via integration, assuming that the sphere has constant density, $\rho$.

1. Remember the formula (no actual integration): \begin{eqnarray} M &=& \int \, dm \\ &=& \int\rho\, dV \nonumber\\ &=& \rho\int dV \nonumber\\ &=& \rho {1\over 8} \left({4\over 3}\pi R^3\right)\nonumber\\ &=& {\pi\rho R^3\over 6}\nonumber \end{eqnarray} The notation in the first line $dm$ is used by some physicists to denote cutting up a big mass $M$ into a bunch of little masses $dm$ and adding up those little masses. Some folks find this notation to be a little misleading, so we will avoid it. By always cutting up space, it can be easier to understand how mass and charge integrals are mathematically the same.

2. Pieces of Spherical Shells: Imagine cutting up the space into parts of spherical shells with surface area ${1\over 8}(4\pi r^2)$ and thickness $dr$. Then: \begin{eqnarray} M &=& \rho\int_0^R {1\over 8}(4\pi r^2)\, dr\\ &=& \left.\rho{\pi\over 2}{r^3\over 3}\right\vert_0^R \nonumber\\ &=& {\pi\rho R^3\over 6} \nonumber \end{eqnarray}

3. Pieces of Disks: Imagine cutting up the space into quarters of disks, with thickness $dz$ and surface area ${1\over 4}\pi r^2$ where $r$ is the radius of the circle at height $z$, i.e. $r=\sqrt{R^2-z^2}$. Then: \begin{eqnarray} \label{diskpieces} M &=& \rho\int_0^R{1\over 4}\pi (R^2-z^2)\, dz \\ &=& \left.\rho{\pi\over 4}\left(R^2z-{z^3\over 3}\right)\right\vert_0^R \nonumber\\ &=& {\pi\rho\over 4}\left(R^3-{R^3\over 3}\right) \nonumber\\ &=& {\pi\rho R^3\over 6} \nonumber \end{eqnarray}

4. Pieces of Cubes: Imagine cutting up the region into tiny cubes, with volume $dx\, dy\, dz$. The integrand in this method is trivial; the tricky part is getting the limits right. You need to decide the order in which you will do the integrals before you can set up the limits. Any order will work, but the limits look different in each case. As an example, let's do the $x$-integral first, then the $y$-integral, and finally the $z$-integral. It helps to draw a picture, and to put explicit parentheses into the calculation to make the order of integration clear. As you do the $x$-integral, you are adding up infinitesimal cubes from $x=0$ out to $x$ on the edge of the sphere, which, for particular values of $y$ and $z$ is at $\sqrt{R^2-y^2-z^2}$. Then you are adding up little one dimensional “rows” running from $y=0$ out to the edge of a quarter disk, which for fixed $z$ lies at $y=\sqrt{R^2-z^2}$. Lastly, you are adding up these quarter disks, as $z$ runs from 0 to $R$. Notice that at the stage of the last integral, this calculation reduces to the previous method, Pieces of Disks, using (\ref{diskpieces}). Thus: \begin{eqnarray} M &=& \rho\int \, dV \\ &=& \rho\int_0^R \left(\int_0^{\sqrt{R^2-z^2}} \left(\int_0^{\sqrt{R^2-y^2-z^2}} dx\right) dy\right) dz \nonumber\\ &=& \rho\int_0^R\left(\int_0^{\sqrt{R^2-z^2}}\sqrt{R^2-y^2-z^2} \, dy\right) dz \nonumber \end{eqnarray} To do the $y$ integral, it is necessary to perform a trig substitution. Note that the need to perform a trig substitution almost always indicates that it would have been wiser to be working in curvilinear coordinates to begin with. (Whether or not it's a good idea to switch halfway through is a more difficult question to answer—watch how much longer this calculation is than the other methods and decide for yourself.) Let \begin{eqnarray} y &=& \sqrt{R^2-z^2}\sin\phi\\ dy &=& \sqrt{R^2-z^2}\cos\phi\, d\phi\\ \phi &:&0\rightarrow{\pi\over 2}\\ \end{eqnarray} Therefore: \begin{eqnarray} M &=& \rho\int_0^R\left(\int_0^{\pi\over 2} \sqrt{R^2-z^2}\sqrt{1-\sin^2\phi} \sqrt{R^2-z^2}\cos\phi \, d\phi\right)dz \\ &=& \rho\int_0^R\left(\int_0^{\pi\over 2} (R^2-z^2)\cos^2\phi \, d\phi\right)dz \nonumber\\ &=& \rho\int_0^R\left(\int_0^{\pi\over 2} (R^2-z^2){1\over 2}(1+\cos 2\phi) \, d\phi\right)dz \nonumber\\ &=& \rho\int_0^R\left. (R^2-z^2){1\over 2}(\phi+{1\over 2}\sin 2\phi)\right\vert_0^{\pi\over 2}dz \nonumber\\ &=& {\pi\rho\over 4}\int_0^R(R^2-z^2)\, dz \nonumber \end{eqnarray} We have now added up the contributions to each of the quarter disks in the $x$, $y$-plane and, as advertised, this integral is the same one as in the previous method, Pieces of Disks.

5. Pieces of Spheres: Imagine cutting up the region into spherical pieces, i.e. cut along surfaces of constant $r$, $\theta$, and $\phi$, respectively. If necessary, review how to obtain $dV=r^2\sin\theta\, dr\, d\theta\, d\phi$ in spherical coordinates and remember that there is a difference between mathematicians' and physicists' conventions for the names of the angles. Because these pieces completely respect the symmetry of the problem, the integrand is trivial to write down and the limits are constant. The integrals split up nicely into three separate integrals, so there is no need to keep track of the order of integration. \begin{eqnarray} M &=& \rho\int\, dV \\ &=& \rho\int_0^{\pi\over 2}\int_0^{\pi\over 2}\int_0^R \,r^2\sin\theta\, dr\, d\theta\, d\phi \nonumber\\ &=& \rho\left.{r^3\over 3}\right\vert_0^R \left.(-\cos\theta)\right\vert_0^{\pi\over 2} \left.\phi\right\vert_0^{\pi\over 2} \nonumber\\ &=& {\pi\rho R^3\over 6} \nonumber \end{eqnarray}

6. Pieces of Cylinders: Imagine cutting up the region into cylindrical pieces, i.e. cut along surfaces of constant $r$, $\phi$, and $z$, respectively. If necessary, review how to obtain $dV=r\sin\phi\, dr\, d\phi\, dz$ in cylindrical coordinates. These coordinates respect the axial symmetry of the problem, but not the full symmetry of the problem; the integrand is not too easy and not too hard, but not all the limits are constants. So, again it is necessary to think carefully about the order of integration. Any order will work. If you do the $r$ integral first and then $\phi$, then, part way through the problem, you'll be back to Pieces of Disks. Instead, let's try adding up the pieces to make pieces of cylindrical shells and then adding up the shells. To make a (quarter of a) cylindrical shell, you need to do the $z$ integral first, then $\phi$, and finally add up the cylindrical shells needed to do an $r$ integral. The integrals and limits are: \begin{eqnarray} M &=& \rho\int\,dV \\ &=& \rho\int_0^R\left(\int_0^{\pi\over 2}\left(\int_0^{\sqrt{R^2-r^2}} r\, dz\right) d\phi\right) dr \nonumber\\ &=& \rho\int_0^R\left(\int_0^{\pi\over 2}\sqrt{R^2-r^2}\, r\, d\phi\right) dr \nonumber\\ &=& {\pi\rho\over 2}\int_0^R\sqrt{R^2-r^2}\,r\, dr \nonumber \end{eqnarray}

   The final integral can be evaluated using a simple substitution—try it.

   The last line is the one you would have started with if you had constructed (quarters of) cylindrical shells to begin with. The height of each shell is $\sqrt{R^2-r^2}$; the circumference of each shell ${\pi r\over 2}$ and the thickness is $dr$. (Draw a picture!).

Embellishments:

Suppose the mass density $\rho$ is not constant. For each of the position-dependent mass densities below, list the methods of finding the total mass above that still work.

1. $\rho=k r^2$
2. $\rho=k z$
3. $\rho=k x$
4. $\rho=k (x^2+y^2)$
5. $\rho=k (x+y)$

Challenge:

It might seem strange to solve a spherical problem in cylindrical coordinates, as in the Pieces of Cylinders example; but, remarkably, it can be a very helpful strategy. As a challenge problem, consider a sphere placed symmetrically inside a cylinder, chosen so that the cylinder just touches the entire equator of the sphere. Show that if you cut a chunk out of the cylinder, and a related chunk out of the sphere, with any two planes that are parallel to the equator of the sphere, that the surface areas of these two chunks are the same. 1) What are the implications of this geometric fact for cartographers?