Region W

For region W (image), the angle at the vertex is $2\pi/3$ , and the height is $1/ \sqrt 3$ . I need to express the triple integral of $\int _W dV$ in cartesian, spherical and cylindrical coordinates.

I'm quite stuck on what to do with cartesian, but here is what I got so far for the other two:

Cylindrical: $\int _0 ^{1/\sqrt 3} \int _0 ^{2\pi} \int _0 ^? r dr d\theta dz$

Spherical: $\int _0 ^{\pi/2 - 2\pi/3} \int _0 ^{2\pi} \int _0 ^? \rho ^2 sin(\phi) d\rho d\theta d\phi$

Not quite sure how to calculate the radius of the cone, and also my angle phi seems to not make sense (negative value for limits on the third integral). I'm assuming that the vertex is the angle from the xy plane to the object, is this incorrect?

  • $\begingroup$ The Cartesian description of a cone is $z^2 = x^2 + y^2$, now you just need to figure out the limits for x and y. Also how did you obtain your upper bound for $\phi$? $\endgroup$
    – Triatticus
    Commented Apr 9, 2017 at 8:00
  • $\begingroup$ Measure your angle for $\phi$ relative to the vector $(0,0,1)$. This means $0 \leq \phi \leq \pi/3$. See the graphic here for a visual: en.wikipedia.org/wiki/Spherical_coordinate_system $\endgroup$
    – Kaj Hansen
    Commented Apr 9, 2017 at 8:46
  • $\begingroup$ Also, if we're looking at the cone face-on, we'll see a triangle (it's very close to this perspective in the image you've posted). I believe the vertex angle is simply the angle measure of the vertex of the triangle touching the origin. $\endgroup$
    – Kaj Hansen
    Commented Apr 9, 2017 at 8:50
  • $\begingroup$ Originally I thought that $2\pi /3$ was the angle from xy plane to the edge of the cone (If we are looking at the cone face on), and so $\phi$ would be $\pi /2 - 2\pi/3$ ? Could you elaborate on how you got $ \pi/3 $ Edit: But $2\pi /3$ is more than 90 degrees, so I guess my original interpretation was incorrect? $\endgroup$
    – JC1
    Commented Apr 9, 2017 at 9:06
  • $\begingroup$ Ah, ok I understand now where $ \pi /3 $ comes from $\endgroup$
    – JC1
    Commented Apr 9, 2017 at 9:12

2 Answers 2


Finding unknown bounds for the cylindrical and spherical cases:

Cylindrical coordinates:

Notice that how far we must go out in the "$r$" direction (in $(r,\theta, z)$ space) is dependent on the value of $z$. If we're at the origin, $r$ has a maximum of $0$ because the vertex is a mere point, but at $z = 1/\sqrt{3}$, $r$ can go all the way out to the edge on the flat top of the cone. Imagine drawing a right triangle with height $z$ and hypotenuse along the outer edge of the cone. Knowing one of the angles of this triangle is $\pi/3$ allows us to "solve the triangle", giving us $\displaystyle r = \frac{\sqrt{3}}{z}$, meaning $r$ is going to run between $0$ and this $z$-dependent value.

Spherical coordinates:

We can take an approach similar to the above to find the bounds for $\rho$. Notice that the upper bound on $\rho$ for given values of $\theta$ and $\phi$ is dependent only on the value of $\phi$. For example, when $\phi = 0$, the maximum value $\rho$ can attain is at its shortest of them all ($0 \leq \rho \leq 1/\sqrt{3}$), but when $\phi = \pi/3$, we have $\rho$ able to attain its longest possible value for the whole cone. Again, draw a triangle as we did above. This time, the triangle will have a height of $1/\sqrt{3}$, and a known angle, allowing you to solve for the hypotenuse, which is $\rho$.

A tip for both of the above:

We are allowed, in this scenario, to rearrange the order of integration. As an example, $\displaystyle \iiint \text{ stuff } \ dr \ dz \ d\theta$ will be the same as $\displaystyle \iiint \text{ stuff } \ dz \ d \theta \ dr$. Let's take advantage of this to the fullest extent possible. In particular, in both cases above, theta will run between $0$ and $2 \pi$, and this is entirely independent of whatever the other variables are doing. So if you think about it, you'll notice that ordering the integrals so that the $d \theta$ is outermost, we'll have: $$\displaystyle \iiint \text{ stuff }\ dz \ dr \ d\theta = 2 \pi \iint \text{ same stuff } dz \ dr$$

And just like that we have only $2$ integrals to worry about instead of $3$.

  • $\begingroup$ I was never very good at setting up triple integrals, so others please let me know if I've messed something up here. $\endgroup$
    – Kaj Hansen
    Commented Apr 9, 2017 at 8:35
  • $\begingroup$ For the value of $\rho$, I've drawn a right angled triangle like so, and using sohcahtoa, it comes out to be $cos(\pi/3) = {(1/\sqrt 3)} / {r} $. Why is this incorrect? $\endgroup$
    – JC1
    Commented Apr 9, 2017 at 9:34

Note that the base radius of the cone is $\frac1{\sqrt3}\tan \frac\pi3= 1$. Then

\begin{align} &V_{cylindrical}=\int _0^{2\pi} \int_0^1\int _{\frac r{\sqrt3}}^{1/\sqrt 3} r \ dzdr d\theta\\ &V_{spherical}=\int _0 ^{2\pi} \int _0 ^{\frac\pi3} \int _0 ^{\frac1{\sqrt3\cos\phi}}\rho ^2 \sin\phi \ d\rho d\phi d\theta \\ &V_{cartesian}=\int _{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int _{\frac {\sqrt{x^2+y^2}}{\sqrt3}}^{1/\sqrt 3} dz dy dx \end{align} Carry out the three integrals to get $$V_{cylindrical}= V_{spherical}= V_{cartesian}=\frac\pi{3\sqrt3}$$


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