Week 4 Activity

Section 4.3 Week 4 Activity

Subsection 4.3.1 Spherical and Cylindrical Coordinates

Activity 4.3.1.

Integrate \(f(x,y,z) = 2x^2 + y^2 +z^2\) on the sphere of radius \(2\) centred at the origin.

The sphere is described nicely in spherical coordinates as \(r \in [0,2]\text{,}\) \(\theta \in [0, 2\pi]\) and \(\phi \in [0, \pi]\text{.}\) The integrand can be seperated as \(x^2 + (x^2 + y^2 + z^2)\) and written as \(r^2 \sin^2 \phi \cos^2 \theta + r^2\text{.}\) I can write the integral in spherical coordinates and split it up into two integrals. The Jacobian is \(r^2 \sin \phi\text{.}\)

\begin{align*} \amp \int_0^2 \int_0^{2\pi} \int_0^{\pi} (r^2 \sin^2 \phi \cos^2 \theta + r^2) r^2 \sin \phi d\phi d\theta dr \\ \amp = \int_0^2 \int_0^{2\pi} \int_0^{\pi} r^3 \sin^3 \phi \cos^2 \theta d\phi d\theta dr + \int_0^2 \int_0^{2\pi} \int_0^{\pi} r^4 \sin \phi d\phi d\theta dr \\ \amp = \int_0^2 r^4 dr \int_0^{2\pi} \cos^2 \theta d \theta \int_0^{\pi} \sin^3 \phi d\phi + \int_0^2 r^4 dr \int_0^{2\pi} d\theta \int_0^{\pi} \sin \phi d\phi \\ \amp = \left( \frac{32}{5} \right) (\pi)\left( \frac{4}{3} \right) + \left( \frac{32}{5} \right) (2\pi) (2) = \frac{128\pi}{15} + \frac{128\pi}{5} = \frac{512 \pi}{15} \end{align*}

Activity 4.3.2.

Integrate \(f(x,y,z) = x^2 + y^2 - 4z\) on the cylinder of radius \(4\) about the \(z\) axis restricted to the height \(z \in [-1,3]\text{.}\)

Solution.

The cylinder is described nicely in cylindrical coordinates as \(r \in [0,4]\text{,}\) \(\theta \in [0, 2\pi]\) and \(z \in [-1,3]\text{.}\) The integrand becomes \(r^2 - 4z\text{.}\) The Jacobian is \(r\text{.}\)

\begin{align*} \amp \int_0^4 \int_0^{2\pi} \int_{-1}^3 (r^2 - 4z) r dz d\theta dr = \int_0^4 \int_0^{2\pi} \int_{-1}^3 r^3 dz d\theta dr - \int_0^4 \int_0^{2\pi} \int_{-1}^3 4rz dz d\theta dr \\ \amp = \int_0^4 r^3 dr \int_0^{2\pi} d\theta \int_{-1}^3 dz - 4 \int_0^4 r dr \int_0^{2\pi} d\theta \int_{-1}^3 z dz \\ \amp = (16)(2\pi)(4) - 4 (8)(2\pi)(4) = 128 \pi - 256 \pi = 128 \pi \end{align*}

Activity 4.3.3.

Integrate \(f(x,y,z) = x^2 + y^2\) on he sphere of radius \(1\) centred at the \((0,0,1\text{.}\)

Solution.

Even though the region is a sphere, the integrand looks better in cylindrical coordinates, so I'll try that. The integrand simply becomes \(r^2\text{.}\) The sphere has cartesian equation \(x^2 + y^2 + (z-1)^2 = 1\text{.}\) In cylindrical coordinates, this becomes \(r^2 + (z-1)^2 = 1\text{.}\) I can take \(z \in [0,2]\) and \(r\) ranging from \(0\) to \(r = \sqrt{1-(z-1)^2} = \sqrt{2z-z^2}\text{.}\) The Jacobian is \(r\text{.}\)

\begin{align*} \amp \int_0^{2\pi} \int_0^2 \int_0^{\sqrt{2z-z^2}} r^2 r dr dz d\theta = \int_0^{2\pi} d\theta \int_0^2 \int_0^{\sqrt{2z-z^2}} r^3 dr dz \\ \amp = (2\pi) \int_0^2 \frac{r^4}{4} \Bigg|_0^{\sqrt{2z-z^2}} dz = \frac{\pi}{2} \int_0^2 (2z-z^2)^2 dz \\ \amp = \frac{\pi}{2} \int_0^2 z^4 - 4z^3 + 4z^2 dz = \frac{\pi}{2} \left( \frac{z^5}{5} - z^4 + \frac{4z^3}{3} \right) \Bigg|_0^2 \\ \amp = \frac{\pi}{2} \left( \frac{32}{5} - 16 + \frac{64}{3} \right) = \frac{88\pi}{15} \end{align*}

Activity 4.3.4.

Find the volume of the sphere of radius \(3\) with the parabaloid \(z = (x^2 + y^2) - 3\) cut out of it.

Solution.

Though I am dealing with a sphere, the parabaloid is easier to describe in cylindrical coordinates, so I will use that system. The whole system is symetrical about the \(z\) axis, so \(\theta\) can simply range from \(0\) to \(2\pi\text{.}\) The sphere is the outside range of \(r\) and the paraboloid is the inside range of \(r\text{,}\) but both depend on \(z\text{.}\) I need equations for both. The sphere is \(x^2 + y^2 +z^2 = 9\text{,}\) which changes to \(r^2 + z^2 = 9\text{.}\) Solving for \(r\) gives \(r = \sqrt{9-z^2}\text{.}\) The parabaloid is \(z = r^2 - 3\text{,}\) which simplifies to \(r = \sqrt{z+3}\text{.}\) The range on \(z\) starts at \(-3\) and ends at the point where the paraboloid and the sphere meet. In the two equations, I can isolate the replace \(r^2\) to get \(z+3+z^2 = 9\text{,}\) which is the quadratic \(z^2 + z - 6 = (z-3)(z+2)\text{.}\) Therefore, the parabaloid and sphere meet at the \(z=2\) plane and \(z=2\) is the upper bound for \(z\text{.}\) The Jacobian is \(r\text{.}\)

\begin{align*} \amp \int_0^{2\pi} \int_{-3}^2 \int_{\sqrt{z+3}}^{\sqrt{9-z^2}} r dr dz d\theta \\ \amp = \int_0^{2\pi} d\theta \int_{-3}^2 \frac{r^2}{2} \Bigg|_{\sqrt{z+3}}^{\sqrt{9-z^2}} dz \\ \amp = 2\pi \int_{-3}^2 \frac{1}{2} \left( 9-z^2 - (z+3) \right) dz\\ \amp = \pi \int_{-3}^2 6 - z - z^2 dz \\ \amp = \pi \left( 6z - \frac{z^2}{2} - \frac{z^3}{3} \right) \Bigg|_{-3}^2\\ \amp = \pi \left( 6(2-(-3)) - \frac{4 - 9}{2} - \frac{8 - (-27)}{3} \right) = \pi \left(30 + \frac{5}{2} - \frac{35}{3} \right) = \frac{125\pi}{6} \end{align*}

Activity 4.3.5.

Find the volume of the portion of the cylinder about the \(z\) axis of radius \(3\) which is above the \(z = -1\) plane but below the \(x + y + z = 5\) plane.

Solution.

I can work in cylindrical coordinates. The planes \(z = -1\) remains \(z=-1\text{.}\) The plane \(x + y + z = 5\) is trickier. In changed into \(r \cos \theta + r \sin \theta + z = 5 = r (\cos \theta + \sin \theta) + z = 5\text{.}\) I can use a trig identity to write this as \(r \left( \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \right) + z = 5\text{.}\) This is tricky, since the equation involves all of the variables. I choose to set \(z\) first, then \(\theta\text{,}\) then \(r\text{.}\) For \(z\text{,}\) I need to calculate the maximum value; the highest point where the cylinder meets the plane. This happens when \(x = y = -3\text{,}\) so \(z=11\) is the highest value and I can take \(z \in [-1,11]\text{.}\) Then at each value of \(z\text{,}\) I have the relationship \(r \left( \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \right) + z = 5\) between \(r\) and \(\theta\text{.}\) If I solve for \(r\text{,}\) I get

\begin{equation*} r = \frac{5-z}{\left( \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \right)} \end{equation*}

This gets particularly tricky. I can't choose either \(r\) or \(\theta\) to be the outside variable for the whole cylinder; I need two different cases for \(z \lt 5\) and \(z \gt 5\text{.}\) On the bottom half of the cylinder, I can use the previous equation to bound \(r\) for \(\theta \in \left[\frac{-\pi}{4} \frac{3\pi}{4} \right]\text{,}\) but \(r\) goes from \(0\) to \(3\) for the rest of the range. For the tope half of the cylinder, the bound of \(r\) in terms of \(\theta\) works for \(\theta in \left[ \frac{3\pi}{4}, \frac{7\pi}{4} \right]\text{.}\) For the rest of the range of \(\theta\text{,}\) there isn't anything at all. This gives three integrals.

\begin{align*} V \amp = \int_{-1}^5 \int_{\frac{-\pi}{4}}^{\frac{3\pi}{4}} \int_0^{\frac{5-z}{\left( \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \right)}} r dr d\theta dz + \int_{-1}^5 \int_{\frac{3\pi}{4}}^{\frac{7\pi}{4}} \int_0^3 r dr d\theta dz\\ \amp \hspace{2cm} + \int_5^{11}\int_{\frac{3\pi}{4}}^{\frac{7\pi}{4}} \int_0^{\frac{5-z}{\left( \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \right)} }r dr d\theta dz \end{align*}

This is a miserable set of integrals, to be sure. Luckly for us, there is another way to approach this entire question. If we look at the full cylinder for \(z \in [-1,11]\text{,}\) the plane cuts this cylinder into two equal pieces. Therefore, the desired volume should be one half of the volume of the cylinder of height \(12\) and radius \(3\text{,}\) which is \(72\pi\text{.}\)

Subsection 4.3.2

Activity 4.3.6.

Let \(S\) be the solid cone about the \(z\) axis with height \(5\) and base radius \(1\text{.}\) The point of the cone is at the origin and it opens upwards, readhing ther radius of \(1\) at \(z = 5\text{.}\) Let \(\rho(x,y,z) = \frac{z^2}{10}\) be the density of \(S\text{.}\) Calculate the mass, centre of mass, the moments of intertia, and the raddi of gyration.

Solution.

Cylindrical coordiantes are the best system to describe this cone, since it is about the \(z\) axis. I can set constant bounds \(\theta \in [0,2\pi]\) and \(r \in [0,1]\text{.}\) Then the boundary of the cone can be described as \(z \in \left[ 5r,5 \right]\text{.}\) With these bounds, I can move on to do the many integrals to calculate mass, first moments, centre of mass, and second moments.

\begin{align*} m \amp = \int_S \rho dV = \int_0^1 \int_0^{2\pi} \int_{5r}^5 \frac{z^2}{10} rdz d\theta dr \\ \amp = 2\pi \int_0^1 r\frac{z^3}{30} \Bigg|_{5r}^5 dr = \frac{\pi}{15} \int_0^1 r (125 - 125 r^3) dr \\ \amp = \frac{125\pi}{15} \int_0^1 r - r^4 dr = \frac{25\pi}{3} \left( \frac{r^2}{2} - \frac{r^5}{5} \right) \Bigg|_0^1 = \frac{25\pi}{3} \left( \frac{1}{2} - \frac{1}{5} \right) = \frac{25\pi}{3} \frac{3}{10} = \frac{5\pi}{2} \end{align*}

I do not need to calculate the first moments \(M_{yz}\) and \(M_{xz}\text{.}\) These moments measure the balance of the object in the \(x\) and \(y\) axes. The cone is completely symmetric in these axes, and the density function is also symmetric. Therefore, the mass is perfectly balances in the \(x\) and \(y\) axes and the moments must be zero. I only need to calculate the moment \(M_{xy}\text{.}\)

\begin{align*} M_{xy} \amp = \int_S z\rho dV = \int_0^1 \int_0^{2\pi} \int_{5r}^5 z\frac{z^2}{10} r dz d\theta dr = \frac{1}{10} 2\pi \int_0^1 \frac{z^4}{4} \Bigg|_{5r}^5 rdr \\ \amp = \frac{\pi}{5} \frac{1}{4} \int_0^1 (625r - 625r^5) rdr = \frac{625\pi}{20} \left( \frac{r^2}{2} - \frac{r^6}{6} \right) \Bigg|_0^1 \\ \amp = \frac{125\pi}{4} \left( \frac{1}{2} - \frac{1}{6} \right) = \frac{125\pi}{4} \frac{1}{3} = \frac{125\pi}{12} \end{align*}

With the mass and first momemnts calculated, I can calculate the coordinates of the centre of mass.

\begin{align*} \bar{x} \amp = \frac{M_{yz}}{m} = 0 \\ \bar{y} \amp = \frac{M_{xz}}{m} = 0 \\ \bar{z} \amp = \frac{M_{xy}}{m} = \frac{\frac{125\pi}{12}}{\frac{5\pi}{2}} = \frac{25}{6} \end{align*}

Note that this is a reasonable answer. The cone is located in the range \(z \in [0,5]\text{.}\) This \(z\) coordinate for the centre of mass is in that range, as should be expected. Now I continue with the moments of intertia.

\begin{align*} I_x \amp = \int_S (y^2 + z^2) \rho dV = \int_0^1 \int_0^{2\pi} \int_{5r}^5 (r^2 \sin^2 \theta + z^2) \frac{z^2}{10} dz rd\theta dr = \\ \amp = \frac{1}{10} \int_0^1 \int_0^{2\pi} \int_{5r}^5 r^3 \sin^2 \theta z^2dz d\theta dr + \frac{1}{10} \int_0^1 \int_0^{2\pi} \int_{5r}^5 z^4 rdz d\theta dr\\ \amp = \frac{1}{10} \int_0^{2\pi} \sin^2 \theta d\theta \int_0^1 r^3 \frac{z^3}{3} \Bigg|_{5r}^5 dr + \frac{1}{10} 2\pi \int_0^1 \frac{z^5}{5} \Bigg|_{5r}^5 rdr\\ \amp = \frac{1}{30} \pi \int_0^1 (125) (r^3 - r^6) dr + \frac{\pi}{25} \int_0^1 (3125) (r-r^6) dr\\ \amp = \frac{25\pi}{6} \left( \frac{r^4}{4} - \frac{r^7}{7} \right) \Bigg|_0^1 + 125\pi \left( \frac{r^2}{2} - \frac{r^7}{7} \right) \Bigg|_0^1 \\ \amp = \frac{25\pi}{6} \left( \frac{1}{4} - \frac{1}{7} \right) + 125\pi \left( \frac{1}{2} - \frac{1}{7} \right)\\ \amp = \frac{25\pi}{6} \frac{3}{14} + 125\pi \frac{5}{14} = \frac{2875\pi}{84} \end{align*}

By symmetry, the moments about the \(x\) and \(y\) axis are the same.

\begin{align*} I_y \amp = I_x = \frac{2875\pi}{84} \\ I_z \amp = \int_S (x^2 + y^2) \rho dV = \int_0^1 \int_0^{2\pi} \int_{5r}^5 r^2 \frac{z^2}{10} rdz d\theta dr\\ \amp = \frac{1}{10} 2\pi \int_0^1 r^3 \frac{z^3}{3} \Bigg|_{5r}^5 dr = \frac{\pi}{15} \int_0^1 125(r^3-r^6) dr = \frac{25\pi}{3} \left( \frac{r^4}{4} - \frac{r^7}{7} \right) \Bigg|_0^1\\ \amp = \frac{25\pi}{3} \left (\frac{1}{4} - \frac{1}{7} \right) = \frac{25\pi}{28}\\ \bar{\bar{x}}^2 \amp = \frac{I_{x}}{m} = \frac{\frac{2875\pi}{84}}{\frac{5\pi}{2}} = \frac{(2875)(2)}{(84)(5)} = \frac{575}{42} \implies \bar{\bar{x}} = \sqrt{\frac{575}{42}}\\ \bar{\bar{y}} \amp = \bar{\bar{x}} = \sqrt{\frac{575}{42}} \\ \bar{\bar{z}}^2 \amp = \frac{I_{z}}{m} = \frac{\frac{25\pi}{28}}{\frac{5\pi}{2}} = \frac{50}{140} = \frac{5}{14} \implies \bar{\bar{z}} = \sqrt{\frac{5}{14}} \end{align*}

Activity 4.3.7.

Let \(S\) be the solid hemisphersphere of radius \(2\) above the \(xy\) plane excluding the ocstant where all the variables are positive Let \(\rho(x,y,z) = \frac{1}{\sqrt{x^2 + y^2 + z^2}}\) be the density function on the sphere. Calculate the the mass, centre of mass, the moments of intertia, and the radii of gyration.

Solution.

This object is a portion of the sphere, so spherical coordinates are obviously appropriate. The hemisphere about the \(xy\) is the hemisphere where the colatitude \(\phi\) is limited to \(\left[ 0, \frac{\pi}{2} \right]\text{.}\) Then, to exclude the octants where all variables are positive, I restrict the longitute to \(\theta \in \left[ \frac{\pi}{2}, 2\pi \right]\text{.}\) The radius bounds are still \(r \in [0,2]\text{.}\) Those bound describe the region, so I can progress to do all the integrals in spherical coordinates.

\begin{align*} m \amp = \int_S \rho dV \\ \amp = \int_0^2 \int_0^{\frac{\pi}{2}} \int_{\frac{\pi}{2}}^{2\pi} \frac{1}{r} r^2 \sin \phi d\theta d\phi dr \\ \amp = \int_0^2 r dr \int_0^{\frac{\pi}{2}} \sin \phi d\phi \int_{\frac{\pi}{2}}^{2\pi} d \theta = (2)(1)\left( \frac{3\pi}{2} \right) = 3\pi \end{align*}

Now I calculate the centre of mass. By symmetry, I expect \(M_{yz}\) and \(M_{xz}\) to be the same; I only need to calculate one of them.

\begin{align*} M_{yz} \amp = \int_S x\rho dV = 0 \\ \amp = \int_0^2 \int_0^{\frac{\pi}{2}} \int_{\frac{\pi}{2}}^{2\pi} r \sin \phi \cos \theta \frac{1}{r} r^2 \sin \phi d\theta d\phi dr \\ \amp = \int_0^2 r^2 dr \int_0^{\frac{\pi}{2}} \sin^2 \phi d\phi \int_{\frac{\pi}{2}}^{2\pi} \cos \theta d\theta \\ \amp = \left( \frac{r^3}{3} \Bigg|_0^2 \right) \left( \frac{x}{2} - \frac{\sin 2x}{4} \Bigg|_0^{\frac{\pi}{2}} \right) \left( \sin \theta \Bigg|_{\frac{\pi}{2}}^{2\pi} \right) \\ \amp = \frac{8}{3} \left( \frac{\pi}{4} - 0 - 0 + 0 \right) \left( 0 - 1 \right) = \frac{-\pi}{6} \\ M_{xz} \amp = \frac{-\pi}{6} \\ M_{xy} \amp = \int_S z\rho dV = 0 \\ \amp = \int_0^2 \int_0^{\frac{\pi}{2}} \int_{\frac{\pi}{2}}^{2\pi} r \cos \phi \frac{1}{r} r^2 \sin \phi d\theta d\phi dr \\ \amp = \int_0^2 r^2 dr \int_0^{\frac{\pi}{2}} \cos \phi \sin \phi d \phi \int_{\frac{\pi}{2}}^{2\pi} d\theta \\ \amp = \left( \frac{r^3}{3} \Bigg|_0^2 \right) \left( \frac{-\cos 2x}{4} - \frac{1}{4} \Bigg|_0^{\frac{\pi}{2}} \right) \left( \theta \Bigg|_{\frac{\pi}{2}}^{2\pi} \right) \\ \amp = \left( \frac{8}{3} - 0 \right) \left( \frac{1}{4} - \frac{1}{4} + \frac{1}{4} + \frac{1}{4} \right) \left( \frac{3\pi}{2} \right) \\ \amp = \frac{8}{3} \frac{1}{2} \frac{3\pi}{2} = \frac{\pi}{32} \end{align*}

Then I can calculate the coordinates of the centre of mass.

\begin{align*} \bar{x} \amp = \frac{M_{yz}}{m} = \frac{\frac{-\pi}{6}}{3\pi} = \frac{-1}{18} \\ \bar{y} \amp = \frac{M_{xz}}{m} = \frac{\frac{-\pi}{6}}{3\pi} = \frac{-1}{18} \\ \bar{z} \amp = \frac{M_{xy}}{m} = \frac{\frac{\pi}{32}}{3\pi} = \frac{1}{96} \end{align*}

Now I proceed with calculating the moments of inertia.

\begin{align*} I_x \amp = \int_S (y^2 + z^2) \rho dV \\ \amp = \int_0^2 \int_0^{\frac{\pi}{2}} \int_{\frac{\pi}{2}}^{2\pi} (r^2 \sin^2 \phi \sin^2 \theta + r^2 \cos^2 \phi) \frac{1}{r} r^2 \sin \phi d\theta d\phi dr \\ \amp = \int_0^2 \int_0^{\frac{\pi}{2}} \int_{\frac{\pi}{2}}^{2\pi} r^3 \sin^3 \phi \sin^2 \theta d\theta d\phi dr + \int_0^2 \int_0^{\frac{\pi}{2}} \int_{\frac{\pi}{2}}^{2\pi} r^3 \cos^2 \phi \sin \phi d\theta d\phi dr\\ \amp = \int_0^2 r^3 dr \int_0^{\frac{\pi}{2}} \sin^3 \phi d\phi \int_{\frac{\pi}{2}}^{2\pi} \sin^2 \theta d\theta + \int_0^2 r^3 dr \int_0^{\frac{\pi}{2}} \cos^2 \phi \sin \phi d \phi \int_{\frac{\pi}{2}}^{2\pi} d\theta\\ \amp = (4) \left( \frac{2}{3} \right) \left( \frac{3\pi}{4} \right) + (4)\left( \frac{1}{3} \right) \left( \frac{3\pi}{4}\right) = 2\pi + \pi = 3\pi \end{align*}

By symmetry, I expect \(I_y\) to be the same as \(I_x\text{.}\)

\begin{align*} I_y \amp = 3\pi \end{align*}

Finally, I need to calculate the last moment of inertia.

\begin{align*} I_z \amp = \int_S (x^2 + y^2) \rho dV \\ \amp = \int_0^2 \int_0^{\frac{\pi}{2}} \int_{\frac{\pi}{2}}^{2\pi} (r^2 \sin^2 \phi \cos^2 \theta + r^2 \sin^2 \phi \sin^2 \theta) \frac{1}{r} r^2 \sin \phi d\theta d\phi dr \\ \amp = \int_0^2 \int_0^{\frac{\pi}{2}} \int_{\frac{\pi}{2}}^{2\pi} r^3 \sin^3 \phi \cos^2 \theta d\theta d\phi dr + \int_0^2 \int_0^{\frac{\pi}{2}} \int_{\frac{\pi}{2}}^{2\pi} r^3 \sin^3 \phi \sin^2 \theta d\theta d\phi dr\\ \amp = \int_0^2 r^3 dr \int_0^{\frac{\pi}{2}} \sin^3 \phi d\phi \int_{\frac{\pi}{2}}^{2\pi} \cos^2 \theta d\theta + \int_0^2 r^3 dr \int_0^{\frac{\pi}{2}} \sin^3 \phi d \phi \int_{\frac{\pi}{2}}^{2\pi} \sin^2 d\theta\\ \amp = (4) \left( \frac{2}{3} \right) \left( \frac{3\pi}{4} \right) + (4)\left( \frac{2}{3} \right) \left( \frac{3\pi}{4}\right) = 2\pi + 2\pi = 4\pi \end{align*}

Now I can calculate the radii of gyration.

\begin{align*} \bar{\bar{x}}^2 \amp = \frac{I_{x}}{m} = \frac{3\pi}{3\pi} = 1 \implies \bar{\bar{x}} = 1 \\ \bar{\bar{y}}^2 \amp = \frac{I_{y}}{m} = \frac{3\pi}{3\pi} \implies \bar{\bar{y}} = 1 \\ \bar{\bar{z}}^2 \amp = \frac{I_{z}}{m} = \frac{4\pi}{3\pi} \implies \bar{\bar{z}} = \frac{2}{\sqrt{3}} \end{align*}

Subsection 4.3.3 Conceptual Review Questions

What are first and second moments? What do they do?
How are density/mass calculations similar to probability calculations?
What are the advantages of curvilinear coordinate systems?