Assignment 10
Problems
Calculating Air Pressure
Our atmosphere - the thin layer of air surrounding our planet making all life possible - is held to the surface via gravity. There is no barrier holding the atmosphere in, but instead it thins out rapidly from its density at the surface to a near perfect vaccuum up in space. Using some classical physics, one learns that in fact the atmospheric density decreases exponentially, following the formula \[\rho(x,y,z)=\rho_0 e^{-z/H}\] Where \(\rho_0\) is the density of air at sea level and \(H\) is the characteristic height of the atmosphere: a number controlling how quickly the atmosphere thins out with altitude. These numbers are different on each planet with an atmosphere: on Earth we find that \(\rho_0=0.0752\) pounds per cubic foot, and \(H=6.21\) miles.
Usually, we can happily ignore air pressure in day-to-day life, as it presses on all sides of us equally, and our bodies have evolved to support this pressure. However, air pressure is very important to take into account in certain engineering situations: especially when designing vaccuum chambers. A vaccuum chamber is an enclosure (ranging from a small box, to an entire room) from which all of the air can be pumped out for experiments requiring an air-free environment (such as testing rocket engine parts).
As part of your job back at the rocket factory, you are tasked with helping the engineering team design a new vaccuum chamber. This chamber is to be a perfect cube, 10 feet on each side.
Specifically we can model the top face of this vaccuum chamber as a square region \(R\) in the plane with \(0\leq x\leq 10\) and \(0\leq y\leq 10\).
Above this square \(R\), there is a column of air extending all the way into space: we denote this region as \[E=\{(x,y,z)\mid (x,y)\in R, z\geq 0\}\]
To calculate the weight of the atmosphere pressing down on the vaccuum chamber, we need to do a of the density of the atmosphere inside of this column of air:
\[M=\iiint_E \rho(x,y,z)dV\]
How much weight is pressing down on the top of the vaccuum chamber from the air? Hint: the range of \(z\) in the region \(E\) will go all the way into space, so you’ll end up with an improper integral from Calc II. It’s easiest to calculate things leaving \(\rho_0\) and \(H\) as constants, and only plugging in their numerical values at the end: remember to convert everything to the same units (say, pounds and feet!)
Optimizing Volume of a Wineglass
We so far have been using triple integrals to compute the total amount of some function \(f\) over a region \(E\). Remember that when this function \(f\) is just the constant \(1\), you end up with an integral of the form \[\iiint_E dV\] and this just computes the volume of the region E! This is very useful in practice, as the volumes of complicated regions can be calculated using integration techniques.
As an example here, imagine you are an industrial engineer working for a company that manufactures sets of high-end drinkware: wine glasses, champagne flutes, and so on. The designers of the new collection have come up with a parametric model: a single model with several tune-able parameters. Their model describes the interior of the glass by a region \(E_{a,b}\), which can be tuned by changing two parameters \(a,b\): \[E_{a,b}=\{(x,y,z)\mid b(x^2+y^2)\leq z\leq a^2\}\]
The units here are centimeters, so the volume of the region is given in \(\mathrm{cm}^3\), better known as milliliters. The goal of this parametric is to be able to define different glass shapes by choosing different relationships between the parameters.
- Wine Glasses are shorter and fatter, so to design a wine glass we set the width and height parameters equal, \(a=b\) and get a model that depends on a single parameter.
- Champagne Flutes are taller and skinnier: to design one of these, we set the height parameter \(b\) equal to three times the width parameter, so \(b=3a\).
If the company later wants to expand its glassware collection, it does not need to go through the process of redesigning new models from scratch, it can just change the relationship between \(a\) and \(b\) to get a new look and feel.
After doing a review of several cups across the industry, your manager finds the average volume of a red wine glass is 360mL. They then task you with taking the model from the design team, and constructing a wine glass as part of their collection that holds this same volume. What are the values of the parameters \(a,b\) that you need to report back to manufacturing, to have a glass of the right size?
Hint: You know you are dealing with a wineglass, so you know how \(a\) and \(b\) are related. Can you set up a triple integral that measures the volume of the wineglass (use the equation given to get bounds for \(z\) and radius)? Then solve for when this volume is 360ml (feel free to use a calculator!)
The Mass of the Earth
The interior structure of the Earth is very difficult to probe: we live on the very surface of the crust, a rocky layer over 20 miles thick, and to date the deepest humans have ever drilled is the Kola Superdeep Borehole, dug to a depth of 7.69 miles by the USSR (and stopped when the heat and pressure was melting through all drillbits).
Thus, it is a true wonder of modern science that we are able to calculate the density of the earth at various depths without directly retrieving samples: instead, we use the vibrations of earthquakes (and a lot of sophisticated mathematics) to deduce the density deep below the crust. To a very good approximation, the earth is spherically symmetric , and so we may record the density of the earth as a function \(D(r)\) of the *radius$ \(r\) from the center. With the density function \(D\) in hand, the mass of the earth \(E\) is given by the triple integral \[M=\iiint_E D\, dV\]
While the true density function of the earth is rather complicated, its general behavior is well-approximated by the following function: its more dense in the center at the metal core, with density dropping off to an approximate constant level in the mantle and surface (where the earth is mostly rock, not metal):
\[D(r)=a + \frac{b}{1+r^2}\]
Where \(r\) is measured in thousands of kilometers and \(a\) is the density of rock (\(a=6,000kg/m^3\)), and \(b\) is the difference in density between rock and metal, or \(b=4,000kg/m^3\). For the coming calculations, its useful to compute these to the units of kilograms per cubic 1000km to match the units on \(r\): That is \(a=6\cdot 10^{12}\) and \(b=4\cdot 10^{12}\).
Approximating the earth as a sphere of radius \(R=6,000\) kilometers, set up a triple integral in spherical coordinates that finds the mass of the earth.
Solve this integral (don’t plug in the numbers until the end, as the formula will be much cleaner if you keep \(R,a,b\) as variables), and then plug in the relevant constants. What is the mass of the earth in kilograms? (Plug the numbers in using a calculator or computer: the result will be a very large number, best reported in scientific notation!)
Solutions
Calculating Air Pressure
We are looking to calculate the air pressure of the atmosphere on the square region \[R=\{0\leq x\leq 10,0\leq y\leq 10\}\] That means we need to calculate the mass of the entire atmosphere over this region. Approximating the atmosphere as extending all the way up into space (to infinity) we can write this region as
\[E=\{(x,y,z)\mid (x,y)\in R, 0\leq z\leq \infty\}\]
To find the mass we need to perform an integral over \(E\) of the density of the atmosphere, which we were given \(\rho(x,y,z)=\rho_0 e^{-z/H}\).
We can express the triple integral as an iterated integral given our bounds
\[M=\iiint_E\rho\,dV = \int_0^{10}\int_0^{10}\int_0^\infty \rho_0 e^{-z/H}dzdydx\]
This allows us to just focus on the \(z\)-integral first: remembering that \(\rho_0\) and \(H\) are just constants,
\[\int_0^\infty \rho_0 e^{-z/H}dz\] \[=\rho_0 e^{-z/H}\frac{1}{-1/H}\Bigg|_0^\infty\] \[=-H\rho_0 e^{-z/H}\Bigg|_0^\infty\] \[=-H\rho_0(0-1)=H\rho_0\]
Now we just need to integrate over \(x\) and \(y\), which is easy since \(H\) and \(\rho_0\) are constant:
\[\int_0^{10}\int_0^{10}H\rho_0 dydx=H\rho_0\int_0^{10}\int_0^{10}dydx=100H\rho_0\]
We have the values of the constants: \(\rho_0=0.0752\) pounds per cubic foot, and \(H=6.21\) miles, but to use them we need to make sure everything is in the same units. Our integral \(x,y,z\) were all in feet, and so is the constant \(\rho_0\), so thats’ good. But \(H\) was given in miles, so we need to convert it to feet:
\[H=6.21 \textrm{miles}\cdot \frac{5280 \textrm{feet}}{\textrm{mile}}=32788.8\] Plugging these two constants in, we find that the weight of the atmosphere pushing down on our vaccuum chamber is
\[100\cdot 32,788\cdot 0.0752=246,571\]
Thus we need to build things so that they can support \(246,561\) pounds of atmospheric pressure.
Optimizing Volume of a Wineglass
For a wineglass we know that \(a=b\) so we can describe our region as
\[E=\{(x,y,z)\mid a(x^2+y^2)\leq z\leq a^2\}\]
We are given \(z\) bounds so we will do that integral first,
\[\mathrm{Slice}=\int_{a (x^2+y^2)}^{a^2} dz\]
The region we integrate this slice over can be found by setting the \(z\) bounds equal (this tells us where the length of the slices go to zero). That is \(a(x^2+y^2)=a^2\) or, cancelling a factor of \(a\), \(x^2+y^2=a\). Thus our integral in \((x,y)\) is over the circle of radius \(\sqrt{a}\), so we should switch these variables to polar.
\[\iint_R\mathrm{Slice}\,\,dA = \int_0^{2\pi}\int_0^{\sqrt{a}} \mathrm{Slice}\,\,rdrd\theta\]
Finally, we can convert the \(x^2+y^2\) occurring in the slices’ bound to polar, yielding an iterated integral in \(r,\theta,z\):
\[\mathrm{Volume}=\iiint_E dV = \int_0^{2\pi}\int_0^{\sqrt{a}}\int_{a r^2}^{a^2} dz \, rdr\,d\theta\]
Computing this integral, we find the \(z\) integral first
\[\int_{a r^2}^a dz = z\Bigg|_{a r^2}^{a^2} =a^2-ar^2\]
Then the \(r\) integral
\[\int_0^{\sqrt{a}} (a^2-ar^2)rdr=\int_0^{\sqrt{a}} (a^2r -ar^3)dr=\left(a^2\frac{r^2}{2}-a\frac{r^4}{4}\right)\Bigg|_0^{\sqrt{a}}\] \[= a^2\frac{(\sqrt{a})^2}{2}-a\frac{(\sqrt{a})^4}{4}=\frac{a^3}{2}-\frac{a^3}{4}=\frac{a^3}{4}\]
and finally the \(\theta\) integral.
\[\int_0^{2\pi}\frac{a^3}{4}d\theta = 2\pi\frac{a^3}{4}=\frac{a^3\pi}{2}\]
Now we know the volume of an arbitrary wineglass with parameter \(a\). To find the wineglass of the right volume, we just set equal to 360 and solve:
\[\frac{a^3\pi}{2}=360 \,\,\implies a^3 = \frac{720}{\pi}\]
\[a=\sqrt[3]{\frac{720}{\pi}}\approx 6.11\]
The Mass of the Earth
We are given the density as a function of radius which suggests we convert to spherical coordinates. The angle bounds are \(0\leq\theta\leq 2\pi\) and \(0\leq \phi\leq \pi\) as we are considering the entire sphere, and the radius bound runs \(0\leq \rho\leq 6000\) in kilometers, as stated in the problem. Recalling the spherical volume element \(dV = \rho^2\sin\phi \,d\rho d\phi d\theta\), the mass is
\[M=\iiint_E D dV = \iint_0^{2\pi}\int_0^\pi\int_{0}^{6000}\left(a+\frac{b}{1+\rho^2}\right) \rho^2\sin\phi \,d\rho d\phi d\theta\]
Since we have written this as an iterated integral, we can proceed one integral at a time. The innermost integral is with respect to \(\rho\) where \(\sin\phi\) is a constant, so we can pull it out front and just consider
\[\int_0^R \left(a+\frac{b}{1+\rho^2}\right) \rho^2 d\rho\]
Where I will write \(R=6000\) and plug in the numerical value at the end, to avoid clutter. This integral can be broken into two integrals
\[=a\int_0^R \rho^2d\rho + b\int_0^R\frac{\rho^2}{1+\rho^2}d\rho\]
The first of these is immediate with the power rule, yielding \(aR^3/3\). The second requires a bit of thought from Calc I and II: we can simplify by adding zero in a clever way to the numerator and regrouping:
\[\frac{\rho^2}{1+\rho^2}=\frac{\rho^2+1-1}{1+\rho^2}=\frac{1+\rho^2}{1+\rho^2}-\frac{1}{1+\rho^2}=1-\frac{1}{1+\rho^2}\] We can now integrate either by remembering that \(1/(1+x^2)\) is the derivative of arctangent, or by doing a trigonometric substitution \(\rho=\sec x\) and using that \(1+\sec^2x =\tan^2 x\). In any case, the result is
\[b(\rho-\arctan\rho)\Bigg|_0^R = b(R-\arctan R)\]
Alltogether the \(\rho\) integral then gives
\[a\frac{R^3}{3}+b(R-\arctan R)\]
This is a constant which we can evaluate plugging in \(a,b,\) and \(R\):
\[(6\cdot 10^{12})\frac{(6000)^3}{3}+(4\cdot 10^{12})(6000-\arctan 6000)\approx 4.32\cdot 10^{23}\]
Given this constant, it is particularly easy to integrate with respect to \(\phi\) and \(\theta\). The \(\phi\) integral (after pulling this constant out) becomes
\[\int_0^\pi \sin\phi d\phi = 2\]
and the \(\theta\) integral (again, pulling the constants out) is \[\int_0^{2\pi} d\theta = 2\pi\]
Putting it all together, we have
\[M = (2\pi)(2)(4.32\cdot 10^{23})\approx 5.43\cdot 10^{24}\mathrm{kg}\]