Assignment 8
Problems
The Mass of Steel Sheets
Integrals measure the area under a curve, but appear in a much wider context than simply geometry: often this area is encoding the total amount of some physical property distributed along an interval, rectangle or surface.
One common example is mass: if you have an object of a given fixed density \(\rho\) and volume \(V\), its possible to find the mass as just the density times volume. But if you have an object whose density varies, then the total mass is the integral of its density! This may be familiar already in one dimension from Calc I/II or introductory physics/engineering, where the mass of a 1-dimensional rod from \(x=a\) to \(x=b\) is given by \[m=\int_a^b \rho(x)dx\]
Multivariate integrals bring these techniques much closer to real world application, as real-world objects are often much more than lines! In this problem we will learn how to calculate the mass of a corrugated steel sheet, the kind used in many manufacturing and engineering designs due to its good ratio of strength to weight.
The density of a flat steel sheet is constant (at 2mm thick, its 15kg per square meter). However, a corrugated sheet has a non-constant density across its surface, due to the fact that it bends in one direction (and there is more sheet, hence more steel, in the bends). The density (in kilograms / square meter) of a corrugated sheet is well modeled by the simple expression \[\rho(x,y)=15+0.5\sin^2(5\pi x)\] (The variable \(y\) does not show up in the formula as the sheet is only corrugated in one direction).
You are on a construction team that is looking to plate the outside of a new building in corrugated steel sheets. You are told that a single panel of such sheeting is a rectangle running from \(x=0\) to \(x=3\) meters, and \(y=0\) to \(y=5\) meters.
- Set up a double integral that gives the mass of this sheet. How much does it weigh?
- The building you are looking to cover in panels has an outside surface area of 800 square meters. How many panels do you need to order to complete the project? If each truck bringing in panels has a maximum capacity of one ton (1,000kg) how many truck loads will be required on construction day?
Building Roads
You are back working in the city planners office, this time consulting on a new road construction project. The planned road is to follow a relatively straight path through a hilly terrain, and to prepare for construction you have recommended bringing in a team of steam shovels and bulldozers to cut a path into the hillside so the road can run straight.
You consult a GPS company to get the elevation data for your county, and learn that the area through which the road passes has a height (in meters) \(h(x,y)\) above the bedrock: \[h(x,y)=0.01xy+0.001x^2\]
In these coordinates the road is to be approximately 6 meters wide, centered on the path \(y=3x-1\). That is, the lower edge of the road is described by \(y=3x-4\) and the upper edge of the road is described by \(y=3x+2\). You are responsible for the construction of this road between its start at \(x=0\) and its first intersection \(1\) kilometer to the east, at \(x=1000\).
The main thing the city needs to know is how much earth needs to be removed during construction. They own a fleet of dumptrucks, each of which can carry \(100m^3\) of dirt at a time, but they would like an estimate of the total number of trips this will take, as that will help plan an accurate timeline for the project.
- Set up a double integral that measures the total amount of land that needs to be removed from the current site of the road.
- Evaluate this double integral. How many truckloads should the city plan for?
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!)
Solutions
Mass of Metal Sheets
We are given the mass density \(\rho(x,y)=15+0.5\sin^2(5\pi x)\) and the size of the sheet \(x\in[0,3],y\in[0,5]\). So, finding its mass just amounts to calculating a double integral:
\[M=\iint_R \rho dA = \int_0^5\int_0^3 15+\frac{1}{2}\sin^2(5\pi x)\,dx\,dy\]
We evaluate this as an iterated integral, so we can first focus just on the \(x\) integral. This has \(\sin^2\) in it, so we use a trigonometric identity to simplify
\[\sin^2(5\pi x)=\frac{1-\cos 10\pi x}{2}\]
\[\int_0^3 15+\frac{1}{2}\frac{1-\cos (10 pi x)}{2}dx\] \[=\int_0^3 15dx+\int_0^3\frac{1}{4}(1-\cos(10\pi x))dx\] \[= 45+\frac{1}{4}\left(x-\frac{\sin(10\pi x)}{10\pi}\right)\Bigg|_0^3\] \[=45+\frac{1}{4}\left(3-\frac{0}{10\pi}\right)\] \[=45+\frac{3}{4}=45.75\]
Now, we integrate the result with respect to \(y\):
\[M=\int_0^5 45.75dy=45.75*5=228.75\mathrm{kg}\]
The building has a surface area of \(800m^2\) and each panel is \(3\times 5=15m^2\). Thus we need \(800/15= 53.33\ldots\) panels, so in reality we need 54 panels to make sure we can cover it all.
Each truck can carry \(1000kg\) total which is \[\frac{1000}{228.75}=4.37\textrm{ panels}\]
Thus, every truck can carry \(4\) panels (as 5 would be too many for its weight). We need 54 panels total, so the number of truck loads required is
\[\frac{54}{4}=13\]
Building Roads
The amount of earth to be cleared away is the double integral of the height of the ground \(h(x,y)\) over the region defined by the future road. The problem gives us \(h\), and describes the region as
\[R=\{0\leq x\leq 1000, 3x-4\leq y\leq 3x+2 \}\]
Thus, the integral which determines the volume of rock to be cleared is
\[V = \iint_R h\,dA = \int_0^{1000}\int_{3x-4}^{3x+2}0.01 xy+0.001x^2\,dydx\]
We compute this as an iterated integral, focusing first only on the innermost integral:
\[\int_{3x-4}^{3x+2}\frac{1}{100}xy+\frac{1}{1000}x^2 dy\] \[=\frac{1}{200}xy^2+\frac{1}{2000}x^2y\Bigg|_{3x-4}^{3x+2}\] \[=\frac{1}{200}x\left((3x+2)^2-(3x-4)^2\right)+\frac{1}{2000}x^2\left((3x+2)-(3x-4)\right)\]
This can be simplified with some algebra: \[=\frac{1}{200}x(36x-12)+\frac{1}{2000}x^2(6)\] \[=\frac{12}{200}(3x^2-x)+\frac{6}{2000}x^2\] \[=\frac{183}{1000}x^2-\frac{3}{50}x=0.189x^2-0.06x\]
Next, we integrate with respect to \(x\):
\[\int_0^{1000}0.189x^2-0.06x\,dx\] \[=0.183\frac{x^3}{3}-0.06\frac{x^2}{2}\Bigg|_0^{1000}\] \[=0.061 (1000)^3-0.03(1000)^2\] \[60,970,000\,m^3\]
Since each truckload can only carry 1000 cubic meters, the city should plan for \(60,970\) truckloads! No wonder road construction takes so long.
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.