Do my Calc 2 project for me /sci/! :D

Math 107 — Spring 2009 — Class Project

Project Description

You work at a gas station, and one of your jobs is to measure the amount of gas left in the
underground storage tanks. To do this, you take a stick specially designed for each of the various
tanks, dip it into a hole in the top of the tank and push it all the way to the bottom. When you
pull the stick out, it is wet with gasoline, and tick-marks on the stick inform you how many gallons
remain in the tank.

One day, the special measuring sticks are stolen. You go the hardware store, but can only find
a stick with inch marks on it. In this project, you will find a formula to compute the amount of gas
in the tanks using only the inch measuring stick.

You do know that each gas tank is a perfect cylinder, lying on its side, as pictured below. Since
the tanks vary in size, let’s consider a tank of radius R inches and length L inches. Suppose that
when you dip the stick into the tank, the wet mark on it indicates the tank is full to a depth of h
inches. (Of course, h will be somewhere between 0 and 2R.)

1. Set up and evaluate a definite integral that gives the number of gallons of gas left in the
storage tank. Be sure to explain all your work, both with words and by drawing appropriate
figures. Tip: You may find it useful to introduce coordinates for the circular cross section of
the cylinder by placing the origin at its center.

2. Now find a way to compute the same thing, using only basic geometry and trigonometry, no
calculus.

3. Show that the answers you got in the previous parts are the same.

4. You decide to mark a stick to be used for future use for one of the tanks. This particular
tank has radius 100 inches and length 400 inches. You put marks indicating 50, 100, 150, etc.
gallons. Indicate how the tick-marks are drawn on the stick.

5. Now suppose the tank is not a circular cylinder, but rather it is an elliptical cylinder. (See
the figure below.) That is, assume the cross section taken perpendicular to the central axis of
the cylinder is an ellipse whose major axis is parallel to the ground. The width of the tank
(i.e., the length of the major axis of the cross section) is 220 inches, the height of the tank
(i.e., the length of the minor axis of the cross section) is 180 inches and the length of the tank
is 400 inches. Use a definite integral to find a formula for the amount of gasoline in the tank
when the stick indicates a depth of h inches.

No.

in ur /sci/, doin ur homework

1)

Put the circle on a coordinate axis as the hint suggests. 

If h \le R slice off a sector of this circle of width h by drawing the line y = -R+h.  The equation of the circle is x^2+y^2 = R^2, which intersects the line at the points where x^2+(-R+h)^2 = R^2 or x = \pm \sqrt{2Rh-h^2}  The area of the region underneath the line and bounded by the circle is then

\int_{-\sqrt{2Rh-h^2}}^{\sqrt{2Rh-h^2}} (-R+h) - \sqrt{R^2-x^2} dx

This multiplied by the length $L$ of the cylinder gives the volume.

If h > R, the area of the sector bounded by the circle and the line y=-R+h is equal to the area \pi R^2 of the circle minus the area of the region above the line bounded by the circle.  So basically \pi R^2 subtract the same integral as above, but with 2R-h instead of h.

Cant be assed to do the rest of them atm.

\pi

Anon delivers:

1) Refer to the following picture: http://www.imagedump.com/index.cgi?pick=get&tp=547398

The volume is the length of the tank times the area of the yellow shaded area of the circular cross section.  To find the area, we're going to integrate the length of a horizontal slice running vertically from the bottom of the tank to the top of the water.  The length marked as "dist" is \sqrt{R^2-(R-y)^2}, so we have to integrate twice this distance from 0 to h, and multiply by L.  The formula for the volume is:

2L \int_0^h \sqrt{R^2-(R-y)^2} dy

Substitute R-y = R \sin \theta, dy = -R\cos \theta d\theta, so that \theta = \sin^{-1} \frac{R-y}{R}

2L \int_{\pi/2}^{\sin^{-1} \frac{R-h}{R}} \sqrt{R^2-(R \sin \theta)^2} (-R\cos \theta) d\theta = -2LR^2 \int_{\pi/2}^{\sin^{-1} \frac{R-h}{R}} \cos^2 \theta d\theta
= -2LR^2 \int_{\pi/2}^{\sin^{-1} \frac{R-h}{R}} \frac{1+\cos 2\theta}{2} d\theta
= -2LR^2 \left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1} \frac{R-h}{R}}
= -LR^2 \sin^{-1} \frac{R-h}{R} - LR^2 \sin \sin^{-1} \frac{R-h}{R} \cos \sin^{-1} \frac{R-h}{R}
= -LR^2 \sin^{-1} \frac{R-h}{R} - LR^2 \frac{R-h}{R} \sqrt{1-\left(\frac{R-h}{R}\right)^2}
= LR^2\frac{\pi}{2}-LR^2 \sin^{-1} \frac{R-h}{R} - L (R-h) \sqrt{R^2-(R-h)^2}


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The rest of the answers will be posted tonight and/or tomorrow.

UNL SUX

THE TWO GUYS SITTING ACROSS THE COMPUTER LAB FROM ME RIGHT NOW ARE FAGS

I LOVE ASHLEY S.

Problem 2 and 3)  Refer to this picture:  http://www.imagedump.com/index.cgi?pick=get&tp=547413


Again we want the area of the yellow section.  Assume first that h<R, so the water level is below the horizontal line through the center of the circle.  The area of the yellow region is equal to half the area of the circle (i.e. \frac{\pi R^2}{2}) minus the area of the pink and blue areas.

The angle 90-x is equal to \cos^{-1} \frac{R-h}{R}, so the angle x is equal to \frac{\pi}{2} - \cos^{-1} \frac{R-h}{R} = \sin^{-1} \frac{R-h}{R}.  This shows that the area of the two pink sectors is equal to

2 \left[ \frac{\sin^{-1} \frac{R-h}{R}}{2\pi} \cdot \pi R^2\right] =  R^2 \sin^{-1} \frac{R-h}{R}.

By the formula for the area of triangles, the combined area of the two blue triangles is

2 \left[\frac{1}{2} (R-h)\sqrt{R^2-(R-h)^2}\right] = (R-h)\sqrt{R^2-(R-h)^2}

Subtracting these two areas from half the area of the circle, and multiplying by L, we find the exact same formula for the volume as we did in problem 1.

Now if h > R and the water level is above midway, the picture is a little different, and you're going to be *adding* an area which is above the midline to the area of the bottom half of the circle.  It's almost completely similar to what I just did, and I'll let you work it out.



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I MISS YOU ASHLEY!!! D:

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OOPSIE:  The problem asks for volume in gallons, but the values R,h are given in inches.  So the formulas above give values in cubic inches, not in gallons.  So since there are 231 cubic inches in a gallon, the formula derived in parts 1 and 2 must be divided by 231.

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Problem 4)  Setting R = 100, L=400, the formula for the volume becomes

V(h) = \frac{1}{231}\left[2000000 \pi-4000000 \sin^{-1} \left(1-\frac{h}{100}\right) - (40000-400h) \sqrt{200h-h^2}\right]
= 27199.94-17316.02 \sin^{-1} \left(1-\frac{h}{100}\right) - (173.16-1.73h)\sqrt{200h-h^2}

We want to know the values of h such that V(h) = 50, V(h)=100, etc.  The total capacity of the tank is equal to the volume of the tank divided by 231, which is \pi * 100^2 * 400 / 231 = 54399.87, so we don't need to actually make a list here.  The process we would go through, though, is to numerically (i.e. with a computer) find the value of h such that V(h)=50, and then make a mark on the stick h inches from the end, with "50" next to it, and repeat for all multiples of 50 from 0 to 54,400. (what a pain in the ass :P)

I'll do problem 5 tomorrow.

YOU LOSERS BETTER START WRITING THIS SHIT UP NOW.  YOU'VE ONLY GOT LIKE TWO WEEKS, AMIRITE?

EDIT: In problem #1 in the last integral, I forgot to put the LR^2\frac{\pi}{2} term in the second to last and third to last lines.  It's in the last line, as it should be.

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Problem 5)

Here's the picture: http://www.imagedump.com/index.cgi?pick=get&tp=547449

We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated.  Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture.  If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes.  If not, there's always wikipedia.  Anyway, the equation of this ellipse turns out to be

\left(\frac{x}{110}\right)^2 + \left(\frac{y}{90}\right)^2 = 1

Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices.  Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation

\left(\frac{x}{110}\right)^2 + \left(\frac{y-90}{90}\right)^2 = 1

If we solve for x, we get

x = 110 \sqrt{1-\left(\frac{y}{90}-1\right)^2}

and the length of the line we want to integrate is twice this.  Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes

\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
=\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy

Substituting 1-\frac{y}{90} = \sin \theta, dy = -90 \cos \theta d\theta, gives

=\frac{880000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \sqrt{1-\sin^2 \theta} (-90 \cos \theta) d\theta
=-\frac{7920000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \cos^2 \theta d\theta
=-\frac{7920000}{231}\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} (we did the integral of \cos^2 \theta in problem 1)
=-\frac{7920000}{231} \left[\frac{\sin^{-1}\left(1-\frac{h}{90}\right)}{2} - \frac{\pi}{4} + 2 \sin \sin^{-1}\left(1-\frac{h}{90}\right) \cos \sin^{-1}\left(1-\frac{h}{90}\right)\right]
=-\frac{7920000}{231} \left[\frac{\sin^{-1}\left(1-\frac{h}{90}\right)}{2} - \frac{\pi}{4} + 2 \left(1-\frac{h}{90}\right) \sqrt{1-\left(1-\frac{h}{90}\right)^2}\right]
Problem 5)

Here's the picture:

We're going to do basically the same thing as in problem 1, except now it's an ellipse, so it's more complicated.  Put the cross-section on an x-y coordinate axis, and set up the ellipse as in the picture.  If you look in your book, there should be something somewhere about how to find the equation of an ellipse given the major and minor axes.  If not, there's always wikipedia.  Anyway, the equation of this ellipse turns out to be

\left(\frac{x}{110}\right)^2 + \left(\frac{y}{90}\right)^2 = 1

Just like in problem 1, we're going to set up an integral over y from 0 to h, and integrate the length of the horizontal slices.  Looking at the picture, the point (x,-90+y) lies on the ellipse, and so satisfies the equation

\left(\frac{x}{110}\right)^2 + \left(\frac{y-90}{90}\right)^2 = 1

If we solve for x, we get

x = 110 \sqrt{1-\left(\frac{y}{90}-1\right)^2}

and the length of the line we want to integrate is twice this.  Then our integral for the volume, including the factor of 400 for the length and dividing by 231 to convert to gallons, becomes

\frac{400}{231}\int_0^h 220\sqrt{1-\left(\frac{y}{90}-1\right)^2} dy
=\frac{88000}{231}\int_0^h \sqrt{1-\left(\frac{y}{90}-1\right)^2} dy

Substituting 1-\frac{y}{90} = \sin \theta, dy = -90 \cos \theta d\theta, gives

=\frac{880000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \sqrt{1-\sin^2 \theta} (-90 \cos \theta) d\theta
=-\frac{7920000}{231}\int_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} \cos^2 \theta d\theta
=-\frac{7920000}{231}\left[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_{\pi/2}^{\sin^{-1}\left(1-\frac{h}{90}\right)} (we did the integral of \cos^2 \theta in problem 1)

=-\frac{7920000}{231} \left[\frac{\sin^{-1}\left(1-\frac{h}{90}\right)}{2} - \frac{\pi}{4} + 2 \sin \sin^{-1}\left(1-\frac{h}{90}\right) \cos \sin^{-1}\left(1-\frac{h}{90}\right)\right]
=-\frac{3960000}{231} \sin^{-1}\left(1-\frac{h}{90}\right) + \frac{1980000 \pi}{231} - \frac{1600}{189} \left(90-h\right) \sqrt{90-\left(90-h\right)^2}
=-\frac{3960000}{231} \sin^{-1}\left(1-\frac{h}{90}\right) + \frac{1980000 \pi}{231} - \frac{1600}{189} \left(90-h\right) \sqrt{180h-h^2}
=26927.937 - 17142.857 \sin^{-1}\left(1-\frac{h}{90}\right) - 8.466 \sqrt{180h-h^2}

To check that this answer is right, we can find the capacity of the entire tank by plugging in h=180, which gives 53,855 gallons.  Remember that in problem 3 we found that if the tank was circular with radius 100, it held 54,400 gallons.  Since an ellipse with major and minor axes of 90 and 110 isn't much different than a circle with radius 100, it makes sense that these two numbers should be close.

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YOUR WELCOME YOU UNGRATEFUL PRICKS.

Fuck, double post. (why the hell can't you delete your posts on this board??) From the part where it says "Problem 5)" for the second time, skip down to the third to last equation near the bottom and continue from there.