Finding the Volume of a Partially Filled Cylinder

[mario99]

A cylindrical oil tank with radius 2 feet is lying on its side. A measuring stick shows that the oil is 1.8 feet deep. What percentage of a full tank is left?

There is nothing better than using a cylindrical coordinate when working with a cylinder.

Let [imath]V[/imath] be the full volume of the tank and [imath]V_p[/imath] be the partial volume.

[imath]\displaystyle V = \int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{h} r \ dz \ dr \ d\theta = h\int_{0}^{2\pi}\int_{0}^{2} r \ dr \ d\theta = 2h\int_{0}^{2\pi} d\theta = 4\pi h[/imath]


[imath]\displaystyle V_p = \frac{1}{2}\int_{0}^{2\pi}\int_{0.2}^{2}\int_{0}^{h} r \ dz \ dr \ d\theta = \frac{1}{2}h\int_{0}^{2\pi}\int_{0.2}^{2} r \ dr \ d\theta = \frac{1}{4}h(2^2 - 0.2^2)\int_{0}^{2\pi} d\theta = \frac{1}{2}\pi h(2^2 - 0.2^2) = 1.98\pi h[/imath]


[imath]\displaystyle \frac{V_p}{V} \times 100 = \frac{1.98\pi h}{4\pi h} \times 100 = 49.5[/imath]% (It says [imath]43.6[/imath]%)

I am sure that my calculations are so precise. Why am I getting a wrong answer?

I don't want to work in the Cartesian coordinate because it will increase the difficulty of the problem.

 

[topsquark]

A cylindrical oil tank with radius 2 feet is lying on its side. A measuring stick shows that the oil is 1.8 feet deep. What percentage of a full tank is left?

There is nothing better than using a cylindrical coordinate when working with a cylinder.

Let [imath]V[/imath] be the full volume of the tank and [imath]V_p[/imath] be the partial volume.

[imath]\displaystyle V = \int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{h} r \ dz \ dr \ d\theta = h\int_{0}^{2\pi}\int_{0}^{2} r \ dr \ d\theta = 2h\int_{0}^{2\pi} d\theta = 4\pi h[/imath]


[imath]\displaystyle V_p = \frac{1}{2}\int_{0}^{2\pi}\int_{0.2}^{2}\int_{0}^{h} r \ dz \ dr \ d\theta = \frac{1}{2}h\int_{0}^{2\pi}\int_{0.2}^{2} r \ dr \ d\theta = \frac{1}{4}h(2^2 - 0.2^2)\int_{0}^{2\pi} d\theta = \frac{1}{2}\pi h(2^2 - 0.2^2) = 1.98\pi h[/imath]


[imath]\displaystyle \frac{V_p}{V} \times 100 = \frac{1.98\pi h}{4\pi h} \times 100 = 49.5[/imath]% (It says [imath]43.6[/imath]%)

I am sure that my calculations are so precise. Why am I getting a wrong answer?

I don't want to work in the Cartesian coordinate because it will increase the difficulty of the problem.

You aren't using the right volume. Look at your angular variable: you are saying you are integrating over the whole circle.

Shetch a diagram. The cylinder is without a complete circle so any cylindrical symmetry is gone.

The simplest way is to put the axis of the cylinder along the z axis, but where the z axis is horizontal. Then you want to integrate over a volume that has an area of a circle with the top horizontally cut off at h = 1.8 ft and a width dz.

The area of the cut off circle can be found from simple geometry, so cylindrical coordinate integration is actually a more difficult way to do it.

Sketch a picture of the volume again and see what you can do with this. If you are still having problems, feel free to post your new attempt and we'll give you more hints.

-Dan

 

[Dr.Peterson]

[imath]\displaystyle V_p = \frac{1}{2}\int_{0}^{2\pi}\int_{0.2}^{2}\int_{0}^{h} r \ dz \ dr \ d\theta = \frac{1}{2}h\int_{0}^{2\pi}\int_{0.2}^{2} r \ dr \ d\theta = \frac{1}{4}h(2^2 - 0.2^2)\int_{0}^{2\pi} d\theta = \frac{1}{2}\pi h(2^2 - 0.2^2) = 1.98\pi h[/imath]

This integral gives the volume of a cylindrical shell, with inner radius 0.2 and outer radius 2. That clearly is not what the problem is asking for.

Did the problem say you had to use integration? If you do choose to use it, you'll need to integrate separately over two different intervals of theta.

I am sure that my calculations are so precise. Why am I getting a wrong answer?

Have you learned the difference between precision and accuracy?

 

[khansaheb]

I am sure that my calculations are so precise.

Precision comes from repeated "action". How many times did you try to calculate this volume - and made precisely the same mistake?

 

[khansaheb]

A cylindrical oil tank with radius 2 feet is lying on its side. A measuring stick shows that the oil is 1.8 feet deep. What percentage of a full tank is left?

There is nothing better than using a cylindrical coordinate when working with a cylinder.

Let [imath]V[/imath] be the full volume of the tank and [imath]V_p[/imath] be the partial volume.

[imath]\displaystyle V = \int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{h} r \ dz \ dr \ d\theta = h\int_{0}^{2\pi}\int_{0}^{2} r \ dr \ d\theta = 2h\int_{0}^{2\pi} d\theta = 4\pi h[/imath]


[imath]\displaystyle V_p = \frac{1}{2}\int_{0}^{2\pi}\int_{0.2}^{2}\int_{0}^{h} r \ dz \ dr \ d\theta = \frac{1}{2}h\int_{0}^{2\pi}\int_{0.2}^{2} r \ dr \ d\theta = \frac{1}{4}h(2^2 - 0.2^2)\int_{0}^{2\pi} d\theta = \frac{1}{2}\pi h(2^2 - 0.2^2) = 1.98\pi h[/imath]


[imath]\displaystyle \frac{V_p}{V} \times 100 = \frac{1.98\pi h}{4\pi h} \times 100 = 49.5[/imath]% (It says [imath]43.6[/imath]%)

I am sure that my calculations are so precise. Why am I getting a wrong answer?

I don't want to work in the Cartesian coordinate because it will increase the difficulty of the problem.

The 'elementary' volume for integration should be in the shape of ~1/2 moon . Draw a sketch first.

 

[mario99]

You aren't using the right volume. Look at your angular variable: you are saying you are integrating over the whole circle.

Shetch a diagram. The cylinder is without a complete circle so any cylindrical symmetry is gone.

The simplest way is to put the axis of the cylinder along the z axis, but where the z axis is horizontal. Then you want to integrate over a volume that has an area of a circle with the top horizontally cut off at h = 1.8 ft and a width dz.

The area of the cut off circle can be found from simple geometry, so cylindrical coordinate integration is actually a more difficult way to do it.

Sketch a picture of the volume again and see what you can do with this. If you are still having problems, feel free to post your new attempt and we'll give you more hints.

-Dan

You gave me two ideas. I will use them below.

It is funny and awkward that a cylinder is more difficult in a Cylindrical coordinate. I have thought the opposite is always true.

This integral gives the volume of a cylindrical shell, with inner radius 0.2 and outer radius 2. That clearly is not what the problem is asking for.

Did the problem say you had to use integration? If you do choose to use it, you'll need to integrate separately over two different intervals of theta.

This problem is related to Polar coordinate. Since it is a volume, doesn't it mean we should work in a Cylindrical coordinate?

Have you learned the difference between precision and accuracy?

Yes. They even may have the same meaning. It all depends on the context.

Precision comes from repeated "action". How many times did you try to calculate this volume - and made precisely the same mistake?

Not always. It depends on the context. In my case it means correct.


My first idea to work in this partial volume is that I use half of a circle and a rectangle.

[imath]\displaystyle A_R: \ [/imath] Rectangle area.

[imath]\displaystyle A_{HC}: \ [/imath] Half circle area.

[imath]\displaystyle h: \ [/imath] Height of the cylinder.


[imath]\displaystyle A_R = 2 \times 0.2[/imath]

[imath]\displaystyle A_{HC} = \frac{\pi 2^2}{2}[/imath]

Then, the partial volume is just

[imath]\displaystyle V_p = h(A_{HC} - A_R)[/imath]

The second idea is to work on a vertical circle ignoring the volume. First, setting the integral in the Cartesian coordinate then move it to the polar one.

[imath]\displaystyle V_p = \int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{-0.2} \ \ dz \ dx[/imath]

[imath]\displaystyle V_p = \int_{-\pi}^{0} \int_{0}^{2} \ \ r \ dr \ d\theta[/imath]


What do you think

 

[Dr.Peterson]

It is funny and awkward that a cylinder is more difficult in a Cylindrical coordinate. I have thought the opposite is always true.

The cylinder is easy in cylindrical coordinates. But this is not all of a cylinder; the surface of the liquid is a plane cutting the cylinder, and that part is harder.

This problem is related to Polar coordinate. Since it is a volume, doesn't it mean we should work in a Cylindrical coordinate?

I didn't say not to. You still can. (I would probably try that first.)

But volumes can be worked out in whatever system you choose.

When you say "related to polar coordinates", are you saying it is in a section about that, or you were told to use that, or just that that feels appropriate to you.

Yes. They even may have the same meaning. It all depends on the context.

This was just a side comment, but did you look at my link? I was making a specific point, which you seem to have missed: You can be very precise, but be precisely wrong! Your integral is wrong, so it doesn't matter how carefully you integrated it.

My first idea to work in this partial volume is that I use half of a circle and a rectangle.

A picture (which I asked for) would be really helpful! Please draw one and give us an image of it.

What half circle? What rectangle? How do they relate to the problem?

 

[limiTS]

[imath]\displaystyle V_p=\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{-0.2} dz\,dx[/imath]

Your Cartesian double integral's [imath]x[/imath] would go from -2 to 2 if the oil in the cylinder were 2 feet deep, because that would be the diameter of the circular cross-section. BUT, the oil is only 1.8 feet deep. That means the [imath]x[/imath] goes from [imath]-\sqrt{2^2-0.2^2}[/imath] to [imath]+\sqrt{2^2-0.2^2}[/imath]

To get the right area, integrate:

[imath]\displaystyle\int_{x=-\sqrt{2^2-0.2^2}}^{\sqrt{2^2-0.2^2}}\int_{z=-\sqrt{2^2-x^2}}^{-0.2} dz\,dx[/imath]

However, I think it's much easier to find the above area using geometry.

 

[mario99]

The 'elementary' volume for integration should be in the shape of ~1/2 moon . Draw a sketch first.

But I am not a beginner.

When you say "related to polar coordinates", are you saying it is in a section about that, or you were told to use that, or just that that feels appropriate to you.

The problem is related to the section of Polar Coordinate.

This was just a side comment, but did you look at my link? I was making a specific point, which you seem to have missed:

I didn't even realize it was a link until you mentioned it. I am looking now. So I have to be both accurate and precise or precisely accurate. Got the point.

A picture (which I asked for) would be really helpful! Please draw one and give us an image of it.

My math level is not beginner, so why do I need a picture, a sketch, or a drawing? Don't take this as oh Mario is always ignoring suggestions and don't listen to the seniors. I didn't mean to do that, I just don't understand why a High Level Math noob needs a sketch?

In fact, the sketch is drawn inside my mind. Trust me the whole picture of the cylindrical oil tank is inside my brain completely.

What half circle? What rectangle? How do they relate to the problem?

The two ends of the cylinder are circles. If an observer is looking at one end of the cylinder (assuming the oil can be seen from outside the cylinder.), he will see that the oil is filling less than half the circle. That missing oil which fills the rest of the half circle is a rectangle. If we know the area of that rectangle which is supposed to be the easiest part, the problem is solved. Although, this method looks logical, it produces wrong results.

Your Cartesian double integral's [imath]x[/imath] would go from -2 to 2 if the oil in the cylinder were 2 feet deep, because that would be the diameter of the circular cross-section. BUT, the oil is only 1.8 feet deep. That means the [imath]x[/imath] goes from [imath]-\sqrt{2^2-0.2^2}[/imath] to [imath]+\sqrt{2^2-0.2^2}[/imath]

To get the right area, integrate:

[imath]\displaystyle\int_{x=-\sqrt{2^2-0.2^2}}^{\sqrt{2^2-0.2^2}}\int_{z=-\sqrt{2^2-x^2}}^{-0.2} dz\,dx[/imath]

However, I think it's much easier to find the above area using geometry.

You are so right. I missed that part. Shame on me.

The Polar version of the integral should be,

[imath]\displaystyle V_p = \int_{-\pi}^{0} \int_{0}^{\sqrt{2^2-0.2^2}} \ \ r \ dr \ d\theta[/imath]


It still produces a wrong result.

 

[Dr.Peterson]

The problem is related to the section of Polar Coordinate.

That can be a hint as to what method you are expected to use. But it doesn't matter; as I said, it's a reasonable way to work the problem. Just don't expect it to be automatically easy, as you seemed to expect.

My math level is not beginner, so why do I need a picture, a sketch, or a drawing? Don't take this as oh Mario is always ignoring suggestions and don't listen to the seniors. I didn't mean to do that, I just don't understand why a High Level Math noob needs a sketch?

In fact, the sketch is drawn inside my mind. Trust me the whole picture of the cylindrical oil tank is inside my brain completely.

I draw sketches, and I have lots of experience! I don't trust mental images to hold all the necessary information; a drawing lets me label the variables so I don't forget which is which, or make wrong assumptions. (That's part of the humility I strongly recommend to students: distrust yourself enough to check every step you write -- and write more than you think you'll need, because what you don't write can easily be wrong.)

So, no, I don't trust you, any more than I trust myself to get everything right.

In particular, your story about a rectangle and a half circle is easily seen to be utterly wrong if you draw a picture. But most important at the moment, we can't see inside your head; a picture of what you have in mind is important for communicating. If you really want help, you need to be willing to do that.

The two ends of the cylinder are circles. If an observer is looking at one end of the cylinder (assuming the oil can be seen from outside the cylinder.), he will see that the oil is filling less than half the circle. That missing oil which fills the rest of the half circle is a rectangle. If we know the area of that rectangle which is supposed to be the easiest part, the problem is solved. Although, this method looks logical, it produces wrong results.

No, that doesn't make sense. The missing oil in the side view forms a segment of a circle, just as the oil itself does. There is no rectangle except the surface of the oil, which is not relevant to the volume.

So you are either visualizing something incorrectly, or not saying what you mean. (Since neither of your integrals is the area/volume of a rectangle, so I'm guessing you didn't really mean rectangle.)

Again, a labeled picture would help us be sure what your work means, so we could point out the error.

The Polar version of the integral should be,

[imath]\displaystyle V_p = \int_{-\pi}^{0} \int_{0}^{\sqrt{2^2-0.2^2}} \ \ r \ dr \ d\theta[/imath]

It still produces a wrong result.

Make a sketch of what region this integral actually covers, and perhaps you can catch you own error.

 

[limiTS]

The Polar version of the integral should be,

[imath]\displaystyle V_p = \int_{-\pi}^{0} \int_{0}^{\sqrt{2^2-0.2^2}} \ \ r \ dr \ d\theta[/imath]


It still produces a wrong result.

Something got lost in translation into polar.

If you flip your mental picture upside down so that the oil is contained between 1) the horizontal line [imath]y=0.2[/imath], which in polar is [imath]r=0.2\csc\theta[/imath], and 2) the edge of the circle [imath]r=2[/imath], then the polar double integral would be:

[imath]\displaystyle\int_{\theta=\arcsin\left(\frac{0.2}{2}\right)}^{\pi-\arcsin\left(\frac{0.2}{2}\right)}\int_{r=0.2\csc\theta}^{2}r\,dr\,d\theta\approx5.48452[/imath]

 

[mario99]

That can be a hint as to what method you are expected to use. But it doesn't matter; as I said, it's a reasonable way to work the problem. Just don't expect it to be automatically easy, as you seemed to expect.

I draw sketches, and I have lots of experience! I don't trust mental images to hold all the necessary information; a drawing lets me label the variables so I don't forget which is which, or make wrong assumptions. (That's part of the humility I strongly recommend to students: distrust yourself enough to check every step you write -- and write more than you think you'll need, because what you don't write can easily be wrong.)

So, no, I don't trust you, any more than I trust myself to get everything right.

In particular, your story about a rectangle and a half circle is easily seen to be utterly wrong if you draw a picture. But most important at the moment, we can't see inside your head; a picture of what you have in mind is important for communicating. If you really want help, you need to be willing to do that.

No, that doesn't make sense. The missing oil in the side view forms a segment of a circle, just as the oil itself does. There is no rectangle except the surface of the oil, which is not relevant to the volume.

So you are either visualizing something incorrectly, or not saying what you mean. (Since neither of your integrals is the area/volume of a rectangle, so I'm guessing you didn't really mean rectangle.)

Again, a labeled picture would help us be sure what your work means, so we could point out the error.


Make a sketch of what region this integral actually covers, and perhaps you can catch you own error.

I don't want to argue with you about an unnecessary sketch, instead, I will be a gentleman and will provide three pictures. But I am not an artist and you have been warned.



Above is the main idea of the problem. As you can see in the picture, the blue line is dividing the cylinder and the circle into two equal halves where below it resides the oil.




Above is my idea about the rectangle. Its width is slightly curvy, but it is still a rectangle.




The above picture shows you how the Polar integral was created. Final segments of the differential element [imath]dr[/imath] (orange is its path) moved from the center till hitting the circumference of the circle ([imath]\theta = 0[/imath]). Before that, the differential element [imath]d\theta[/imath] (light blue is its path) starts its journey from [imath]-\pi[/imath] to [imath]0[/imath], with each [imath]r \ d\theta \times dr[/imath] all the required area of the oil is covered.

Something got lost in translation into polar.

If you flip your mental picture upside down so that the oil is contained between 1) the horizontal line [imath]y=0.2[/imath], which in polar is [imath]r=0.2\csc\theta[/imath], and 2) the edge of the circle [imath]r=2[/imath], then the polar double integral would be:

[imath]\displaystyle\int_{\theta=\arcsin\left(\frac{0.2}{2}\right)}^{\pi-\arcsin\left(\frac{0.2}{2}\right)}\int_{r=0.2\csc\theta}^{2}r\,dr\,d\theta\approx5.48452[/imath]

Thanks limiTS.

I will study your integral and I will get back to you.

 

[Dr.Peterson]

I don't want to argue with you about an unnecessary sketch, instead, I will be a gentleman and will provide three pictures. But I am not an artist and you have been warned.

Artistry is not a requirement for communication. What I was hoping for, as I mentioned in the end, was a labelled drawing, to show what the variables are, and so on.

Thanks for your condescension.

Above is my idea about the rectangle. Its width is slightly curvy, but it is still a rectangle.

Well, no, it's not a rectangle. It's something you're choosing to call a rectangle. But now, because of the picture, I have an idea what you meant by that word. This is why I said a picture was necessary for communication.

The above picture shows you how the Polar integral was created. Final segments of the differential element dr (orange is its path) moved from the center till hitting the circumference of the circle (θ=0). Before that, the differential element dθ (light blue is its path) starts its journey from −π to 0, with each r dθ×dr all the required area of the oil is covered.

There are much clearer ways to show what dθ and dr are, as shown in typical textbooks; they don't really "move along those paths". That may be where you are having trouble making the right integral. Here are two random examples: this link and this link. Those are the best I could find in a quick search, but I may find more tomorrow.

Hopefully as you reverse-engineer @limiTS' integral, you will see how it actually works.

 

[mario99]

Thanks limiTS.

I will study your integral and I will get back to you.

I have studied your integral. I think that I understand now where was my mistake. By looking at the engineering of your integral and the websites suggested by Dr.Peterson, I caught my miss. The differential element [imath]dr[/imath] starts from the origin (center) of the circle while the differential element [imath]d\theta[/imath], goes around the origin (center) of the circle. I think that I know this Polar coordinate behavior before, but I mixed it a little bit with the Cartesian version.

Precision comes from repeated "action". How many times did you try to calculate this volume - and made precisely the same mistake?

And I went too far to do the integral so precisely wrong.

Thanks for your condescension.

You're welcome Dr.

Well, no, it's not a rectangle. It's something you're choosing to call a rectangle. But now, because of the picture, I have an idea what you meant by that word. This is why I said a picture was necessary for communication.

So, my assumption has failed and now I understand why. Thanks.

Artistry is not a requirement for communication. What I was hoping for, as I mentioned in the end, was a labelled drawing, to show what the variables are, and so on.

Thanks for your condescension.


Well, no, it's not a rectangle. It's something you're choosing to call a rectangle. But now, because of the picture, I have an idea what you meant by that word. This is why I said a picture was necessary for communication.

There are much clearer ways to show what dθ and dr are, as shown in typical textbooks; they don't really "move along those paths". That may be where you are having trouble making the right integral. Here are two random examples: this link and this link. Those are the best I could find in a quick search, but I may find more tomorrow.

Hopefully as you reverse-engineer @limiTS' integral, you will see how it actually works.

I have looked at the websites. They're helpful. You don't need to send more.

That genius limiTS has flipped my oil tank upside down, so he can get rid of the annoying negative signs (this is just my assumption). I admit that he has done a brilliant job, but I insisted to get the same result without flipping.

Here is my cooking.

[imath]\displaystyle \int_{-\pi + \sin^{-1}(0.2/2)}^{-\sin^{-1}(0.2/2)}\int_{0.2\csc\theta}^{2}r\,dr\,d\theta\approx5.48452[/imath]


[imath]\displaystyle \frac{V_p}{V} = \frac{5.48452}{4\pi} \times 100 \approx 43.6[/imath]%

You aren't using the right volume. Look at your angular variable: you are saying you are integrating over the whole circle.

Shetch a diagram. The cylinder is without a complete circle so any cylindrical symmetry is gone.

The simplest way is to put the axis of the cylinder along the z axis, but where the z axis is horizontal. Then you want to integrate over a volume that has an area of a circle with the top horizontally cut off at h = 1.8 ft and a width dz.

The area of the cut off circle can be found from simple geometry, so cylindrical coordinate integration is actually a more difficult way to do it.

Sketch a picture of the volume again and see what you can do with this. If you are still having problems, feel free to post your new attempt and we'll give you more hints.

-Dan

If I have listened carefully to what Dan said in the first post, I would have quickly fixed my error. His genius idea of starting the integral with a circle instead of a volume was the start of igniting the spark. I hope that he will not see this post as he was right in other posts when he said "You think that you know something in Math, but really you know nothing".

I thought that I was a very professional noob in finding areas and volumes, but this problem taught me that whatever the quantity of math I know, I know zero math.

FYI: I could have done this problem in the cartesian coordinate correctly, but the question was purely in the Polar coordinate. When I failed, I had to try. I had to understand where was my mistake because if this simple volume problem have beaten the Airy equation guy, it would be shame on Mario to open a math book again.

I would like to thank Dan, Dr.Peterson, khansaheb, and limiTS for contributing in this problem. And Steven for not contributing in this problem.

 

[topsquark]

If I have listened carefully to what Dan said in the first post, I would have quickly fixed my error. His genius idea of starting the integral with a circle instead of a volume was the start of igniting the spark. I hope that he will not see this post as he was right in other posts when he said "You think that you know something in Math, but really you know nothing".

You "do not know nothing" but in your push to keep moving forward you are missing a lot more than you seem to think that you are. As I've repeatedly told my Intro level Physics students, there are probably problems at the end of the chapters that can take me. (Not many, but I occasionally run into one that will take even me more than a week.). What do I do about that? I don't say, "Hey! I'm the guy that does QFT, so it doesn't matter.". No, I go back and review where the problem was. If I need to take some time out of my research to learn something I missed, or relearn something I have forgotten them that's what I need to do. I like doing that, as it improves my understanding of Physics.

There is no shame in making a mistake. The shame is when you refuse to fix it. As I've told you many many times in the past: you need to go back to (at least) Calculus I and do a stiff review: you are missing basic concepts and are making mistakes on a level no one taking PDEqs should be making.

Stop pushing forward so fast and go back to the basics. It won't take you forever to do so, just a little more time that will be well spent.

-Dan

Addendum: Problem solving skills should probably be added to your list to review. The net is overflowing with suggestions. The following comments don't always work, but they help to give you overall structure. I'm sure you do at least some of this, but you might find something you can add.

1. Always draw a diagram and label all information you can gather from the problem statement. This adds a visual context, which your mind processes differently than logic or verbal.

2. Make sure to have your references handy, so that you don't have to get up in the middle of a problem and find a book.

3. If you are working on a topic, possibly even put a bookmark in certain sections of the chapter the problem comes from. Odds are that a text will give you some clue as to how to proceed.

4. When you ask questions, give your tutor everything that you have: the full problem statement and any work that you've done, even if that work is wrong. The less work you make your tutor do, the better they will be able to help you, and the faster they might be able to spot where you may have taken a wrong turn.

This one is specifically for you, though I've seen it in many other students:
5. Put your ego on the shelf: it only gets in the way. I don't really care that you have solved the Airy equation, that doesn't have anything to do with the problem at hand. Your pride is getting in the way of listening to what we are suggesting you do to solve the problem. It's not just the complaint about drawing a diagram from this thread; you have repeatedly ignored advice about how to do a problem because you insist that it can be done your way. Maybe it actually can't be done your way? Maybe there's a reason we are making the suggestion? Give it full consideration. After all, you came to us for help!

6. When you do your work, write out every step, no matter how small. It'much easier to catch your mistakes that way.

7. If you see that are having a specific difficulty during the solution, stop what you are doing and review the section or method that is troubling you, not just what immediately pertains to what you think you are having a problem with. You never know, the solution to your difficulty might be something that you haven't considered yet, not just that you might be screwing up the calculation somehow.

8. Try to keep it simple. (I admit that I have issues with this one. If there's a complicated approach to a problem, trust me, that's my first instinct!). The more involved your logic chain, the harder it will be to apply the concepts. Your impulse here was to write an integral. Certainly this can be (and was) done. It's good to know how to do it this way. But it's a lot simpler to just use some basic circle geometry.

 

[khansaheb]

I have studied your integral. I think that I understand now where was my mistake. By looking at the engineering of your integral and the websites suggested by Dr.Peterson, I caught my miss. The differential element [imath]dr[/imath] starts from the origin (center) of the circle while the differential element [imath]d\theta[/imath], goes around the origin (center) of the circle. I think that I know this Polar coordinate behavior before, but I mixed it a little bit with the Cartesian version.



And I went too far to do the integral so precisely wrong.



You're welcome Dr.



So, my assumption has failed and now I understand why. Thanks.



I have looked at the websites. They're helpful. You don't need to send more.

That genius limiTS has flipped my oil tank upside down, so he can get rid of the annoying negative signs (this is just my assumption). I admit that he has done a brilliant job, but I insisted to get the same result without flipping.

Here is my cooking.

[imath]\displaystyle \int_{-\pi + \sin^{-1}(0.2/2)}^{-\sin^{-1}(0.2/2)}\int_{0.2\csc\theta}^{2}r\,dr\,d\theta\approx5.48452[/imath]


[imath]\displaystyle \frac{V_p}{V} = \frac{5.48452}{4\pi} \times 100 \approx 43.6[/imath]%




If I have listened carefully to what Dan said in the first post, I would have quickly fixed my error. His genius idea of starting the integral with a circle instead of a volume was the start of igniting the spark. I hope that he will not see this post as he was right in other posts when he said "You think that you know something in Math, but really you know nothing".

I thought that I was a very professional noob in finding areas and volumes, but this problem taught me that whatever the quantity of math I know, I know zero math.

FYI: I could have done this problem in the cartesian coordinate correctly, but the question was purely in the Polar coordinate. When I failed, I had to try. I had to understand where was my mistake because if this simple volume problem have beaten the Airy equation guy, it would be shame on Mario to open a math book again.

I would like to thank Dan, Dr.Peterson, khansaheb, and limiTS for contributing in this problem. And Steven for not contributing in this problem.

To recap:
Assuming:

What percentage of a full tank is left? - means the volume of oil is left​

i) calculate the AREA of the "maroon" part (preferably using polar co-ordinates). - call it A

ii) Calculate the volume by MULTIPLYING the length (L) of the tube (V = A * L)

iii) Calculate ratio [ = V/(pi*r² * L) = A/(pi * r²)] .....Express as % ..... the second step was NOT necessary

Yes. They even may have the same meaning. It all depends on the context.

To a scientist, those DO NOT have the same meaning

But I am not a beginner.

You must have realized by this time that sketching is a tool - just like "integration". This tool is specially used by experts - to avoid mistakes.

 

[khansaheb]

Problem solving skills should probably be added to your list to review.

An excellent reference would be:

How to solve it - G. Polya​

 

[khansaheb]

A cylindrical oil tank with radius 2 feet is lying on its side. A measuring stick shows that the oil is 1.8 feet deep. What percentage of a full tank is left?

If the same tank is lying on hill with slope \(\displaystyle \theta\) - the complexity of the problem increases many-fold. Of course the problem has to be restated - so that the meaning of the dip-stick measurement is clarified.

 

[khansaheb]

And Steven for not contributing in this problem.

This is extremely rude. I would have to report your post and I would recommend "ban" for you again.

 

[stapel]

Three-week time-out.