calculating work by using integral, pumping water out of a tank, calculus 2 tutorial

Application of integration. Calculating the work it needs to pump the water out of a conical tank. Calculus 2 tutorial.
Check out my 100 calculus 2 problems to help you with your calc 2 final: • 100 calculus 2 problem...
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[@blackpenredpen]

Bunny Burrito makes lives better!

[@duggydo]

Was the original video messed up? Why redo?

[@1234Daan4321]

blackpenredpen the cat was cute too :D

[@blackpenredpen]

1234Daan4321 There's a debate, which oreo is cuter? P.s. Both the bunny and the cat are named "Oreo" but owned by different people

[@iHATEbigots666]

this is true!!

[@leonardobarrera2816]

Burrito is a word of my lenguage (Spanish) I am so happy!!!

[@joseumanawills8523]

Love his enthusiasm when teaching and explaining step-by-step process (like using the point-slope form rather than using slope-intercept form)

[@timehrensberger6479]

You manage to make this problem easily comprehensible for a highschool student knowing only the basics of calculus. Compliments!

[@pythagoras31416]

sir you don't know how much u helped me. i couldn't sleep for 2 nights to get a proper theory on this topic. i was so close and now i completely understand this. thank you so much from the core of my heart.

[@mariobezrukovs2547]

Thanks! You made it look a million times easier than my calc 2 professor thank you!

[@jaggysmf1860]

Currently taking Calc II, I could not understand this whatsoever from my textbook, this video makes the concept so clear. Thank you.

[@feezy243]

Why is it so easy when you watch videos but so hard when doing homework or a test? :(

[@ChrisMMaster0]

Cause you are not doing it, he is doing it

[@whiteskeleton9453]

@@ChrisMMaster0 lol

[@AarmOZ84]

Thanks to graphing calculators being able to solve definite integrals, the set up is the most important part to getting the problem right. Thanks for helping this old guy who is taking Calc II online. :)

[@Smokeybear69420]

I'm surprised, this really helped me understand how to do this problem a lot better. You earned a sub :)

[@ChaosPod]

I think saying you need to multiply the density by 9.8 is still confusing. Here is how I would have said it: work done (joule, J) = distance (metre, m) x force (newton, N) F (force) = m (mass) * a (acceleration) in this case the acceleration is gravity so Fw (weight) = m (mass) * g (gravity, 9.8 m/s^2) Fw = m * 9.8 m/s^2. Mass = Density x Volume Then I would have put them together: W=dF=d(gm)=dg(Density x Volume) where g = 9.8

[@ZipADeeeDoooDaaa]

I totally agree. I wanted to write exactly the same.

[@montald001]

And density for water is 1kg per 1L, and the volume of the body is 1/3(pi)x(radius)^2x(height), but hey, if it is a calculus 2 test, I don’t think you are allowed to use other methods aside from calculus, but you can use it to check your answer since it is easier and less time-consuming eh.

[@DanielInfrangible]

super smart. I like it.

[@tomatrix7525]

I agree was confused by him saying that. I made sense of it myself soing in my head exactly what you did. Remember aha he is no physics teacher lmao he is math

[@cosmicxistence3428]

thank you this helped me a lot

[@picorrow]

Beautifully crafted video! The bunny allowed me to ease into the complication. :)

[@PlayTheMind]

*π,* bro, *π*

[@bananabananae]

PlayTheMind I see you on every science video

[@shambosaha9727]

@@bananabananae Me too

[@thespicymeatball5625]

work = 3362827.80 joules love your videos so much keep up the good work. Maybe i can be first this time last video i was second by a minute

[@blackpenredpen]

TheSpicy MeatBall yes you are!! Congrats!!!

[@blackpenredpen]

TheSpicy MeatBall and thanks!!!!!!!

[@joeli8409]

I get about 3,362,203

[@diamonddave2622]

Joel I - You are x a big number by pi so it depends on your value for pi. I got the same as you.

[@guitarttimman]

The numerical value isn't important. The concept is important. Once you complete the integration, just plug everything into a calculator, and you'll get the right answer.

[@scharpmeister]

This man swaps his pens with absolute STEEZ

[@GhostyOcean]

Good job on correcting and explaining your mistake! This is how I see it: units like Lbs already have the force of gravity factored into them while SI needs to be multiplied by the force of gravity because mass is a measurement of how much matter is there. For example: something that weighs 100lbs on Earth does not weigh 100lbs on the moon, but the mass of the object stays the same.

[@JensenPlaysMC]

mass is not a measure of just matter. bit more complicated

[@ClumpypooCP]

This REALLY helped me, thank you

[@ZipADeeeDoooDaaa]

I just wanted to share this. I found a surprisingly simple formula for this problem. W = F * (H - 3/4*h), where H = 10, h = 8 and of course F = m * g, where m is the mass of the water and g = 9.8 Calculating the mass of the water is a bit tricky. We need the Volume of the water cone and the density of water. V = 1/3 * π * r^2 * h The mass of the water is about m = 85786.42 kg Plugging all the numbers into the formula above gives you W = 3362828 Joules

[@blackpenredpen]

ZipADeeeDoooDaaa Wow nice. What's the name of the formula? And why H-3/4h?

[@ZipADeeeDoooDaaa]

I derived the formula by myself, so I don't know if it has name. And about the H-3/4h, I have no clue. It is kind of cool that's why I posted it. Maybe someone else has a good explanation.

[@ZipADeeeDoooDaaa]

I thought about why H-3/4h and now I understand. It is the distance from the center of mass up to the top of the container. The center of mass of an upside down cone is at 3/4 h and so the distance to move that center of mass to the top is H-3/4h.

[@blackpenredpen]

ZipADeeeDoooDaaa Ahhhh I see!!! And we can also do integral to get the center of mass as well!! Cool!!!!!!!

[@theaclairy8992]

if the water is full would the h be 10?

[@jesusreignonhigh6732]

Nostalgia. In college we were advised to use trig functions on the x, y relation for the conic shape.

[@spacej0e]

Fantastic video, thank you so much! I think you just saved my Calculus 2 grade!

[@HettyPatel]

bro, he is dual wielding dry erase board markers LOL never have I ever seen a person more badass.

[@redpandazzz8421]

thx for getting right to point your great!

[@daniellaz8172]

Yes very clear I understood everything! 👍🏼👍🏼

[@jamesleroy572]

Thanks! This may be way late as far as a comment is concerned, but this helped me tremendously!

[@courtneysumner7504]

Thanks! And the bunny is so cute!

[@BruderAdrian]

You're saving my grade! Thanks boss, you clutch!

[@rockyboy5545]

Thank you. Thank you.

[@laurenayara]

The way he switch the markers tho. Anyway, thank you sir.

[@jmich7]

Wonderful, as usual.

[@ayyanali8477]

A horizontal cylindrical water tank has lenght 26 ft and circulare face radius is 4 ft. Suppose water is leaking from the base where the radius of hole is 10 inches. If the tank is initally full of water then find the formula for h(t). Also find the time when water level become equal to half of the tank height.

[@putriamalia9710]

Thank you, this video saved my life!!

[@natthidat5251]

This video is helpful, Thank you!

[@louisthurston3067]

Correct language is "work done against gravity." The pumping process will necessarily eject the fluid with some KE and there will be some viscous or thermal effects. If you only apply an upward force equal to the weight of a fluid element, it will not move. You must apply a larger force to accelerate it giving it KE.

[@nunya5027]

I know this comment is 3 years old however it really did kinda make the idea of having to add gravity to mass for weight make sense . The thing that always screws me up is how colloquially weight /mass are somewhat interchangeable. And getting weighed at the dr in kilos .

[@user-jz5vn5jj5h]

Tanks for the effort brotha

[@guitarttimman]

Density times volume is just another way of representing the mass. Force X Distance = work Density, which is mass/volume, times the volume = mass F = mass x g

[@guitarttimman]

Okay. I caught my error. It is as follows: There's an alternative way to get the function of X expressed in terms of Y. I would've utilized similar triangles. 4/10 as X/Y. Solve for X in terms of Y and you'll get X = 2/5 Y. When you solve the problem, you'll get 1000*pi*g [ 8/15Y^3 -1/25 Y^4] evaluate at 8. ( you don't need to worry about subtracting the 0). g = 9.8 m/s^2. My numerical answer was around 3.4 X 10^6 Joules. ha ha

[@AlphaNumeric123]

Great dude, idk why I was suggested this video but it was an insta-sub. Love to help if you have a patreon

[@joshharris9788]

Nice job..thank you!

[@justkarl2922]

Addicted to find the general formula to this case of a cylindrical tank, setting the water density p to p= 10³ kg/m³ g=9.81 m/s² and labeling H=10m R=4m and x=2m we get work as W=πpgR²(H-x)³(H+3x)/(12H²) which in this case is W=3.366259 MJ

[@ryaneakins7269]

This is the minimum work required, right? Applying a force equal and opposite to gravity gives a net force, and thus net acceleration, of zero. So unless the disc of water is already moving, it stays exactly where it is (in the static situation, before we start inputting work, this force comes from the buoyancy of the water itself). Since work has no explicit time dependence, it must be possible to have any arbitrary amount of work above the calculated value - it just means the water has a higher velocity at the top.

[@androidtech1501]

Thanks for the video man

[@noelforde237]

Couldn't you have gotten the radius with similar triangles? Or does it not work for all work problems?

[@mikerophone9618]

Thx blackpenredpen :D I am learning many things from you. May I ask you one thing about math? To improve mathematic comprehension, what method would be good for it? Is there a series of proper procedure? (I mean, how to improve mathematic comprehension to be a mathematician?)

[@blackpenredpen]

Mike rophone Hi Mike, I will say you will have to start with learning proof and learning how to love to prove theorem. To do so, oftentimes we need to read lots of the famous proofs, then try many exercises. Also, the proofs in different fields of math are also very different. Pick a field first, such as number theory, PDE or analysis. And you can go from there. Btw, Dr. Peyam is also a great person to ask this question to. I did an interview with him in the summer on my channel. Check that out. :)

[@viobandolininstrumentals5936]

but how can you calculate the x(the one you got from the point slope form) if the shape is a frustrum of a cone, a hemisphere or other shapes.

[@ArKriEvRDu]

thats a nice marker you have there, what do you call that

[@tomatrix7525]

Thanks this is interesting

[@wwDango_]

tysm

[@AidenP361]

Bruh this was on my AP calc test

[@gabrialpetersen1926]

Thats a lot of Work.

[@mohan153doshi]

Absolutely awesome explanation. You are the best

[@rodbhar6522]

I think it is much easier to just calculate the average height of the water and the volume. Then you know how far you need to raise all the water to get to the top.

[@JensenPlaysMC]

that will yeild the wrong result

[@SuperMtheory]

The value of x/y (from the picture) is proportional to 4/10. Therefore, x = 0.4y

[@tonilynn1391]

I have a couple of questions. This example assumes a constant force, correct? So what if the force is NOT constant? Another question is what if we "tip" the cone and it's inclined?

[@larryxiang822]

i dont know if you still care, but: the force is the force of gravity, and for our purposes, is always going to be constant. Gravity is not constant, and weakens as you go further from the center of the Earth (aka higher in the cone), but that difference is so small that we ignore it. Also, tipping the cone will make the "slices of water" have a different shape, therefore a different formula to calculate their volume, and it will change the height of the top relative to the lowest point of the cone, so those two things change. If you have the formula for the area of each slice of a tilted cone (if you have the point facing down, each slice should just be an ellipse rather than a circle in a cone facing straight down), and you can use some trigonometry to find the new height to the top, the problem is still solved the same way.

[@carlospitcher4335]

Too bad you had to change the video. It took me some quick research to understand that weight is a force... having grown in the metric world, we offen use the mass. We were able to see anyways that the calculus were right and you even explained them when you added the 9.8... Anyways, great video!

[@Lily-of1gi]

You’re the GOAT

[@bensnodgrass6548]

This is a rather elementary solution which does get a different answer, but what if you were to work out the "average" height of the water? Then, as gravitational potential energy is directly proportional to height (E=mgh), you could multiply the total mass by the the gravitational field strength and then by the average height gain needed? It gets an answer reasonably close to his one but why is it wrong?

[@mikeviermann4488]

Final tomorrow and you just saved my a!

[@jeffreystockdale8292]

What is the cats name? He’s precious!!

[@guitarttimman]

Here's a good physics point: Mass and Weight are not the same thing. They are used interchangeably because we are on earth, but the weight of an object varies from planet to planet. Mass is constant. Density is always the same. Let's not forget what Kepler proved.

[@wouterfransen9771]

Why is the radius x instead of 4-x?

[@ArtsFracture1.0]

YOU'RE A GOOD MEN.

[@tvaughn61]

So if the density is cubed, do you have to do the volume of the cone rather than the area of the cylinder to cancel out the cubed units?

[@tvaughn61]

I am not sure if I am confused or not. Maybe you don't need the volume of a cylinder a just need the area.

[@austingulotta9817]

Why don't we have to account for the last two meters? It seems like we only counted the work to move all the water up to the 8 meter marker where the water ends. But we still have to lift that another two meters.

[@JensenPlaysMC]

that is factored into the (10-y). this is the thing that ensures it will be lifted to the very top. the 0 to 8 means that he needs to lift up all of the water between those regions. if he put 0 to 10 it would mean that there was water for those last 2 metres

[@thugken]

lifesaver

[@GueVonez]

Is there a way of doing this using multivariable calc? I wana see double/tripple integrals hehe

[@GueVonez]

I guess you could use triple integral to calculate the volume by using equation of a cone...

[@xtremeskill744]

why wasn't there more bunny burrito

[@thomasg6830]

As i wrote on the previous version of this video, there is simply no way to answer physics questions and make everybody happy. Don't worry. That's just how it works. It will always be wrong (for some).

[@anthonyvallejo9127]

savior

[@edeltovilla2773]

why dy for volume?. anyone can explain that, please. i did not get that.

[@averydrago]

(9pi)/25*mass density*gravity*(8^4)/6

[@roys139]

i almost clicked away by the intro cuz i thought its completely irrelevant from the title

[@heliocentric1756]

The US units never made any sense. If a rock is put in a zero gravitational field, does this make its "density" zero???

[@jimstiles26287]

The US units make a great deal of sense. The problem should have used specific weight instead of density. The specific weight of water is 9810 N/m^3 or 62.4 lbs/ft^3. The density of water is 1000 kg/m^3 or 1.94 slugs/ft^3.

[@heliocentric1756]

For your information, gravitational acceleration is not the same on our planet -which by the way includes countries other than "Merica" !! (Mount Nevado Huascarán in Peru has the lowest gravitational acceleration, at 9.7639 m/s2, while the highest is at the surface of the Arctic Ocean, at 9.8337 m/s2)

[@PurpleFiiilth]

The only difference between the SI and Imperial system is that in the SI system weight and mass are equal on earth. Anywhere else and you have to make the same adjustments for the acceleration due to gravity than you would in the Imperial system. For example if I weighed 100 kg on earth and I went to the moon, I would now "weigh" about 16.5kg. However my mass would still be 100 kg. Both systems make sense just the SI system is more convenient for us here on earth. By the way density is independent of gravity, and weight is a force it is different than mass.

[@jimstiles26287]

Density and weight are never the same. Weight changes from planet to planet and from location to location on the same planet. Density is always the same. One cubic foot of water always has a density of ~1.94 slugs, and one cubic meter of water always has a density of ~1000 kg.

[@PurpleFiiilth]

People express weight using kilograms in the SI system. If you ask someone in Europe how much they weight they will give you an answer in kilograms, which is normally a unit of mass. The reason this makes sense is because in the SI system weight and mass are approximately equal on earth. So yes they can be the same actually.

[@stephenphelps920]

it's ok

[@blackpenredpen]

Stephen Phelps 👌

[@thiagoalves9189]

please, help me, integral of x^4sin(x^3+4)dx

[@rudolfgyorkei9558]

faster than a calculator

[@ichigo_nyanko]

What happened to the last video?

[@heliocentric1756]

Here it's kzbin.info/www/bejne/ooHUlqRvbamtfbs&lc

[@blackpenredpen]

I just made it unlisted

[@blackpenredpen]

Jdm1891 I thought I made that too long anyway (it was over 15 min)

[@lorenzobarbolla7988]

3,4 MJ i think

[@wyong6198]

answer : 3,362,827.797 or 1,070,421.333π

[@nabihat.1426]

how did you get the joules answer ?

[@diegoboi2662]

blackpenredpenbluepen this time

[@brandonklein1]

I know this is a math video but I feel like just a little bit more respect to the physics of what's going on here should be said, maybe show in dimensional analysis how the units work or how 9.8 is an acceleration and density is mass/volume, so through Newton's second law f=ma, beautifully done, I love your videos, I just like the physics that's going on here too:)

[@melihkaganozge1210]

In Physics W=F.x 😊

[@euyin77]

Just one correction: u r calculating potential energy, not work.

[@TimeTraveler-hk5xo]

But the work needed to drain all the water is equal to the potential energy, by conservation of energy.

[@0dayExploit]

Why the reupload?

[@0dayExploit]

Ah okay, so reading the other comments and yours I think I understand. He didn't use the correct force for the work eq. because of the imperial system using weight, which is already a force specific to Earth's gravity, instead of mass. I don't know how I didn't catch that on the first watch; I must have not been paying enough attention and went too fast. Thanks!

[@glennschexnayder3720]

About 3.4 MJ?

[@brunoeigenmann1667]

yes👍🏻

[@geniscordoba3821]

Why don't you start: W= F*x? - then substitute F=weight=m*g - then substitute m with m=d*v Its more clear this way i think.

[@marinamaged962]

i love you

[@rafyreyes3797]

isn't pi(r^2) an area formula, not volume?

[@ViniciusTeixeira1]

yes, but he used π*(r^2)*thickness, which is the volume of a cylinder

[@DanBurgaud]

I'll drill a hole at the bottom... least work of all...

[@user-bs2dv5po8i]

it should be 130.690 foot-pounds

[@nikheetisnotcool2951]

Why do we need to have 2 measurement systems???!!!!!!!!!!!!!

[@rezaetemad591]

200704(3.14) right ? and by the way, million thanks for these awesome videos ...love your channel, keep on the good work man.

[@Mr.Jack_learning]

what a legend

[@68933964]

可以创造跟多华文视频吗？我觉得蛮有趣的！

[@blackpenredpen]

Wu Meng Hui 像是什麼題目呢？

[@68933964]

积分题，区别题，矢量题。。。什么都行！我的老师用英文解释数学题目，所以我觉得听您用华语讲解题目很有趣

[@blackpenredpen]

我儘量!!!

[@wenhanzhou5826]

Thank God you are using meter and not feet lol.

[@ayyanali8477]

Plz solve this

[@onetwo8037]

Wait, why is thickness dy?

[@brandonklein1]

One Two it is some "difference of y" along the noted axis, this is what "Dy" stands for, just some tiny nudge in the y direction.

[@ormsucher]

To extend Brandon's answer: The thickness of a disk is DELTA(y) = y(top) - y(bottom). We want a pile of infitely thin disks. That means for every disk DELTA(y) has to go to zero. And this is the transition from DELTA(y) to dy.

[@onetwo8037]

Ah ok

[@williamadams137]

x = (5/2)y not (5/2)x

[@SeriousApache]

Is Bunny Burito dinner for that cat?

[@adambrucker819]

6

[@LeonMakoto]

3362827.797 Joules