Intro to Integrals - The area of the wall calculation

Mark Misin
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INTUITION MATTERS! - Applied Calculus for Engineers-Complete

Calculus + Engineering + PID: Functions, Limits, Derivatives, Vectors, differential equations, integrals: BEST CALCULUS

34:53:37 of on-demand video • Updated October 2020

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English [Auto] We'll come back. And now we have a problem. We have a building here that looks like a parabola and our problem is to find the area of the wall on that building. So it's some kind of hanger. Right. Kind of looks like a hanger. If you think about it if you look at it in three dimensions so some kind of hanger maybe some kind of military hanger or anger's like that you can find them in airports and then you have airplanes inside. Right. And this side of the hangar has this parabolic shape. OK. So of course hangar. This hangar in three dimensions would look something like this. But then we're interested in this side here. OK. And we want to calculate the area of this side and we can model this site exactly. With a parabolic function because it was built using the parabolic function. Now it's one thing that this hangar the zero zero point is not here the zero zero point is here. How would this function look like if it was symmetric about zero zero point. Well the parabola would be something like this right. But now it's shifted it's it's here. So everything starts from here. Here you have the zero zero point. So you have to shift the function in this way. Right. In other words this point needs to come here. So remember how we shift the functions. If this would be original function and you can see that why here the highest point of the hangar is 16 meters. Right. And then the other dimension with the hangar is eight metres. Then if we started counting our hangar from here if this was our 00 then our function would be y equals 16 minus x square right. That would be this. But now we need to shift the function right. So this for of here we have to shift it in this direction. And we would do that is we would take the X like we have learned before we would take the X only X and then we would subtract 4 from the X. So we would have y equals 60 minus and then X minus 4 and then all that squared. And now you would have y equals 16 minus X minus 4 square. And I can see that. Fine I'll take my 8. Right then you can see that my y here is 0 and I can check it like this. Y equals 16 minus 8 minus four square I will get 16 minus 4 squared. 16 minus 16 equals zero. So it works. And if I take for example for I need to get 16 so I can say why equal 16 minus and then four minus four squared I would get 16 minus zero and I would get 60 and I would get 16 here. So everything works. But now you have to calculate the area of the wall of the hanger one approach that I could take is to discrete twice this parabola like this with levels right. I have these different levels here. And then once I have done that I can calculate the area underneath those levels so I can calculate this area and I can calculate this area and this area. So you can see that I have rectangles here. So if I calculate the areas of these rectangles then I will approximate the area of this hangar size so I can approximate the area of this war. OK. And I can also number these rectangles one two three and four five six seven and eight. OK. And the way I would approximate them would be like this. Now remember this interval here is Delta X.. OK. And this is equal everywhere and considering equal intervals right now. So this would also be delta x and this would be delta x right. This would also be delta x. Now to calculate the area of the wall I first of all need to calculate the area of each rectangle and then I have to sum those rectangles up. So the way I would do it is like this I could do it like this area and this is no approximated area. OK. Approximated area this approximated area would equal Delta X times Y one. So for the rectangle one the level is why one for the rectangle to the level is why two for the rectangle three. The level is why three and then why four here have why five why six Why seven and why eight and then the area would approximately be dealt X times while one plus delta x times Y to. Right because if you think about it you take delta x this interval and you multiplied by the height of the rectangle by y 2. If you do that if you multiply delta x by y two you will get the area of the rectangle so it can continue plus delta X times Y 3 and I can't just say that plus three dots plus delta X times Y A. So that's how you would approximately calculate your area. Now of course I can also write it like this. I can simply factor out my delta x right and then I can have dealt X times and in the parenthesis I will have y one plus y to plus Y three plus three dots up until Y 8 and I can also write it down like this. Or there is even a more compact version of that I could say that my area approximately equal to delta x times. And now I'm going to put this sign here which is a summation sign. This is a summation sign here. And then I'm going to just write here why I and I'm going to say that this I go from one up until and equals a so. OK thanks. Here is the same like this. And this entire thing. Right. Is just this entire thing in parenthesis. All right. So why one plus why two plus Why three. Up until I was eight. So this is a compact way to write this. But remember this equation here only gives me an approximated area. It's not the exact area I have errors here. You see I have errors. These are errors here. So this I'm not taking into account this part and this part and this part and this poor and there is no reason to think that OK this cancels out this or this cancels out this. So it's these are just errors and no matter what I do here in this formulation I don't get my exact theory. I get an approximate area. So that means that I have some kind of error. Now what should I do to decrease my errors and to get an area that is more and more precise.