Every calculus teacher I know skips this!! 

That's a ingenious idea. This makes sense👌

That was my first thought. That parabola is still a "nice" function in a rotated coordinate system (x' = y - x, y' = y + x). Trivial substitution.

Yeah, it is actually the same thing (just less convoluted). The change of variable is (u,v)=1/sqrt(2) (y-x , x+y) and you get the curve by solving sqrt(2) (u+v)=(v-u)^2. Then you can apply the usual technique on the interval u in [0,sqrt(2)].

@@quintonpierre Thing is, it becomes multivariable when you let u = (y-x) and v = (y + x). Then you have to deal with Jacobians, which you did implicitly.

@@emanuellandeholm5657 things are nicer if you include a factor 1/sqrt(2) in the new variables. otherwise you need to think about scaling factors in the integral limit and differential

Yes i noticed the same thing

It's compensated around 13:20 when Delta - y is calculated for the distance (?)

@@thomashoffmann8857 No that's where it changes the problem entirely to make it far easier to solve than it otherwise would have been. This error persisted throughout the video, and I'm not sure whether the answer that he got is incorrect or whether he just pulled it from mathematica without checking it through the calculations.

​@@sweetcornwhiskey No, the value (4*xi+1-sqrt(8*xi+1))/2 us the value that makes the problem easier to solve. xi - (2*xi+1-sqrt(8*xi+1))/2 = (-1+sqrt(8*xi+1))/2 which removes the xi term so the two parts would not be the same.

Yeah that honestly bothered me a lot lol.

Much simpler method. The volume is an onion made of cones which slant height is h(x) = x - x^2. The radius is just h(x) / sqrt(2) (simple isocele 45° triangle) Thus, the surface of this cone is s(x) = pi * h(x)^2 / sqrt(2) =pi * (x - x^2) ^2 / sqrt(2) = pi * (x^4 - 2x^3 + x^2) / sqrt(2) The thickness of each cone is dx Then V = / s(x) dx in interval (0,1) V = pi * (F(1) - F(0)) / sqrt(2) with primitive F(x) = x^5 / 5 - x^4 / 2 + x^3/3 F(0) = 0 F(1) = 1/30 " 1/5 - 1/2 + 1/3 = 1/30 " V = pi / (30* sqrt(2) ) CQFD NOTE : Démonstration rapide de la surface d'un cône de longueur latérale L et de rayon R. Un cône déplié fait un secteur d'angle Alpha et sa surface est le ratio Alpha/ 2*pi du cercle de rayon L (pi*L^2) Sa surface est donc S = pi*L^2 * Alpha/ 2*pi = L^2* Alpha/ 2 La circonférence de la base du cône fait 2*pi*R mais aussi Alpha*L Donc Alpha = 2*pi*R /L donc Alpha/ 2 = pi*R /L Finalement S = L^2 * pi*R/L = pi*R *L la surface d'un cône de longueur latérale L et de rayon R vaut S = pi*R *L Conic onion peels characteristics : PRIMO : Each cone has a constant angle of 45° with its axis (base radius and height are equals) SECUNDO : dx is the thickness of the peel because x axis is perpendicular to the cone generating line Then s(x) is something simple proportional to the square of the slant height and slant height is also simple y(top of the conic peel ) - y(parabola matching lower point of the peel)

Pretty much any other method is simpler than his

|\ |\ \ | \ \ | \ So the infinitesimal volume (V_conearea) is the volume between those 2 slanted lines rotated around the vertical line (vertically extruded lateral cone area), where the vertical line represents y=x.

Except districts are worried about test scores. I love my subject but my district may have a hard time keeping me.

@@UltraMaXAtAXX How backasswards is it that funding should depend on test scores. Low test scores should indicate students and teachers need _more_ funding not less.

Pappus theorem.

+1 I was going to comment on the same idea: Calculate the center of gravity and use the Pappo-Guldin Theorem. Depending of the problem, calculating the center of gravity can be done in different way, for a parabola segment is known since Archimedes times, sometime symmetries can help, or there is a general formula in worse cases.

The point is that there is not ONE way to arrive at a correct answer. That the answer is correct because there is agreement of the solution INDEPENDENT of the method of its derivation. You can integrate disks or revolve curves, it's the same outcome. If you only ever confine yourself to revolving curves, you may not include integrating disks into your thinking, and will be subsequently stopped because there are no appropriate curves to revolve. Remember, everything looks like a nail when all you have is a hammer. But if you never learn to use a screwdriver, screws just appear as poor performing nails.

@@abenedict85 For this, I prefer integrating cones. (Take the segment from x^2 to x and revolve it around the x=y line.)

Can you explain how you calculated the centre of gravity in step 4 please? I don't understand what function you are rotating about the axes to get the volumes in parts 2 and 3

(P.S. Your "Mr. Rogers mannerisms" are really nice; I should try to emulate that more)

what would it even mean to rotate sth about a curve? afaik that doesnt make sense. take a random point in the shaded area. can you tell me unambiguously where it would rotate to?

@@jakobunfried2669 Hmm... how about using the curvature of a point on the parabola to generate a circle of inversion and then invert the point on the line?

@@derendohoda3891 only 90% sure i understand what you mean, but assuming i do: a point on the line could then be on several different such circles and it would be unclear, where it should rotate to

@@jakobunfried2669 there's only one circle perpendicular to the curvature with all points equidistant from a particular point of the curve. The team problem we would run into is overlapping circles -- a naive integral would double count some volumes when rotating about a curve

Maybe next year

This situation happened to me on accident when our prof made an error in the homework. Me and another guy couldn't solve it so we asked, but he just fixed the error instead of telling us how to solve it.

How is revolution around y=mx+b more involved? You just move everything down by b and you're back at y=mx, don't you? Well, technically, that one simple thing makes it a little more involved. But you know what I mean.

Why?

Instead of relying on that magic, you can avoid the sqrt altogether by not solving for x in terms of xi. Instead turn x^2+x=2xi into: xi=x(x+1)/2 dxi=(x+1/2)dx and substitute accordingly so x is integration variable (and bounds remain at 0 and 1). After much reduction, end up with: sqrt(2)pi/4 integral from 0 to 1 dx [2x^5 - 3x^4 + x^2] which is very easy to integrate.

To clarify, the "much reduction" part involves many but easy reductions since we are simplifying sums and multiplications of small polynomials (no radicals).

The right result shows that you made an even number of errors 😉

I think a combination of both. In general, surfaces of revolution are kind of skimmed over. They just give the integral formula without giving it too much thought, which results in not knowing how to do these, since there's not a standard formula for this problem.

As someone who has taught college calculus a few times I can tell you that the semester tends to be pretty packed. This would work well in something like calculus BC in high school where there's a fair amount of time and you can do some neat projects, but if you tried this in a college calculus class (probably calc 2, but some people do solids of rotation at the end of calc 1 in a semester system; I don't remember where everything goes in quarters) you'd be pressed for time and the students wouldn't get as much out of it as you might want. This would make for a cool extra credit project, though, and if you were looking for an undergrad paper this would probably be an interesting one. It does kind of get trivialized by rotation matrices, though. Another reason I would want to skip this is that I generally think of solids of rotation as a somewhat gentle introduction to applications of integration. They're easy to visualize - the slices for the Riemann sum are literal slices - rather than trying to "slice" time or some other abstract concept. It goes well alongside the classics like lifting objects with ropes of non-zero mass and pumping water out of various troughs and such. If you then make the algebra complicated it defeats the purpose of having a gentle introduction as the students get lost in the algebra and don't comprehend the cool stuff you're showing them.

@@skylardeslypere9909 It's funny, I never even remember the formulas myself because I always want to draw everything. I think my students want or expect the formula but I just keep drawing stuff; the formula is in the book, they can look it up if they want to. I'm just going to keep unrolling cylinders and stacking discs. I mean, who cares about the formula anyway? The whole point of having solids of revolution as part of the introduction to Riemann sums is to help gain intuition about what a Riemann sum really is by having a visualizable application, so why ever worry about the formula? That just skips all the visualizing and jumps to the answer, which defeats the purpose. Also, I'm pretty sure (though I haven't done it) that you could derive a general formula for rotations about lines not square to the axes by using a rotation matrix to rotate your axes, do all the calculations in the new coordinate system, and then another to rotate them back. My guess is that you can probably just run the existing formulas for discs and shells through those rotation matrices to get the formulas straight away. You could probably also do it with standard algebra (like Michael did in the video), but that feels like it would be harder. So I strongly suspect there's a reasonably-derived standard formula, but again, why bother if the whole point is to think about the problem and how Riemann sums work? Now that I think of it, this might make a good project in a linear algebra class when talking about either rotation matrices or change of basis, as this is a really good use of a change of basis matrix. I think I might steal this for exactly that purpose. Thanks for making me think on this topic Skylar, I think my students will really like this :-)

@@jgray2718 You mentioned that it gets trivialized by rotation matrices, and I was thinking something along those lines too while watching the video. Is it actually easier to find the volume of the surface of revolution around any line in the plane, if you apply a change of base? That is, translate your coordinate system so the origin is on that line, and then rotate it so it becomes one of the axes.

@@skylardeslypere9909 In general, yes it's probably easier. Applying a rotation matrix is easy, then you have a standard solid of revolution, which is also easy. The algebra Michael did seemed much harder to me. "Trivialized" was probably an overstatement on my part, but it's not difficult, just a little bit longer than the usual solid of revolution. This particular problem was also chosen specifically to have "easy" algebra since it was rotated around y = x. I'd think a rotation matrix would be the way to go for sure if you wanted to rotate around y = 2x. God help you if you want a general formula for rotating about y = ax + b, though. I'd have to consider how to do that, but I think you're into affine algebra rather than linear algebra at that point, so maybe the approach Michael took in the video would actually be easier in that case. I don't even know what an affine rotation matrix looks like, now that I think of it. Probably a slide then a spin?

That how I would do it.

Without doing it, you would need to change the integration to 0 to sqrt(2) no?

Not entirely trivial to rotate the parabola. Not impossible. But I'm not sure it's easier than this video.

Same, although I think the cones have the same radius and height of r(x)=(x-x^2)/sqrt(2), we want the shells so I did Vshell(x)=pi/3*r(x)^3-pi/3*(r(x)-sqrt(2)dx)^3 I.e. volume outer cone minus volume of inner cone. Note you need sqrt(2)dx because the cone height lies on the diagonal line rather than aligned with the x axis. After cancelling and dropping terms with dx^2 or dx^3, you get Vshell(x)=pi/sqrt(2)(x-x^2)^2 dx and when you integrate this from 0 to 1 you get pi/(30*sqrt(2))

You will be hard pressed to find a better mathematician than an EE grad. I earned my BSEE studying computer chip design. Graduated in the tech bubble burst. No jobs anywhere, world-wide. Ended up doing statistics for the insurance industry.

@@billyoung8118 A better mathematician than an EE grad is pretty easy to find if you've ever heard of a 'mathematician'.

Very cool approach and easier than in the video. I would like to point out that until the line "sum becomes the following integral" everything is correct. The problem comes after that. If u take an area of πr² then u need to multiply the differential element perpendicular to the given area. So multiplying "ds" simply gives the wrong solution. Differential element should be along the line ms+q acc. to notations u have taken which then gives the differential element as ds.√(1+m²). (From pythagoras theorem directly; taking a triangle of ds base and dy =mds as altitude) The entire integral then becomes Integration from 0 to 1[π(s-s²)²/√2 ds] I think this will give the correct value now.

@@Goku_is_my_idol Yes, you're right, very smart fix. Thank you

Yes i solved it this way,it was nice

Me too, check out my solve!

I used Lebesgue integration, and it felt much neater to me! It worked by splitting the volume into hollow cylinders with the central axis being y=x. The radius r cylinder would have width dr and volume 2 pi r len(r) dr, where len(r) is the length of the cylinder of radius r. To find len(r), we just need to find the points such that y = x^2 and y = x - r/sqrt(2) (since that is the parallel line to y=x at distance r from it). The x values of the points are the solutions to x^2 - x + r/sqrt(2) = 0. The difference between the x values is sqrt(b^2 - 4ac), which in this case is sqrt(1 - 2sqrt(2)r). The distance between the points is then sqrt(2)sqrt(1-2sqrt(2)r) (because the slope of the line is 1). So our answer is the integral of 2sqrt(2) pi r sqrt(1-2sqrt(2)r) dr, where r goes from 0 to... 1/2sqrt(2), since that's where len(r) becomes 0. Substitute t=1-2sqrt(2)r, so r = (1-t)/2sqrt(2) and dr = -dt/2sqrt(2). Now we're solving integral of 2sqrt(2) pi (1-t)/2sqrt(2) sqrt(t) (-dt/2sqrt(2)), which is: -pi/2sqrt(2) times integral of sqrt(t)-tsqrt(t) dt, where t goes from 1 to 0. The final integral to solve is super easy, and you get -pi/2sqrt(2) times [2/5 - 2/3], which is 2pi/15sqrt(2), which is different from the answer given in the video. :o I have no idea where I went wrong. I had a silly calculation mistake in my first attempt and it magically matched the video answer with that mistake.

Genius

Great idea! Only the distance between the point of the parabola (x,x^2) and the axis of rotation y=x is equal to L=(x-x^2)/√2, 0≤ x≤ 1. When the parabola rotates, this segment forms a cross-section of the figure perpendicular to the axis of rotation - a circle with an area of π L^2. If dl is an element of the axis of the geometric body , then dl=√2dx. V=∫(from 0 to 1)π*[(x-x^2)^2/2]√2dx=π√2/60.

We calculate the volume of the geometric body as the sum of the volumes of elementary cylinders of differentially small height. In this sense, the cone has nothing to do with it.

@@Vladimir_Pavlov You are right of course, r=(x-x^2)/sqrt(2), since (x^2-x)/sqrt(2) is negative. I think what you call L was r for me. Splitting the volume up into disks as you suggest is very good and simpler. I do like the idea of splitting it into cones though!

It should be noted that this approach to solving the problem makes it easy to get a result if the axis of rotation is an arbitrary straight line passing through the origin, y=tga *x, a≠π/2.)

I see comments regarding forward rotation of the axes. … I am not impressed by the counterintuitive nature of that idea. Rotate the curves instead as you suggest. Det(J) or det(g) =1 so there is no need to do any heroic math. The standard parametric statement of the volume integral definition includes the derivative of x with respect to a parameter t. y’^2(dx’/dx)dx can all be expressed in terms of the original variable x, so the bounds of integration are unchanged.

Yes, but that's not something a second semester calculus student would have access to when first studying solids of revolution

@@williamwolfs4819 It's a shame because Pappus' centroid theorem is easily understood earlier, and it is FAR more general than solids of revolution so restricting it to that is a disservice to the theorem.

@@1ucasvb can you explain to me how to use it in this specific problem?

No! The better example for method is in page 588 ( chapter 9 ) 4th Edition. (Discovery Project). Adrián Irias

@@adrianignacioirias I don't have the 4th edition but I guess you are talking about the Discover project "Rotation about a slant line" in the chapter about Applications of Integration. You don't need this for the example of the video. And, if you read the whole chapter, you'd better use the Theorem of Pappus to compute the volume.

It was delta x squared inside the square root, he just forgot