Volume of the Solid Generated by Rotating An Ellipse Around The x-axis

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Find the volume generated by revolving the top half of the ellipse x^2/a^2 + y^2/b^2 = 1 about the x-axis. This problem is an example the book Integral Calculus written by H.B. Phillips. This book was published in 1917.
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22 сен 2024

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Комментарии : 24

@Boognish64 2 года назад

I was never successful in math classes at any level. In college I was placed in a sub-100 level developmental course where I studied two hours a night, panicked over every single exam, scraped by with a C- before they would let me attempt a college level math course. I passed college Algebra by...transferring Universities and the new place counted my passing Math 80 with a C- as the Math requirement. I work in the trades now, am happy in my career decisions and will never attempt any higher level Math, at least not in any formal capacity. With all that said: I really find these videos FASCINATING to watch. Explanations of unique problems, the pathway of manipulating equations (even though I can't do it myself, its neat to see a pro do it) and the final outcomes are somehow engrossing. I liken myself to a semi-literate man who loved Shakespeare, saying he (the reader) "Just liked the way the words sound".

@audiomac Год назад

Super helpful. I was so confused when I had a question like this because my professor never told me that a > b.

@ahsan4306 2 года назад

Simple proof: Since an ellipse is a flattened circle, the volume generated would be equal to the volume of the sphere of radius 'a' times b²/a². Required volume equals 4/3 *pi* a³ times b²/a² which equals 4/3* pi* a*b²

@4thesakeofitname 2 года назад

Great! I think this's a special application of Papus's Theorem, for volumes of revolutions...

@ahsan4306 2 года назад

@@4thesakeofitname I don't immediately see the connection to Pappus theorem but its somewhat closely related to Cavalieri's theorem

@4thesakeofitname 2 года назад

@@ahsan4306 Hmm yes Cavalieri's theorem is also quite useful, but I instead used Pappus' (second) theorem to find & relate the volume of a sphere and an ellipsoid by considering them as revolutions of a vertical cross sections (in the first octant) about vertical z-axis (x-y being horizontal)... You first find the horizontal centroid *rx* of the vertical cross section of area *A* , then find the distance L traveled by centroid of it during one full revolution : *L = 2.pi.rx* then the volume is V = L . A = 2.pi.rx.A ... It can be seen that the cross sectional area *B* of the ellipsoid is *b/a* of that of sphere, and also its horizontal centroid is the same of that of the sphere, then their volumes are just scaled by (b/a) ... yielding your elegant simple proof... [I think I've mixed "a" and "b"]

@ahsan4306 2 года назад

@@4thesakeofitname That's an excellent line of reasoning. The essential difference in ur method is that ur method involves dividing the ellipsoid into thin cylindrical shells with the common axis of rotation. And then averaging the volumes to get the area times centroid distance formula. In my proof I instead divided the ellipsoid into thin disks normal to the axis. The area of each disk is b²/a² of the corresponding spherical disk since area varies as square of the radius. The results are the same. As they should be. but I still can't clearly understand the MATHEMATICAL reason why dividing the solid in these two different ways MUST give the same result for the factor (b²/a²) . In other words , can u see why the two modes of subdivision give the same factor-- one factor comes from the fact that area is diminished by b/a as is the centroid distance ; the other factor comes just from the fact that the disc's area is diminished by b²/a² which is the same number BUT it arises completely differently ( area alone scaled)

@4thesakeofitname 2 года назад

@@ahsan4306 Well thank you. By the way, unfortunately it's extremely painful to make successful calculus discussions in this plain text format, which would be very simple on paper; That's to say *pen & paper* is still the medium for calculus as of 2022! And for the proofs I think I get what you mean. Eventually all methods should yield the same result of course. In my revolutions approach, however, I didn't explicitly utilized shells as it's implicitly included in the pappus's theorem, which thus can be a simplification for volume calculations provided the centroid is easy to find, or not necessary to explicitly compute as was in this example.

@speedspeed121 2 года назад

I just graduated with a BS in physics. If anybody wants an advantage going into upper-div, learn to solve integrals. I'd say about 75% of the graduates have a hard time solving anything more difficult than a basic u-sub. This really isn;t as bad as it sounds because the professors give you integral tables.

@ermiasawoke192 2 года назад

Thank you, keeping going bro. It is helpful.

@TheMathSorcerer 2 года назад

Thank you, I will!

@NattyPi 2 года назад

This was fairly straightfoward my mistake would have been forgetting that b^2 was a !constant! accidently integrated as 1/3b^3 lol

@urekah37 2 года назад

Both educational and entertaining.

@TheMathSorcerer 2 года назад

thank you:)

@azimuth4850 2 года назад

Really love these calculus problems.

@ussdfiant 2 года назад

Hello friend! My quest to relearn math from the ground up continues with Prealgebra. I put Discrete Math on hold because the “juice wasn’t worth the squeeze”. Anyway I enjoy this and it removed some of the cobwebs from the rules of integration learned 35 years ago!

@TheMathSorcerer 2 года назад

Excellent ! Wow 35 years ! It’s cool you are back at it. Math is one of those things that I feel never dies💪

@swinfwar 2 года назад

possible alternate method: make a coordinate transform using x' = (a/b)x. in the new coordinate system you now have a sphere with radius b so it's volume is (4/3)pi b^3. then you just need to transform this volume back into the original system by dividing by the jacobian which would be J=b/a so V=(4/3) pi a b^2