I'm sorry, I may be confused but, If "H" is a length (lets say measured in metres for ease), then "H^2" is an area in metres squared, meaning the units of the final figure suggest we've calculated an area and not a volume. Have I misunderstood something? genuine question, not a criticism, I'm just curious how this works…

Isn't this just the shell method, in terms of y? So you could've just went right to the integral right? This is just the entire idea behind the cylindrical shell method right? Except that would have a 2pi, so maybe not.

It does look confusing! However, if you do this with a more general equation such as: y = a* x^2, where the coefficient 'a' has dimensions of (1/length), the volume comes out to be: V = (Pi/2) * (H^2 / a). Now the second term has dimensions of (length)^3 which is a volume dimension. In this example, a = 1, so the dimension of the constant '1' is hidden.

Volumes of information here, nicely presented by this directrix who displays excellent focus. When you reflect on these concepts, you become a parabolic reflector, which allows you to see really far out places. It's refreshing to have calculus freely available on the web. May the math be with us!

It does look confusing! However, if you do this with a more general equation such as: y = a* x^2, where the coefficient 'a' has dimensions of (1/length), the volume comes out to be: V = (Pi/2) * (H^2 / a). Now the second term has dimensions of (length)^3 which is a volume dimension. In this example, a = 1, so the dimension of the constant '1' is hidden

Amazingly, the volume of a paraboloid was computed by the Muslim scientist, Thābit ibn Qurra, in the 9th century, i.e., 8 centuries before Newton invented (or reinvented) calculus

## 21 Comments

## fyckthissht

mit you're the best! good looking lecturer as well!

## TV 2 FILM

I Really Like The Video From Your Computing the Volume of a Paraboloid

## TheDaekle

I'm sorry, I may be confused but, If "H" is a length (lets say measured in metres for ease), then "H^2" is an area in metres squared, meaning the units of the final figure suggest we've calculated an area and not a volume. Have I misunderstood something? genuine question, not a criticism, I'm just curious how this works…

## Doggo Willinks

Isn't this just the shell method, in terms of y? So you could've just went right to the integral right? This is just the entire idea behind the cylindrical shell method right? Except that would have a 2pi, so maybe not.

## therealjordiano

very interesting imo, thanks

## Sudhakar Medepalli

It does look confusing! However, if you do this with a more general equation such as: y = a* x^2, where the coefficient 'a' has dimensions of (1/length), the volume comes out to be: V = (Pi/2) * (H^2 / a). Now the second term has dimensions of (length)^3 which is a volume dimension. In this example, a = 1, so the dimension of the constant '1' is hidden.

## Word Sailor ADD INFUNITEMS

Volumes of information here, nicely presented by this directrix who displays excellent focus. When you reflect on these concepts, you become a parabolic reflector, which allows you to see really far out places. It's refreshing to have calculus freely available on the web. May the math be with us!

## Saurabh Gorai

good

## historymine

Thank you so much!

## Isabella Oliveira

Thanks a lot, I was searching for a video about this theme and It was great.

## Ricochet Rob

Very interesting proof. Thank you

## Subarna Subedi

what is the volume of paraboloid z=x^2+Y^2 above triangle y=x,x=0,x+y=2

## ABHIJIT DHERE

final answer pi*H2/2 has unit square meter but volume has unit cubic meter

## ebadullah ahmadzai

It does look confusing! However, if you do this with a more general equation such as: y = a* x^2, where the coefficient 'a' has dimensions of (1/length), the volume comes out to be: V = (Pi/2) * (H^2 / a). Now the second term has dimensions of (length)^3 which is a volume dimension. In this example, a = 1, so the dimension of the constant '1' is hidden

## Ojasee Duble

You can check my math but I think it's correct 😉

## Pranat Sharma

How can I find the volume of hyper paraboloid

## Tareq

Amazingly, the volume of a paraboloid was computed by the Muslim scientist, Thābit ibn Qurra, in the 9th century, i.e., 8 centuries before Newton invented (or reinvented) calculus

## student life

Thank you madam…..I m from India

## NUCLEONS GROUP

This derivation is not satisfying

## shashi shekhar azad

This is not satisfactory answer mam ……becoz last answer is not a volume according to dimensions or unit……

## shashi shekhar azad

Becoz unit of last answer is meter square