Discover MakerZone

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

Learn more

Discover what MATLAB® can do for your career.

Opportunities for recent engineering grads.

Apply Today

Thread Subject:
calculated surface area

Subject: calculated surface area

From: Haris Hameed

Date: 19 Jun, 2010 17:37:05

Message: 1 of 2

The surface area (As) of a taper cylinder is
As= pi*(d1+d2)/2*L
d1= diameter of right face
d2= diameter of left face
L= length of taper cone
Similarly surface area for any type of such surface will be
As= (p1+p2)/2*L
Where, p1=perimeter of right face
              P2= perimeter of left face
In the above two cases the two faces are connected by a tapered (inclined) but a straight line
So my question is how to fine the surface area if the two faces are joined by a curve
For above case we can write in integration form As= int(p(x) dx)
int= integral
what will be P(x) in case of curved surface.

Subject: calculated surface area

From: Roger Stafford

Date: 19 Jun, 2010 18:25:05

Message: 2 of 2

"Haris Hameed" <harishameed_33@hotmail.com> wrote in message <hviv81$o75$1@fred.mathworks.com>...
> The surface area (As) of a taper cylinder is
> As= pi*(d1+d2)/2*L
> d1= diameter of right face
> d2= diameter of left face
> L= length of taper cone
> Similarly surface area for any type of such surface will be
> As= (p1+p2)/2*L
> Where, p1=perimeter of right face
> P2= perimeter of left face
> In the above two cases the two faces are connected by a tapered (inclined) but a straight line
> So my question is how to fine the surface area if the two faces are joined by a curve
> For above case we can write in integration form As= int(p(x) dx)
> int= integral
> what will be P(x) in case of curved surface.
- - - - - - - - -
  This is more properly a question in elementary calculus and not in matlab. However, the area of a surface of revolution is given by

 A = integral 2*pi*f(x)*sqrt(1+(f'(x)^2) dx

where x is the distance along the central axis of revolution and f(x) is the orthogonal distance from that axis to the surface. This f(x) would be equal to your P(x)/(2*pi), and dx would be your dL/sqrt(1+(f'(x)^2)).

Roger Stafford

Tags for this Thread

No tags are associated with this thread.

What are tags?

A tag is like a keyword or category label associated with each thread. Tags make it easier for you to find threads of interest.

Anyone can tag a thread. Tags are public and visible to everyone.

Contact us