Calculating the differential area dA of a disk
mast4as
area of a disl
The differential area dA becomes r dr dtheta
where of course r is the radius distance and theta the angular
distance (0 to 2PI). Could someone explain me please how we get this
formula from the disk area ? I don't get it ?
Many thx -coralie
Kaba
Differential area is related to a parametrization function, not a
geometric shape. The shape of a disk is most naturally parametrized in
polar coordinates. This is because when changing to polar coordinates, a
disk transforms into a rectangle. And rectangular domains are
particularly easy to handle (e.g., w.r.t. integration).
Conversion from polar to cartesian coordinates is given by:
x(r, theta) = r cos(theta)
y(r, theta) = r sin(theta)
Differentiate x and y w.r.t. r and theta to get:
x_r(r, theta) = cos(theta)
y_r(r, theta) = sin(theta)
x_theta(r, theta) = -r sin(theta)
y_theta(r, theta) = r cos(theta)
This gives you two derivative vectors (x_r, y_r) and (x_theta, y_theta)
which "span" a parallelogram. Weight the first by dr and the second by
dtheta. The area of the resulting parallelogram is dA. It is most easily
calculated using the determinant:
dA(r, theta)
= |x_r dr x_theta dtheta|
|y_r dr y_theta dtheta|
= x_r dr y_theta dtheta - x_theta dtheta y_r dr
= r dr cos^2(theta) dtheta + r dtheta sin^2(theta) dr
= r dr dtheta (cos^2(theta) + sin^2(theta))
= r dr dtheta
The idea is to take a differential area in the polar coordinate domain
and see what happens to its area when mapped to cartesian coordinates.
mast4as
idea when you say that using polar coordinates 'transforms' the disk
into a rectangle ? you could as well express the coordinates of a disk
with cartesian coordinates no ? Could you please explain this in a
little bit more detail ?
Thanks again so much for your time and knowledge. -c
Asbjørn
> Thank you so much Kaba. This is great. However I don't understand the
> idea when you say that using polar coordinates 'transforms' the disk
> into a rectangle ? you could as well express the coordinates of a disk
> with cartesian coordinates no ?
Well, if you look at the interior points in cartesian coordinates, then
they're bounded by the circumference of the disk. The points in
cartesian coordinates are then given by
x = -1..1, y = -sqrt(1 - x^2)..sqrt(1 - x^2)
assuming radius = 1, or something equally "ugly".
If you look at the points when you use polar coordinates, then the
points are given by
r=0..1, theta=0..2pi
and all the interior points are contained within this region, which is a
rectangle.
Something like that :)
- Asbjørn
Kaba
> Thank you so much Kaba. This is great. However I don't understand the
> idea when you say that using polar coordinates 'transforms' the disk
> into a rectangle ? you could as well express the coordinates of a disk
> with cartesian coordinates no ? Could you please explain this in a
> little bit more detail ?
Assume vector space R^2 with the usual norm and that
(x, y) = (r cos(theta), r sin(theta))
Then
|p| <= R
<=>
|p|^2 <= R^2
<=>
x^2 + y^2 <= R^2
<=>
(r cos(theta))^2 + (r sin(theta))^2 <= R^2
<=>
r^2 (cos^2(theta) + sin^2(theta)) <= R^2
<=>
r^2 <= R^2
<=>
r <= R
Thus if you think the polar-coordinate space as a Cartesian space of (r,
theta) pairs, with r on the x-axis and theta on the y-axis (say), then
an origin-centered disk is mapped to the set
{(r, theta) : r <= R, thetaBegin <= theta < thetaBegin + 2pi}
which is a rectangle. Here thetaBegin depends on what angle range you
want to use for the polar coordinates, popular ones are thetaBegin = -pi
and thetaBegin = 0.
mast4as
very precise. Many thx. -c