Circled Part Formula in Double Integral: Explaining the Use of dA in Polar Form?

[chetzread]

Homework Statement

can someone explain about the formula of the circled part?
Why dA will become r(dr)(dθ)?

Homework Equations

The Attempt at a Solution

A = pi(r^2)
dA will become 2(pi)(r)(dr) ?
why did 2(pi) didnt appear in the equation ?

 

[SteamKing]

Homework Statement

can someone explain about the formula of the circled part?
Why dA will become r(dr)(dθ)?

Homework Equations

The Attempt at a Solution

A = pi(r^2)
dA will become 2(pi)(r)(dr) ?
why did 2(pi) didnt appear in the equation ?

dA = r ⋅ dr ⋅ dθ as can be seen from the diagram below:

​

2π has nothing to do with converting dA into its polar equivalent.

 

[vela]

A = pi(r^2)
dA will become 2(pi)(r)(dr)?

The dA you calculated here is the infinitesimal increase in the area of a circle if you increase the radius by dr. In other words, it's the area of a ring of radius r and width dr. It's not the same dA that appears in the integral.

 

[BvU]

why did 2(pi) didnt appear in the equation ?

Is it clear to you now that this hasn't happened because the integral reads$$
\displaystyle \iint\limits_R f(x,y)\; dA = \iint\limits_R f(r,\theta)\; r \;dr\;d\theta\quad ?$$Only if $\ f(r,\theta) = f(r)\$ i.e. f does not depend on $\theta$, the integration over $d\theta$ can be carried out (yielding $2\pi$) and the result is$$
\displaystyle \iint\limits_R f(x,y)\; dA = \int\limits_{r_0}^{r_1} f(r)\; r \;dr\;\int\limits_0^{2\pi} d\theta\quad = 2\pi \int\limits_{r_0}^{r_1} f(r)\; r \;dr\;$$

 

[chetzread]

Is it clear to you now that this hasn't happened because the integral reads$$
\displaystyle \iint\limits_R f(x,y)\; dA = \iint\limits_R f(r,\theta)\; r \;dr\;d\theta\quad ?$$Only if $\ f(r,\theta) = f(r)\$ i.e. f does not depend on $\theta$, the integration over $d\theta$ can be carried out (yielding $2\pi$) and the result is$$
\displaystyle \iint\limits_R f(x,y)\; dA = \int\limits_{r_0}^{r_1} f(r)\; r \;dr\;\int\limits_0^{2\pi} d\theta\quad = 2\pi \int\limits_{r_0}^{r_1} f(r)\; r \;dr\;$$

yes, thanks