Converting the integral for the area of a circle from an expression involving $dp$ to one involving $dr$ From the Better Explained website, the integral for calculating the area of a circle with respect to its perimeter is given as $$\int_0^{2\pi r} \frac 12 r dp$$ where $r$ is the radius and $p$ is the perimeter. To play around with my understanding of transforming the expression, I tried to find a value of $dp$ and substitute it.
Given that the perimeter of a circle is $P = 2 \pi r$, I reasoned that  $\frac{dp}{dr} = 2 \pi$ and consequently that $dp = 2\pi dr$. I substituted the value for $dp$ into the original equation and got $$\int_0^{2\pi r} \frac 12 r 2 \pi dr$$ Wolfram Alpha evaluates this integral to be $2 \pi ^3 r^2$, which is not right at all. I must have made a logical error somewhere, but I have no idea where.
Background: I did basic integration about 8 years ago in highschool, and am now doing a course where this is assumed knowledge, so I have been trying to teach myself the assumed knowledge while keeping up with the usual content. I'm aware that $dy$ and other such terms can't be thrown around like they are variables, but through some theorems they can act like regular variables in some situations. I feel like this is an area where my understanding is lacking, and if I had to guess where my error was, I would say that I probably did something invalid with a $d$ somewhere.
 A: What you have can be explained by figure 1.
If you consider a very small sector which has arc length dp and a radius of R then the area dA of the sector will approximately be 1/2*R*dp and if all the dps are of unit length then you'll add them 2πR times. Now, this is a sum of continuous variables which can be written as an Integral.
$$dA = \frac 12*R*dp$$
$$A = \int_0^{2\pi R}dA$$
$$A = \int_0^{2\pi R} \frac 12 R dp$$
$$A = \pi R^2$$
Now, if you want the same integral in form of radius, think of small ring or a circular strip with perimeter of p=2πr, radius r and thickness of dr. (look at figure 2) If you add area of all the small rings you should get total area of the circle. Area of a small ring dA will be 2*πrdr. Since the radius is continuously varying the sum can be expressed in form of following integral, and if all the drs of unit length you will add it R times.
$$dA = 2*\pi *r*dr$$
$$A = \int_0^{R}dA$$
$$A = \int_0^{R} 2\pi rdr$$
$$A = \pi R^2$$
all the approximations work because the fractions like dr, dp are very very small. i.e. are of the sector = 1/2*R*d because shape of the sector is equivalent to a triangle.
Now, as far as I understand you can't inter-substitute in this case because in substitution you change a integral from one variable to another if you have a relation between them. Here dp is related to R and not r so you cant apply substitution rule, rather they are simplification of more general integral $$A = \int_0^{R} \int_0^{2\pi R} \frac rR dp*dr$$.
If you want I can explain this term to you. Both of the individual cases can be achieved after integrating with respect to any one of them.
For more information: https://en.wikipedia.org/wiki/Multiple_integral
Figures 1 & 2
