Derivation of double integration in polar coordinates

Question

I understand the traditional derivation of double integration in polar cords where you see that $\mathrm{d}a = r \mathrm{d}r\mathrm{d}θ$.

However, I thought it would be a fun exercise to try and derive it myself a different way. I wanted to approach it kind of like the derivation of the double integral in rectangular coordinates, where you think of integrating $f(x,y)$ first with respect to $y$ getting a equation of the area for a slice of the region at any $x$ point and the multiplying this by $Δx$ to get a volume and then letting that $Δx$ go to zero to get a $\mathrm{d}v$ and summing all those up.

Here's the reasoning behind my approach. By the way, this is a totally hand wavy proof for I have limited knowledge of writing formal proofs. We know that the area under a curve in polar cords is: $1/2∫r^2 \mathrm{d}θ,$ where $1/2r^2Δθ$ is the area of each little pie slice.

To get a volume we need to multiply this by a $h$ (height). If we integrate $f(r,θ)$ with respect to $r$ we get the area of any slice at a specific value of $θ$, we can divide this by $r$ to get a height at any value of $θ$. Thus our equation for the volume of a specific slice of volume becomes, $$1/2r^2Δθ * \left(\int f(r,θ)\mathrm{d}r\right)/r = 1/2rΔθ * \left(\int f(r,θ)dr\right).$$ If we now let $Δθ$ go to zero and sum up all the slices of volume this tends towards the volume of the region and thus we get this as our resulting formula: $$1/2∫r∫f(r,θ)\mathrm{d}r\mathrm{d}θ.$$

Now, I know this wrong, I just can't see what assumption I made that lead me to the wrong conclusion and that's where I need help. Thanks.

Answer 1

1. It is not true that the integral in polar coordinates is effectively a sum of little pie slices (triangles) $\frac{1}{2}r\Delta r\Delta\theta\,.$ It is rather a sum of little rectangles $r\Delta r\Delta\theta$ which leads to the known volume formula $$ V=\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}f(r,\theta)\,r\,dr\,d\theta\,. $$ There is no factor $1/2$ and you cannot pull the $r$ out of the $dr$-integral. See this tutorial.

2. You can however (under suitable assumptions on $f$) reverse the order of integration: $$ V=\int_{r_1}^{r_2}r\int_{\theta_1}^{\theta_2}f(r,\theta)\,d\theta\,dr\,. $$ Intuitively, $\int_{\theta_1}^{\theta_2}f(r,\theta)\,d\theta$ is an "area" at $r\,.$ If you consider its units however you can see what is going on: this area does have units height times angle. To make it an "area" of height times radius we must multiply it with radius. Then $$ A(r)=r\int_{\theta_1}^{\theta_2}f(r,\theta)\,d\theta $$ can be viewed as the radius dependent area you can integrate to get the volume.