Calculus in Polar Coordinates

We finally move to the section of calculus. Recall the reason we introduced polar coordinates, that is, it can simplify the equations and make problems easier to solve. Of course that includes calculus.

Polar coordinates can greatly simplify the equations of some graphs and that means make them easier to integrate or differentiate.

We will firstly give some preliminaries.

Preliminaries

In the polar coordinates, the area of a sector of ${\displaystyle \theta }$ degrees is ${\displaystyle \frac{1}{2}\theta r^2}$. It is actually similar with the area of triangles, since ${\displaystyle \theta r}$ is the radius, which similar with the height of the triangle.

Integration

We start with finding the area.

Let ${\displaystyle r=f(\theta )}$ determine a curve, then we define its area:

${\displaystyle A=\frac{1}{2}{\int }_b^a[f(\theta ){]}^{2}d\theta }$

Which can be derived from the Riemann sum of many small sectors.

To find the area enclosed by two curves, similar to that in cartesian coordinate, we follow the steps below:

1. Solve two equations simultaneously, find the two interceptions.
2. Find the difference function, using the equation above to minus the equation below.
3. Find the definite integral between those interceptions.

Tangents

This is actually the slope, or the derivative.

Since the definition of slope was defined in the cartesian coordinate, we are still finding the slope of the tangent line in the ${\displaystyle (x,y)}$ coordinate system at the end.

We gave the relationship between ${\displaystyle (r,\theta )}$ and ${\displaystyle (x,y)}$ before insection 5.3, and we use it here to find the derivative, or the slope.

Since:

• ${\displaystyle y=r\sin \theta =f(\theta )\sin \theta }$
• ${\displaystyle x=r\cos \theta =f(\theta )\cos \theta }$

With the chain rule and the rule of finding the derivative of parametric equations, we have:

${\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}}$

And:

• ${\displaystyle \frac{dy}{d\theta }=f^{\prime }(\theta )\sin \theta +f(\theta )\cos \theta }$
• ${\displaystyle \frac{dx}{d\theta }=-f^{\prime }(\theta )\sin \theta +f(\theta )\cos \theta }$

So, we have the slope of the tangent line:

${\displaystyle m=\frac{f^{\prime }(\theta )\sin \theta +f(\theta )\cos \theta }{-f^{\prime }(\theta )\sin \theta +f(\theta )\cos \theta }}$