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17calculus > polar coordinates > calculus

Polar Coordinates and Calculus

On this page we cover the most common calculus problems involving polar coordinates. The two main topics are differentiation and integration.

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This table lists the topics on this page with links to jump to the topic you are interested in.

Differentiation

Integration

Slope and Tangent Lines

Area

Arc Length

Surface Area

As mentioned on the main polar coordinates page, polar coordinates are just parametric equations. If you are familiar with parametric equations, this material should be very intuitive.

Differentiation - Slope and Tangent Lines

In order to find slope and, subsequently, a tangent line to a smooth graph, we need to be able to calculate the derivative of a parametric equation.

We have the graph represented by the parametric equations \( x = r(\theta)\cos(\theta) \) and \( y = r(\theta)\sin(\theta) \). Here we are emphasizing that r is a a function of \(\theta\) by writing \(r(\theta)\). The derivative is given by

\(\displaystyle{ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{r' \sin(\theta) + r \cos(\theta)}{r' \cos(\theta) - r \sin(\theta)} }\)     where \( r' = dr/d\theta \).

Polar Slope Proof

Polar Slope Proof

Theorem: Polar Slope

For a set of polar equations \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \) where \(r\) is a function of \(\theta\), the slope \(dy/dx\) of a smooth function is given by

\(\displaystyle{\frac{dy}{dx} = \frac{r'\sin(\theta) + r \cos(\theta)}{r'\cos(\theta) - r \sin(\theta)}}\)

where \( r' = dr/d\theta \)

To prove this, we use the theorem of parametrics derivative, the result of which is \(\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} }\)

In our case, the parameter is \(\theta\) and our parametric equations are \( x(\theta) = r(\theta)\cos(\theta) \) and \( y(\theta) = r(\theta)\sin(\theta) \). So our parametric derivative is \(\displaystyle{ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} }\)

Using the chain rule, we calculate
\(\displaystyle{\frac{dy}{d\theta} = r'\sin(\theta) + r \cos(\theta)}\) and \(\displaystyle{ \frac{dx}{d\theta} = r'\cos(\theta) - r \sin(\theta)}\)

Putting these results in the parametric derivative equation, we have

\(\displaystyle{ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{r'\sin(\theta) + r \cos(\theta)}{r'\cos(\theta) - r \sin(\theta)} }\)
[qed]

This equation gives us the slope of the graph at every point. Once we have the slope, we can calculate the equation of a tangent line using the technique found on the main tangent page. Before we go on, let's work some practice problems.

Basic Problems

Practice 1

Calculate the slope of the tangent line to the curve \(r=4\) at \(\theta=\pi/4\).

solution

Practice 2

Calculate the slope of the tangent line to the curve \(r=1+5\cos\theta\) at \(\theta=\pi/4\).

solution

Practice 3

Calculate the slope of the tangent line to the curve \(r=\sin(5\theta)\) at \(\theta=\pi/5\).

solution

Practice 4

Find the tangent line to the curve \(r=\cos(2\theta)\) at \(\theta=\pi/4\).

solution

Intermediate Problems

Practice 5

Find the points on the polar curve \(r=3\cos(\theta)\) where the graph of the tangent line is vertical or horizontal.

solution

Integration - Area

To calculate the area defined by a polar curve r, which is a function of \(\theta\), between the rays \(\theta = \theta_0\) and \(\theta = \theta_1\), we use the integral

\(\displaystyle{A = \frac{1}{2} \int_{\theta_0}^{\theta_1}{ r^2 d\theta}}\)

Here is a good in-depth video discussing area using polar coordinates.

MIT OCW - Lec 33 | MIT 18.01 Single Variable Calculus, Fall 2007

Basic Problems

Practice 6

Calculate the area enclosed by the polar curve \(r=2\cos\theta\).

solution

Practice 7

Calculate the area bounded by one loop of \(r=2\cos(4\theta)\).

solution

Practice 8

Calculate the area enclosed by one loop of \(r=\sin(4\theta)\).

solution

Practice 9

Calculate the area enclosed by \(r=3+\sin\theta\).

solution

Intermediate Problems

Practice 10

Calculate the area inside \(r=1+2\cos(\theta)\) and outside \(r=2\).

solution

Practice 11

Calculate the area inside both \(r=\cos\theta\) and \(r=\sin(2\theta)\).

answer

solution

Practice 12

Calculate the region that lies inside \(r=1+\sin\theta\) and outside \(r=1\).

solution

Practice 13

Calculate the area of the region inside both \(r=9\cos\theta\) and \(r=9\sin\theta\).

solution

Practice 14

Calculate the area outside \(r=\sin\theta\) and inside \(r=2\sin\theta\).

solution

Practice 15

Calculate the area of the region inside both \(r=\sin(2\theta)\) and \(r=\cos(2\theta)\).

solution

Integration - Arc Length

To calculate the arc length of a smooth curve, we can use this integral.

\(\displaystyle{s = \int_{\theta_0}^{\theta_1}{\sqrt{r^2 + [r']^2} ~ d\theta}}\)
where r is a function of \(\theta\), \( r' = dr/d\theta \) and we are looking at the arc from \( \theta = \theta_0 \) to \( \theta = \theta_1 \).

Basic Problems

Practice 16

Calculate the arc length of the polar curve \(\displaystyle{r=e^{\theta/2}}\), \(0\leq\theta\leq4\pi\).

solution

Intermediate Problems

Practice 17

Calculate the arc length of the polar curve \(r=\sin^2(\theta/2)\), \(0\leq\theta\leq\pi\).

solution

Integration - Surface Area

To calculate the surface area of a polar curve revolved about an axis, we use these integrals.

The equation of the polar curve is in the form \( x=r(\theta) \cos(\theta) \) and \( y = r(\theta) \sin(\theta) \) and the curve that is being revolved is defined to be from \(\theta_0\) to \(\theta_1\). We also define \( ds = \sqrt{r^2 + [dr/d\theta]^2}~dt \).

surface area

rotation about the x-axis (the polar axis)

\(\displaystyle{ S = 2\pi \int_{\theta_0}^{\theta_1}{y~ds} = }\) \(\displaystyle{ 2\pi \int_{\theta_0}^{\theta_1}{ r(\theta) \sin(\theta) \sqrt{r^2 + [dr/d\theta]^2}~dt } }\)

rotation about the y-axis

\(\displaystyle{ S = 2\pi \int_{\theta_0}^{\theta_1}{x~ds} = }\) \(\displaystyle{ 2\pi \int_{\theta_0}^{\theta_1}{ r(\theta) \cos(\theta) \sqrt{r^2 + [dr/d\theta]^2}~dt } }\)

So, why the \(ds\) term? If you look at the previous section on arc length, you will notice that this term appears in that integral. This is called a differential length and is just a convenient way of writing these integrals.

Intermediate Problems

Practice 18

Calculate the surface area formed by rotating the polar curve \(r=4\sin\theta\), \(0\leq\theta\leq\pi\) about the x-axis.

solution

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