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

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