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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
 
On this page we cover the most common calculus problems involving polar coordinates. The two main topics are differentiation and integration.
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
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 

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

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

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

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

video by PatrickJMT
close solution 
Calculate the slope of the tangent line to the curve \(r=\sin(5\theta)\) at \(\theta=\pi/5\).
Problem Statement 

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

video by PatrickJMT
close solution 
Find the tangent line to the curve \(r=\cos(2\theta)\) at \(\theta=\pi/4\).
Problem Statement 

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

video by Krista King Math
close solution 
Intermediate Problems 

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

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

video by Krista King Math
close 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.
video by MIT OCW
Basic Problems 

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

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

video by Krista King Math
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Calculate the area bounded by one loop of \(r=2\cos(4\theta)\).
Problem Statement 

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

video by Krista King Math
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Calculate the area enclosed by one loop of \(r=\sin(4\theta)\).
Problem Statement 

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

video by PatrickJMT
close solution 
Calculate the area enclosed by \(r=3+\sin\theta\).
Problem Statement 

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

video by PatrickJMT
close solution 
Intermediate Problems 

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

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

video by Krista King Math
close solution 
Calculate the area inside both \(r=\cos\theta\) and \(r=\sin(2\theta)\).
Problem Statement 

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

\(\displaystyle{ \frac{\pi}{4}  \frac{3\sqrt{3}}{16} }\) 
Problem Statement 

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

The final answer at the end of the video is incorrect. At about the 11:26 mark he notes his mistake.
video by PatrickJMT
Final Answer 

\(\displaystyle{ \frac{\pi}{4}  \frac{3\sqrt{3}}{16} }\) 
close solution 
Calculate the region that lies inside \(r=1+\sin\theta\) and outside \(r=1\).
Problem Statement 

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

video by PatrickJMT
close solution 
Calculate the area of the region inside both \(r=9\cos\theta\) and \(r=9\sin\theta\).
Problem Statement 

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

video by PatrickJMT
close solution 
Calculate the area outside \(r=\sin\theta\) and inside \(r=2\sin\theta\).
Problem Statement 

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

video by PatrickJMT
close solution 
Calculate the area of the region inside both \(r=\sin(2\theta)\) and \(r=\cos(2\theta)\).
Problem Statement 

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

video by Krista King Math
close 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 

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

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

video by Krista King Math
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Intermediate Problems 

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

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

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

Calculate the surface area formed by rotating the polar curve \(r=4\sin\theta\), \(0\leq\theta\leq\pi\) about the xaxis.
Problem Statement 

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

video by Krista King Math
close solution 