On this page we cover a common calculus problem involving polar coordinates, determining 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.
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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 \[ A = \frac{1}{2} \int_{\theta_0}^{\theta_1}{ r^2 d\theta} \] Here is a good indepth video discussing area using polar coordinates.
video by MIT OCW 

Practice
Unless otherwise instructed, calculate these areas, giving your answers in exact, simplified form.
Basic 

\(r=2\cos\theta\)
Problem Statement 

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

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one loop of \(r=2\cos(4\theta)\)
Problem Statement 

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

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one loop of \(r=\sin(4\theta)\)
Problem Statement 

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

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\(r=3+\sin\theta\)
Problem Statement 

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

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Intermediate 

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 

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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.
Final Answer 

\(\displaystyle{ \frac{\pi}{4}  \frac{3\sqrt{3}}{16} }\)
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inside \(r=1+\sin\theta\) and outside \(r=1\)
Problem Statement 

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

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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 

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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 

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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 

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Really UNDERSTAND Calculus
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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Practice Instructions
Unless otherwise instructed, calculate these areas, giving your answers in exact, simplified form.