\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Polar Coordinates - Area

Limits

Using Limits


Derivatives

Graphing

Related Rates

Optimization

Other Applications

Integrals

Improper Integrals

Trig Integrals

Length-Area-Volume

Applications - Tools

Infinite Series

Applications

Tools

Parametrics

Conics

Polar Coordinates

Practice

Calculus 1 Practice

Calculus 2 Practice

Practice Exams

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Articles

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.

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 in-depth video discussing area using polar coordinates.

MIT OCW - Lec 33 | MIT 18.01 Single Variable Calculus, Fall 2007 [35min]

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

1383 video

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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

1384 video

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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

1272 video

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

Problem Statement

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

Solution

1276 video

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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

1382 video

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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.

1273 video

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

1274 video

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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

1275 video

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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

1277 video

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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

1278 video

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You CAN Ace Calculus

Trig Formulas

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 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle 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

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

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

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