17Calculus Polar Coordinates - Area
17Calculus
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 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 |
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Practice
Unless otherwise instructed, calculate these areas, giving your answers in exact, simplified form.
Basic |
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\(r=2\cos\theta\)
Problem Statement |
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Calculate the area enclosed by the polar curve \(r=2\cos\theta\).
Solution |
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1383 video
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one loop of \(r=2\cos(4\theta)\)
Problem Statement |
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Calculate the area bounded by one loop of \(r=2\cos(4\theta)\).
Solution |
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1384 video
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one loop of \(r=\sin(4\theta)\)
Problem Statement |
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Calculate the area enclosed by one loop of \(r=\sin(4\theta)\).
Solution |
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1272 video
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\(r=3+\sin\theta\)
Problem Statement |
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Calculate the area enclosed by \(r=3+\sin\theta\).
Solution |
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1276 video
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Intermediate |
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inside \(r=1+2\cos(\theta)\) and outside \(r=2\)
Problem Statement |
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Calculate the area inside \(r=1+2\cos(\theta)\) and outside \(r=2\).
Solution |
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1382 video
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inside both \(r=\cos\theta\) and \(r=\sin(2\theta)\)
Problem Statement |
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Calculate the area inside both \(r=\cos\theta\) and \(r=\sin(2\theta)\).
Final Answer |
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\(\displaystyle{ \frac{\pi}{4} - \frac{3\sqrt{3}}{16} }\)
Problem Statement |
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Calculate the area inside both \(r=\cos\theta\) and \(r=\sin(2\theta)\).
Solution |
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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 |
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\(\displaystyle{ \frac{\pi}{4} - \frac{3\sqrt{3}}{16} }\)
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inside \(r=1+\sin\theta\) and outside \(r=1\)
Problem Statement |
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Calculate the area that lies inside \(r=1+\sin\theta\) and outside \(r=1\).
Solution |
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1274 video
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inside both \(r=9\cos\theta\) and \(r=9\sin\theta\)
Problem Statement |
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Calculate the area of the region inside both \(r=9\cos\theta\) and \(r=9\sin\theta\).
Solution |
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1275 video
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outside \(r=\sin\theta\) and inside \(r=2\sin\theta\)
Problem Statement |
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Calculate the area outside \(r=\sin\theta\) and inside \(r=2\sin\theta\).
Solution |
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1277 video
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inside both \(r=\sin(2\theta)\) and \(r=\cos(2\theta)\)
Problem Statement |
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Calculate the area of the region inside both \(r=\sin(2\theta)\) and \(r=\cos(2\theta)\).
Solution |
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1278 video
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Really UNDERSTAND Calculus
Topics You Need To Understand For This Page
Related Topics and Links
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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
Unless otherwise instructed, calculate these areas, giving your answers in exact, simplified form.
