Calculus BC ยท Unit 9
Square r, multiply by 1/2, and integrate in theta. For a petal, stop where r returns to zero.
Area inside one polar curve
\(A=\dfrac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta\)
The factor \(1/2\) is part of the formula. Square \(r\) before you integrate.
Limits are angles, not x-values
A full circle in \(r\) may use an interval of length \(\pi\) or \(2\pi\), depending on how the curve is traced.
Area between two polar curves
\(A=\dfrac{1}{2}\int_{\alpha}^{\beta}\left(r_{\text{outer}}^2-r_{\text{inner}}^2\right)\,d\theta\)
A petal has its own angle window
Find where \(r=0\) to locate the start and end of a petal. Do not integrate a whole turn if the question asks for one petal.
r is constant, so the integral is a rectangle in theta.
\(A=\dfrac{1}{2}\int_0^{2\pi} 4\,d\theta\)
\(\dfrac{1}{2}\cdot 4\cdot 2\pi\)
\(4\pi\)
This matches \(\pi\cdot 2^2\).
Answer: \(4\pi\)
\(\dfrac{1}{2}\int_0^{2\pi} 25\,d\theta\)
\(25\pi\)
Answer: \(25\pi\)
\(\dfrac{1}{2}\int_{-\pi/2}^{\pi/2} 4\cos^2\theta\,d\theta\)
\(\cos^2\theta=\dfrac{1+\cos 2\theta}{2}\)
\(2\int_{-\pi/2}^{\pi/2}\dfrac{1+\cos 2\theta}{2}\,d\theta\)
\(\int_{-\pi/2}^{\pi/2}(1+\cos 2\theta)\,d\theta\)
\(\pi\)
Answer: \(\pi\)
\(\dfrac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2\,d\theta\)
Answer: \(\dfrac12\int_0^{2\pi}(1+\cos\theta)^2\,d\theta\)
Square each radius, then subtract, then multiply by \(1/2\).
\(\dfrac{1}{2}(9-1)\)
Answer: A. \(\dfrac12(9-1)\)
\(\dfrac{1}{2}\cdot 8\cdot 2\pi\)
\(8\pi\)
Answer: \(8\pi\)
One petal, or a sector between two radii.
\(A=\dfrac{1}{2}\int_0^{\pi/2} 9\sin^2(2\theta)\,d\theta\)
\(\sin^2(2\theta)=\dfrac{1-\cos 4\theta}{2}\)
\(\dfrac{9}{2}\int_0^{\pi/2}\dfrac{1-\cos 4\theta}{2}\,d\theta\)
\(\dfrac{9}{4}\left[\theta-\dfrac{\sin 4\theta}{4}\right]_0^{\pi/2}\)
\(\dfrac{9}{4}\cdot \dfrac{\pi}{2}\)
\(\dfrac{9\pi}{8}\)
Answer: \(\dfrac{9\pi}{8}\)
\(\dfrac{1}{2}\int_0^{\pi} 16\sin^2\theta\,d\theta\)
\(8\int_0^{\pi}\dfrac{1-\cos 2\theta}{2}\,d\theta\)
\(4\left[\theta-\dfrac{\sin 2\theta}{2}\right]_0^{\pi}\)
\(4\pi\)
Answer: \(4\pi\)
\(\dfrac{1}{2}\int_0^{\pi/2}(4-1)\,d\theta\)
\(\dfrac{1}{2}\cdot 3\cdot \dfrac{\pi}{2}\)
\(\dfrac{3\pi}{4}\)
Answer: \(\dfrac{3\pi}{4}\)
\(\sin(2\theta)\) completes two full oscillations as \(\theta\) goes from 0 to \(2\pi\), giving 4 petals.
Each petal has area \(9\pi/8\).
\(4\cdot \dfrac{9\pi}{8}\)
\(\dfrac{9\pi}{2}\)
Answer: 4 petals, total area \(\dfrac{9\pi}{2}\)
The formula has a factor \(1/2\).
The area is \(4\pi\), not \(8\pi\).
Answer: the factor \(\dfrac12\)
\(\dfrac{1}{2}\int_0^{\pi/3}\sin^2(3\theta)\,d\theta\)
Answer: \(\dfrac12\int_0^{\pi/3}\sin^2(3\theta)\,d\theta\)
Evaluate a petal, or subtract two squared radii.
\(\sin^2(3\theta)=\dfrac{1-\cos 6\theta}{2}\)
\(\dfrac{1}{2}\int_0^{\pi/3}\dfrac{1-\cos 6\theta}{2}\,d\theta\)
\(\dfrac{1}{4}\left[\theta-\dfrac{\sin 6\theta}{6}\right]_0^{\pi/3}\)
\(\dfrac{1}{4}\cdot \dfrac{\pi}{3}\)
\(\dfrac{\pi}{12}\)
Answer: \(\dfrac{\pi}{12}\)
Replacing \(\sin(3\theta)\) by \(3\sin(3\theta)\) multiplies \(r^2\) by 9.
One petal: \(9\cdot \dfrac{\pi}{12}=\dfrac{3\pi}{4}\).
Three petals: \(\dfrac{9\pi}{4}\).
Answer: one petal \(\dfrac{3\pi}{4}\), all three \(\dfrac{9\pi}{4}\)
\(\dfrac{1}{2}\int_0^{2\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta\)
\(\cos^2\theta=\dfrac{1+\cos 2\theta}{2}\)
The integrals of \(\cos\theta\) and \(\cos 2\theta\) over a full turn are 0.
\(\dfrac{1}{2}\int_0^{2\pi}\left(1+\dfrac{1}{2}\right)\,d\theta\)
\(\dfrac{1}{2}\cdot \dfrac{3}{2}\cdot 2\pi\)
\(\dfrac{3\pi}{2}\)
Answer: \(\dfrac{3\pi}{2}\)
\(1+\cos\theta=1\) when \(\cos\theta=0\), so \(\theta=\pm\pi/2\) in one trace, or \(\pi/2\) and \(3\pi/2\).
The cardioid is outside \(r=1\) for \(-\pi/2\le\theta\le\pi/2\).
\(\dfrac{1}{2}\int_{-\pi/2}^{\pi/2}\left((1+\cos\theta)^2-1\right)\,d\theta\)
Answer: \(\dfrac12\int_{-\pi/2}^{\pi/2}\left((1+\cos\theta)^2-1\right)\,d\theta\)
\((1+\cos\theta)^2-1=2\cos\theta+\cos^2\theta\)
\(\dfrac{1}{2}\int_0^{\pi/2}\left(2\cos\theta+\dfrac{1+\cos 2\theta}{2}\right)\,d\theta\)
\(\left[\sin\theta\right]_0^{\pi/2}+\dfrac{1}{4}\left[\theta+\dfrac{\sin 2\theta}{2}\right]_0^{\pi/2}\)
\(1+\dfrac{1}{4}\cdot\dfrac{\pi}{2}\)
\(1+\dfrac{\pi}{8}\)
The full area from \(-\pi/2\) to \(\pi/2\) is twice this, because the integrand is even around 0.
\(2+\dfrac{\pi}{4}\)
Answer: \(2+\dfrac{\pi}{4}\)
That equation traces the same circle twice if you let \(\theta\) run a full turn: \(r\) is negative on half of the interval and retraces the circle.
Use \([-\pi/2,\pi/2]\) or \([0,\pi]\) for one trace. The area is \(\pi\), not \(2\pi\).
Answer: the circle is traced twice
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