Area enclosed by a polar curve

Polar coordinates

A-level Further Maths (9FM0) · Core Pure

ORIGINAL

Half the integral of r squared — expand, use the double-angle identity, and integrate over one full turn.

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Question

The curve

C

has polar equation

r = 4 + 3\cos\theta

0 \le \theta < 2\pi

. Find the exact area enclosed by

C

Every step worked, with the reasoning.

1
$A = \dfrac12\displaystyle\int_0^{2\pi} r^2\,d\theta = \dfrac12\int_0^{2\pi} (4 + 3\cos\theta)^2\,d\theta$
Use the polar area formula. $r = 4 + 3\cos\theta \ge 1 > 0$ for every $\theta$ , so one full turn traces the curve exactly once.
2
$(4 + 3\cos\theta)^2 = 16 + 24\cos\theta + 9\cos^2\theta$
Expand the bracket.
3
$= 16 + 24\cos\theta + \tfrac92(1 + \cos 2\theta) = \tfrac{41}{2} + 24\cos\theta + \tfrac92\cos 2\theta$
Use $\cos^2\theta = \tfrac12(1 + \cos 2\theta)$ so every term can be integrated directly.
4
$A = \dfrac12\Big[\tfrac{41}{2}\theta + 24\sin\theta + \tfrac94\sin 2\theta\Big]_0^{2\pi}$
Integrate term by term: $\int \tfrac92\cos 2\theta\,d\theta = \tfrac94\sin 2\theta$ .
5
$A = \dfrac12(41\pi + 0 + 0) - 0 = \dfrac{41\pi}{2}$
$\sin 2\pi = \sin 4\pi = 0$ , and every term is $0$ at the lower limit.

Answer: $\dfrac{41\pi}{2}$