Area in Polar Coordinates

One of the most elegant applications of polar coordinates is computing the area enclosed by a polar curve. The formula differs from the Cartesian one and is a key part of the AQA Further Mathematics specification.

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1 The Area Formula

Derivation

Consider a polar curve r = f(theta). A thin "sector" from angle theta to theta + delta theta has:

• Radius approximately r = f(theta)
• Angle delta theta

The area of this sector is approximately (1/2) r^2 delta theta (using the sector area formula).

Summing all such sectors and taking the limit:

A = (1/2) integral from alpha to beta of r^2 d theta

where alpha and beta are the angular limits.

Key Point: This gives the area swept by the radius vector from theta = alpha to theta = beta. It is the area between the curve and the pole.

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2 Worked Example 1 -- Circle

Find the area enclosed by r = a.

Solution:

A = (1/2) integral from 0 to 2pi of a^2 d theta = (1/2) a^2 [theta] from 0 to 2pi = (1/2) a^2 (2pi) = pi a^2

This confirms the well-known formula for the area of a circle.

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3 Worked Example 2 -- Cardioid

Find the area enclosed by r = a(1 + cos theta).

Solution:

The curve is traced completely for 0 <= theta <= 2pi. By symmetry about the x-axis, the area is twice the area from 0 to pi.

A = 2 * (1/2) integral from 0 to pi of [a(1 + cos theta)]^2 d theta

= a^2 integral from 0 to pi of (1 + cos theta)^2 d theta

= a^2 integral from 0 to pi of (1 + 2 cos theta + cos^2 theta) d theta

For cos^2 theta, use the identity cos^2 theta = (1 + cos 2 theta)/2:

= a^2 integral from 0 to pi of (3/2 + 2 cos theta + cos(2 theta)/2) d theta

= a^2 [3 theta/2 + 2 sin theta + sin(2 theta)/4] from 0 to pi

= a^2 [(3pi/2 + 0 + 0) - (0 + 0 + 0)]

= 3pi a^2/2

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4 Worked Example 3 -- Rose Curve

Find the area of one petal of r = a cos(2 theta).

Solution:

One petal (in the range where r >= 0) extends from theta = -pi/4 to theta = pi/4.

A = (1/2) integral from -pi/4 to pi/4 of [a cos(2 theta)]^2 d theta

= (a^2/2) integral from -pi/4 to pi/4 of cos^2(2 theta) d theta

= (a^2/2) integral from -pi/4 to pi/4 of (1 + cos(4 theta))/2 d theta

= (a^2/4) [theta + sin(4 theta)/4] from -pi/4 to pi/4

= (a^2/4) [(pi/4 + 0) - (-pi/4 + 0)]

= (a^2/4)(pi/2)

= pi a^2/8

Since there are 4 petals, the total area is 4 * pi a^2/8 = pi a^2/2.

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5 Worked Example 4 -- Spiral Region

Find the area swept by r = theta from theta = 0 to theta = 2pi.

Solution:

A = (1/2) integral from 0 to 2pi of theta^2 d theta

= (1/2) [theta^3/3] from 0 to 2pi

= (1/2)(8pi^3/3)

= 4pi^3/3

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6 Area Between Two Polar Curves

If two curves r = f(theta) and r = g(theta) satisfy f(theta) >= g(theta) >= 0 for alpha <= theta <= beta, the area between them is: