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  • Apply the formula for area of a region in polar coordinates.
  • Determine the arc length of a polar curve.

In the rectangular coordinate system, the definite integral provides a way to calculate the area under a curve. In particular, if we have a function y = f ( x ) defined from x = a to x = b where f ( x ) > 0 on this interval, the area between the curve and the x -axis is given by A = a b f ( x ) d x . This fact, along with the formula for evaluating this integral, is summarized in the Fundamental Theorem of Calculus. Similarly, the arc length of this curve is given by L = a b 1 + ( f ( x ) ) 2 d x . In this section, we study analogous formulas for area and arc length in the polar coordinate system.

Areas of regions bounded by polar curves

We have studied the formulas for area under a curve defined in rectangular coordinates and parametrically defined curves. Now we turn our attention to deriving a formula for the area of a region bounded by a polar curve. Recall that the proof of the Fundamental Theorem of Calculus used the concept of a Riemann sum to approximate the area under a curve by using rectangles. For polar curves we use the Riemann sum again, but the rectangles are replaced by sectors of a circle.

Consider a curve defined by the function r = f ( θ ) , where α θ β . Our first step is to partition the interval [ α , β ] into n equal-width subintervals. The width of each subinterval is given by the formula Δ θ = ( β α ) / n , and the i th partition point θ i is given by the formula θ i = α + i Δ θ . Each partition point θ = θ i defines a line with slope tan θ i passing through the pole as shown in the following graph.

On the polar coordinate plane, a curve is drawn in the first quadrant, and there are rays from the origin that intersect this curve at a regular interval. Every time one of these rays intersects the curve, a perpendicular line is made from the ray to the next ray. The first instance of a ray-curve intersection is labeled θ = α; the last instance is labeled θ = β. The intervening ones are marked θ1, θ2, …, θn−1.
A partition of a typical curve in polar coordinates.

The line segments are connected by arcs of constant radius. This defines sectors whose areas can be calculated by using a geometric formula. The area of each sector is then used to approximate the area between successive line segments. We then sum the areas of the sectors to approximate the total area. This approach gives a Riemann sum approximation for the total area. The formula for the area of a sector of a circle is illustrated in the following figure.

A circle is drawn with radius r and a sector of angle θ. It is noted that A = (1/2) θ r2.
The area of a sector of a circle is given by A = 1 2 θ r 2 .

Recall that the area of a circle is A = π r 2 . When measuring angles in radians, 360 degrees is equal to 2 π radians. Therefore a fraction of a circle can be measured by the central angle θ . The fraction of the circle is given by θ 2 π , so the area of the sector is this fraction multiplied by the total area:

A = ( θ 2 π ) π r 2 = 1 2 θ r 2 .

Since the radius of a typical sector in [link] is given by r i = f ( θ i ) , the area of the i th sector is given by

A i = 1 2 ( Δ θ ) ( f ( θ i ) ) 2 .

Therefore a Riemann sum that approximates the area is given by

A n = i = 1 n A i i = 1 n 1 2 ( Δ θ ) ( f ( θ i ) ) 2 .

We take the limit as n to get the exact area:

A = lim n A n = 1 2 α β ( f ( θ ) ) 2 d θ .

This gives the following theorem.

Area of a region bounded by a polar curve

Suppose f is continuous and nonnegative on the interval α θ β with 0 < β α 2 π . The area of the region bounded by the graph of r = f ( θ ) between the radial lines θ = α and θ = β is

A = 1 2 α β [ f ( θ ) ] 2 d θ = 1 2 α β r 2 d θ .

Finding an area of a polar region

Find the area of one petal of the rose defined by the equation r = 3 sin ( 2 θ ) .

The graph of r = 3 sin ( 2 θ ) follows.

A four-petaled rose with furthest extent 3 from the origin at π/4, 3π/4, 5π/4, and 7π/4.
The graph of r = 3 sin ( 2 θ ) .

When θ = 0 we have r = 3 sin ( 2 ( 0 ) ) = 0 . The next value for which r = 0 is θ = π / 2 . This can be seen by solving the equation 3 sin ( 2 θ ) = 0 for θ . Therefore the values θ = 0 to θ = π / 2 trace out the first petal of the rose. To find the area inside this petal, use [link] with f ( θ ) = 3 sin ( 2 θ ) , α = 0 , and β = π / 2 :

A = 1 2 α β [ f ( θ ) ] 2 d θ = 1 2 0 π / 2 [ 3 sin ( 2 θ ) ] 2 d θ = 1 2 0 π / 2 9 sin 2 ( 2 θ ) d θ .

To evaluate this integral, use the formula sin 2 α = ( 1 cos ( 2 α ) ) / 2 with α = 2 θ :

A = 1 2 0 π / 2 9 sin 2 ( 2 θ ) d θ = 9 2 0 π / 2 ( 1 cos ( 4 θ ) ) 2 d θ = 9 4 ( 0 π / 2 1 cos ( 4 θ ) d θ ) = 9 4 ( θ sin ( 4 θ ) 4 | 0 π / 2 = 9 4 ( π 2 sin 2 π 4 ) 9 4 ( 0 sin 4 ( 0 ) 4 ) = 9 π 8 .
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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