When we are defining the double integral of a function over a region R in the xy-plane, we cut R up into rectangles. Now we are using polar coordinates, so we will have polar rectangles whose sides have constant r- and q - values. Suppose that a function f (r, q) is defined over a region R that is bounded by the rays q = a and q = b and the continuous curves r₁ = g₁ (q) and r₂ = g ₂ (q). (See figure 1)

figure 1

Suppose 0 £ g₁ (q) £ g ₂ (q) £ a for all q between a and b. Then R lies in a fan-shaped region Q defined by the inequalities 0 £ r £ a and a £ q £ b. Cover Q with a grid of circular arcs and rays. The arcs are cut from circles centered at the origin, with radii D r, 2D r, . . . mD r, where D r = a / m. The rays are given by q = a, q = a + D q, q = a + 2D q, . . . q = a + m'D q = b, where D q = (b - a )/ m'. The arcs and rays partition Q into small patches called "polar rectangles". Now number the polar rectangles that lie inside R calling their areas D A₁, D A₂, . . . D A _n. Let (r _k, q _k) be the center of the polar rectangle whose area is D A _k. (By "center" we mean the point halfway between the circular arcs on the ray that bisects the arcs.) If I sum up these rectangles, I get the following sum.

If f is continuous throughout R, then this sum will approach a limit as we refined the grid to make D r, D q ® 0 and

figure 2

To evaluate this limit, we must write the sum S _n in a way that expresses D A _k in terms of D r and D q . The radius of the inner bounding arc is r _k - D r/ 2, and the radius of the outer bounding arc is r _k + D r/2. (See figure 2 )

The area of the smaller sector is

 and the area of the large sector is

D A _k = area of large sector - area of small sector

 Therefore,

and

 FACT:

EXAMPLE 1:

Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral.

SOLUTION:

x ² + y ² = 1 ® r = 1 and 0 £ q £ 2p .

EXAMPLE 2:

Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral.

SOLUTION:

x ² = 4 - y ² ® x ² + y ² = 4 ® r = 2, and we are working on the quarter circle. So 0 £ q £ p /2. Also x ² + y ² = r ².

EXAMPLE 3:

Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral.

SOLUTION:

1 - (x - 1) ² = y ² ® y ² + (x - 1) ² = 1 ® y ² + x ² - 2x + 1 = 1 ® r ² - 2r cos q = 0 ® r = 2cos q . When q = 0, r = 2, and when q = p /2, r = 0.

 The area of a closed and bounded region R in the polar coordinate plane is

EXAMPLE 4:

Find the area of the region that lies inside the cardiod r = 1 + cos q and outside the circle r = 1.

1 £ r £ 1 + cos q , so now I have to find the points of intersection.

EXAMPLE 5:

Find the centriod of the region enclosed by the cardiod r = 1 + cos q .

Since is the centriod that we want to find, then the density d = 1. Furthermore, 0 £ r £ 1 + cos q , and 0 £ q £ 2p .

EXAMPLE 6:

Find the average distance from a point P (x, y) in the disk x ² + y ² £ a ² in the xy-plane.

The distance from P (x, y) to the origin is

We are asked to find the average distance on the disk x ² + y ² £ a ², so we are looking at a circle with radius a. Therefore the area is p a ².