Question 2: Explain geometrically how the formula for a single A is obtained. Explain why this area depends on r and not just r. Note that this is about just a single "polar rectangle" and not the Riemann sum as a whole.

Hint 1: Read the subsection "Integrals in Polar Coordinates" (p. 915 - 916).

Hint 2: This extra r appears in the double integral (middle of p. 916) and will notably appear in every double integral in polar coordinates, including the area formula (boxed on p. 917). This extra r factor can be hard for people to remember, but hopefully this exercise helps explain why it's there: the area of these "polar rectangles" depends on how far away you are from the origin and not just its dimensions.

15.4 Double Integrals in Polar Form 915 Integrals in Polar Coordinates When we defined the double integral of a function over a region R in the xy-plane, we began by cutting R into rectangles whose sides were parallel to the coordinate axes. These were the natural shapes to use because their sides have either constant x-values or constant y-values. In polar coordinates, the natural shape is a "polar rectangle" whose sides have constant y- and #-values. To avoid ambiguities when describing the region of integration with polar coordinates, we use polar coordinate points (r, 0) where r 2 0. Suppose that a function f(r, #) is defined over a region R that is bounded by the rays 0 = a and 0 = # and by the continuous curves r = g,(0) and r = $2(0). Suppose also that 0 s g(0) s g(0) s a for every value of of between a and B. Then R lies in a fan- shaped region @ defined by the inequalities 0 = / = a and a S 0 = B, where 0 = B - a = 27. See Figure 15.22. + 240 D = B Ar 3Ar 2Ar = 0 FIGURE 15.22 The region R: 8(8) = = $7(8), a = # = B, is contained in the fan- shaped region Q:0 s r s a, a s 0 s B, where 0 s B - a s 2w. The partition of @ by circular ares and rays induces a partition of R. We cover @ by a grid of circular ares and rays. The ares are cut from circles centered at the origin, with radii Ar. 2Ar. . . . . mar. where Ar = /m. The rays are given by 8 = a + AB. 8 = a + 248. 8 = a + m'Ae = B.where A0 = (8 - a)/m'. The ares and rays partition O into small patches called "polar rectangles." We number the polar rectangles that lie inside R (the order does not matter), calling their areas AA. AAy. .... AA,. We let (n, 6:) be any point in the polar rectangle whose area is AA,. We then form the sum 51 = If f is continuous throughout R, this sum will approach a limit as we refine the grid to make Ar and A0 go to zero. The limit is called the double integral of f over R. In symbols, lim S = Small seclor 1-+DO Large sector To evaluate this limit, we first have to write the sum S,, in a way that expresses AA, in terms of Ar and Ad. For convenience we choose a to be the average of the radii of the inner and arcs bounding the &th polar rectangle AA. The radius of the inner are bounding then a - (Ar/2) (Figure 15.23). The radius of the outer are is FIGURE 15.23 The observation that n + (Ar/2). area of area of The area of a wedge-shaped sector of a circle having radius r and angle @ is AA = large sector small sector leads to the formula AM, = n Ar &0. A = 20-r.916 Chapter 15 Multiple Integrals as can be seen by multiplying w/. the area of the circle, by #/2w, the fraction of the cir- cle's area contained in the wedge. So the areas of the circular sectors subtended by these ares at the origin are Inner radius: 2 2 Outer radius: AO. Therefore. AA, = area of large sector - area of small sector = Combining this result with the sum defining 5, gives R 1/2 (V.V2) S. = Efn. Ban Ar Ad. As n - co and the values of Ar and Ad approach zero, these sums converge to the double integral lim S, = = f(r. 0) r dr do.Leaves at / = 2 A version of Fubini's Theorem says that the limit approached by these sums can be evalu- ated by repeated single integrations with respect to r and d as f(r, 0) r dr do. or R r= V2ocd Enters at = v/2 ca o Finding Limits of Integration The procedure for finding limits of integration in rectangular coordinates also works for [b) polar coordinates. We illustrate this using the region & shown in Figure 15.24. To evaluate (/ f(r, #) dA in polar coordinates, integrating first with respect to r and then with respect Largest O is to 8, take the following steps. 1. Sketch. Sketch the region and label the bounding curves (Figure 15.24a). 2. Find the r-limits of integration. Imagine a ray & from the origin cutting through R in the direction of increasing r. Mark the r-values where /. enters and leaves R. These are the Smallest o is r-limits of integration. They usually depend on the angle o that L makes with the posi- tive x-axis (Figure 15.24b). 3. Find the O-limits of integration. Find the smallest and largest 0-values that bound R. These are the 0-limits of integration (Figure 15.24c). The polar iterated integral is (C) FIGURE 15.24 Finding the limits of integration in polar coordinates. EXAMPLE 1 Find the limits of integration for integrating f(r. @) over the region R that lies inside the cardioid y = 1 + cos # and outside the circle r = 1. Solution 1. We first sketch the region and label the bounding curves (Figure 15.25). 2. Next we find the r-limits of integration. A typical ray from the origin enters R where 7 = 1 and leaves where r = 1 + cose.15.4 Double Integrals in Polar Form 917 3. Finally we find the 0-limits of integration. The rays from the origin that intersect R run from 0 = -w/2 to 0 = w/2. The integral is fir, Or dr do. 2 If f(r. 0) is the constant function whose value is 1, then the integral of f over R is the area of R. Enters Leaves at at p = 1 + cos 8 Area in Polar Coordinates = 1 The area of a closed and bounded region R in the polar coordinate plane is FIGURE 15.25 Finding the limits of integration in polar coordinates for the A = / rar del. region in Example 1. Area Differential in Polar Coordinates This formula for area is consistent with all earlier formulas. dA = r dr do EXAMPLE 2 Find the area enclosed by the lemniscate r = 4 cos 20. Solution We graph the lemniscate to determine the limits of integration (Figure 15.26) and see from the symmetry of the region that the total area is 4 times the first-quadrant portion. V4 cos 20 de Leaves at =V4008 28 = 4 2 cos 20 af) = 4 sin 20 1=/4 = 4. Enters at P= = 4 cus 20 Changing Cartesian Integrals into Polar Integrals The procedure for changing a Cartesian integral JJ, /(x. y) edx dy into a polar integral has FIGURE 15.26 To integrate over two steps. First substitute x = roos # and y = r sin d, and replace dx dy by r dr de in the the shaded region, we run r from 0 to Cartesian integral. Then supply polar limits of integration for the boundary of R. The Car- V4 cos 28 and # from 0 to #/4 tesian integral then becomes (Example 2)f(x, y) dx dy = f(r cos 0, r sin 0) r dr do, G where G denotes the same region of integration, but now described in polar coordinates. This is like the substitution method in Chapter 5 except that there are now two variables to substitute for instead of one. Notice that the area differential dx dy is not replaced by dr do but by r dr d0. A more general discussion of changes of variables (substitutions) in multi- ple integrals is given in Section 15.8. VI-1 EXAMPLE 3 Evaluate = pty dy dx, 8=0 where R is the semicircular region bounded by the x-axis and the curve y = VI - x (Figure 15.27). FIGURE 15.27 The semicircular region Solution In Cartesian coordinates, the integral in question is a nonelementary integral in Example 3 is the region and there is no direct way to integrate e" IT with respect to either x or y. Yet this integral 05151, OSAST. and others like it are important in mathematics-in statistics, for example-and we need