Question 3: Summarize how to find the limits of integration when integrating in order dr dθ. We will typically integrate in this order.

Hint 1: Read the subsection “Finding Limits of Integration” (p. 916 – 917).
Transcribed Image Text:rsin - y=√/2 or r=√√/2csc 0 2 2 √₂ Leaves at r = 2 L R R Enters at r (b) Largest 0 is L √2 csc 0 Smallest is. >X FIGURE 15.24 Finding the limits of integration in polar coordinates. 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 as 8-B pr-gy(0) f(r, 0)r dr do. [[ f(r, 9) dA = Finding Limits of Integration The procedure for finding limits of integration in rectangular coordinates also works for polar coordinates. We illustrate this using the region R shown in Figure 15.24. To evaluate f(r, 0) dA in polar coordinates, integrating first with respect to r and then with respect to 0, 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 L. from the origin cutting through R in the direction of increasing r. Mark the r-values where L enters and leaves R. These are the r-limits of integration. They usually depend on the angle that L makes with the posi- tive x-axis (Figure 15.24b). 3. Find the 0-limits of integration. Find the smallest and largest 9-values that bound R. These are the 0-limits of integration (Figure 15.24c). The polar iterated integral is 8/2 [[ f(r. 0 ) dA = A = 6/4 Jr=√₂ csc 9 f(r, 0) r dr d0. EXAMPLE 1 Find the limits of integration for integrating f(r, 8) over the region R that lies inside the cardioid r = 1 + cos 0 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 r = 1 and leaves where r = 1 + cos 0.
Transcribed Image Text:Enters at r=1 r=1+cos 0 2 Enters at 7 = 0 Leaves at r = 1 + cos FIGURE 15.25 Finding the limits of integration in polar coordinates for the region in Example 1. Area Differential in Polar Coordinates dA = r dr do Leaves at r=V/4 cos 26 X ²-4 cos 20 FIGURE 15.26 To integrate over the shaded region, we run from 0 to √4 cos 20 and 6 from 0 to 7/4 (Example 2). 3. Finally we find the 0-limits of integration. The rays from the origin that intersect & run from 0 = -7/2 to 0 = 7/2. The integral is •/2 pl+c88 f(r, 0) r dr do. If f(r, 0) is the constant function whose value is 1, then the integral of f over R is the area of R. Area in Polar Coordinates The area of a closed and bounded region R in the polar coordinate plane is 15.4 Double Integrals in Polar Form 917 A = 4 - If = 4 A = This formula for area is consistent with all earlier formulas. EXAMPLE 2 Find the area enclosed by the lemniscate 2² = 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. #/4 V4 cos 26 #/4 r=V/4 con 28 " r dr do = 4 -4.**** do r dr do. /4 2 cos 20 d0 - 4 sin 20 1/4 Jo = 4. Changing Cartesian Integrals into Polar Integrals The procedure for changing a Cartesian integral f(x, y) dx dy into a polar integral has two steps. First substitute x = r cos 8 and y = r sin 8, and replace dx dy by r dr do in the Cartesian integral. Then supply polar limits of integration for the boundary of R. The Car- tesian integral then becomes

Question

Question 3: Summarize how to find the limits of integration when integrating in order dr dθ. We will typically integrate in this order.

Hint 1: Read the subsection “Finding Limits of Integration” (p. 916 – 917).

Accepted Answer

Let us summarize how to find the limits of integration when integrating in the order drdθ.