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). rsin - y=√/2orr=√√/2csc 022√₂Leaves at r = 2LRREnters at r(b)Largest 0 isL√2 csc 0Smallest is.>XFIGURE 15.24 Finding the limits ofintegration 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 as8-B pr-gy(0)f(r, 0)r dr do.[[f(r, 9) dA =Finding Limits of IntegrationThe procedure for finding limits of integration in rectangular coordinates also works forpolar coordinates. We illustrate this using the region R shown in Figure 15.24. To evaluatef(r, 0) dA in polar coordinates, integrating first with respect to r and then with respectto 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 thedirection of increasing r. Mark the r-values where L enters and leaves R. These are ther-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 is8/2[[ f(r. 0 ) dA =A =6/4 Jr=√₂ csc 9f(r, 0) r dr d0.EXAMPLE 1 Find the limits of integration for integrating f(r, 8) over the region Rthat lies inside the cardioid r = 1 + cos 0 and outside the circle r = 1.Solution1. 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 wherer = 1 and leaves where r = 1 + cos 0. Entersatr=1r=1+cos 02Enters at7 = 0Leaves atr = 1 + cosFIGURE 15.25 Finding the limits ofintegration in polar coordinates for theregion in Example 1.Area Differential in Polar CoordinatesdA = r dr doLeaves atr=V/4 cos 26X²-4 cos 20FIGURE 15.26 To integrate overthe 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 & runfrom 0 = -7/2 to 0 = 7/2. The integral is•/2 pl+c88f(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 thearea of R.Area in Polar CoordinatesThe area of a closed and bounded region R in the polar coordinate plane is15.4 Double Integrals in Polar Form 917A = 4- If= 4A =This formula for area is consistent with all earlier formulas.EXAMPLE 2Find the area enclosed by the lemniscate 2² = 4 cos 20.Solution We graph the lemniscate to determine the limits of integration (Figure 15.26) andsee from the symmetry of the region that the total area is 4 times the first-quadrant portion.#/4 V4 cos 26#/4r=V/4 con 28" r dr do = 4-4.****dor dr do./42 cos 20 d0 - 4 sin 201/4Jo= 4.Changing Cartesian Integrals into Polar IntegralsThe procedure for changing a Cartesian integral f(x, y) dx dy into a polar integral hastwo steps. First substitute x = r cos 8 and y = r sin 8, and replace dx dy by r dr do in theCartesian integral. Then supply polar limits of integration for the boundary of R. The Car-tesian integral then becomes

Answered: Summarize how to find the limits of…

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

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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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).

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.

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

With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.

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