Case III. Complex conjugate roots are of minor practical importance, and we discuss the derivation of real solutions from complex ones just in terms of a typical example
10.3 Calculus Review: Double Integrals
Optional
This section is optional. Students familiar with double integrals from calculus should skip this review and go on to Sec. 10.4. This section is included in the book to make it reasonably self-contained.
In a definite integral (1), Sec. 10.1, we integrate a function over an interval (a segment) of the x-axis. In a double integral we integrate a function , called the integrand,over a closed bounded region2Rin the xy-plane, whose boundary curve has a unique tangent at almost every point, but may perhaps have finitely many cusps (such as the vertices of a triangle or rectangle).
The definition of the double integral is quite similar to that of the definite integral. We subdivide the region Rby drawing parallels to the x- and y-axes (Fig. 227). We number the rectangles that are entirely within Rfrom 1 to n. In each such rectangle we choose a point, say, in the kth rectangle, whose area we denote by Then we form the sum
Jn⫽ a
n
k⫽1
f(xk, yk) ¢Ak.
¢Ak. (xk, yk)
f(x, y) f(x)
2A region Ris a domain (Sec. 9.6) plus, perhaps, some or all of its boundary points. Ris closedif its boundary (all its boundary points) are regarded as belonging to R; and Ris boundedif it can be enclosed in a circle of sufficiently large radius. A boundary pointPof Ris a point (of Ror not) such that every disk with center P contains points of Rand also points not of R.
y
x
Fig. 227. Subdivision of a region R 11. On Example 4. Show that in Example 4 of the text,
Give examples of domains in which the integral is path independent.
12. CAS EXPERIMENT. Extension of Project 10. Inte- grate over various circles through the points (0, 0) and (1, 1). Find experimentally the smallest value of the integral and the approximate location of the center of the circle.
13–19 PATH INDEPENDENCE?
Check, and if independent, integrate from (0, 0, 0) to (a, b,c).
13. 2ex2(x cos 2y dx⫺sin 2y dy) x2y dx⫹2xy2 dy F⫽grad (arctan (y>x)).
14.
15.
16.
17.
18.
19.
20. Path Dependence. Construct three simple examples in each of which two equations are satisfied, but the third is not.
(6r)
(cos (x2⫹2y2⫹z2)) (2x dx⫹4y dy⫹2z dz) (cos xy)(yz dx⫹xz dy)⫺2 sin xy dz
4y dx⫹z dy⫹(y⫺2z) dz ey dx⫹(xey⫺ez) dy⫺yez dz x2y dx⫺4xy2 dy⫹8z2x dz (sinh xy) (z dx⫺x dz)
R1
R2
Fig. 228. Formula (1)
This we do for larger and larger positive integers nin a completely independent manner, but so that the length of the maximum diagonal of the rectangles approaches zero as n approaches infinity. In this fashion we obtain a sequence of real numbers
Assuming that is continuous in Rand Ris bounded by finitely many smooth curves (see Sec. 10.1), one can show (see Ref. [GenRef4] in App. 1) that this sequence converges and its limit is independent of the choice of subdivisions and corresponding points This limit is called the double integral of over the region R, and is denoted by
or
Double integrals have properties quite similar to those of definite integrals. Indeed, for any functions fand gof (x, y), defined and continuous in a region R,
(kconstant)
(1)
(Fig. 228).
Furthermore, if Ris simply connected (see Sec. 10.2), then there exists at least one point in Rsuch that we have
(2)
where Ais the area of R. This is called the mean value theoremfor double integrals.
冮R冮f(x, y) dx dy⫽f(x0, y0)A, (x0, y0)
冮R冮f dx dy⫽ 冮R
1冮f dx dy⫹ 冮R
2冮f dx dy 冮R冮( f⫹g) dx dy⫽ 冮R冮f dx dy⫹ 冮R冮g dx dy
冮R冮kf dx dy⫽k冮R冮f dx dy 冮R冮f(x, y) dA.
冮R冮f(x, y) dx dy
f(x, y) (xk, yk).
f(x, y)
Jn1, Jn2,Á .
Evaluation of Double Integrals by Two Successive Integrations
Double integrals over a region R may be evaluated by two successive integrations.We may integrate first over yand then over x. Then the formula is
(3) 冮R冮f(x, y) dx dy⫽ 冮abc冮g(x)h(x)f(x, y) dyd dx (Fig. 229).
Here and represent the boundary curve of R(see Fig. 229) and, keeping xconstant, we integrate over yfrom to . The result is a function of x, and we integrate it from to (Fig. 229).
Similarly, for integrating first over xand then over ythe formula is
(4) 冮R冮f(x, y) dx dy⫽ 冮cdc冮p(y)q(y)f(x, y) dxd dy (Fig. 230).
x⫽b x⫽a
h(x) g(x) f(x, y)
y⫽h(x) y⫽g(x)
y
x h(x)
g(x) b a
R
y
x p(y)
q(y) c
d
R
Fig. 229. Evaluation of a double integral Fig. 230. Evaluation of a double integral
The boundary curve of R is now represented by and Treating yas a constant, we first integrate over xfrom to (see Fig. 230) and then the resulting function of yfrom to
In (3) we assumed that Rcan be given by inequalities and
Similarly in (4) by and If a region Rhas no such representation, then, in any practical case, it will at least be possible to subdivide Rinto finitely many portions each of which can be given by those inequalities. Then we integrate over each portion and take the sum of the results. This will give the value of the integral of
over the entire region R.
Applications of Double Integrals
Double integrals have various physical and geometric applications. For instance, the area Aof a region Rin the xy-plane is given by the double integral
The volumeVbeneath the surface and above a region Rin the xy-plane is (Fig. 231)
because the term in at the beginning of this section represents the volume of a rectangular box with base of area ¢Akand altitude f(xk, yk).
Jn f(xk, yk)¢Ak
V⫽ 冮R冮f(x, y) dx dy z⫽f(x, y) (⬎0)
A⫽ 冮R冮dx dy.
f(x, y)
f(x, y) p(y)⬉x⬉q(y).
c⬉y⬉d
g(x)⬉y⬉h(x).
a⬉x⬉b y⫽d.
y⫽c
q(y) p(y) f(x, y)
x⫽q(y).
x⫽p(y)
As another application, let be the density ( mass per unit area) of a distribution of mass in the xy-plane. Then the total massMin Ris
the center of gravityof the mass in Rhas the coordinates , where and
the moments of inertia and of the mass in Rabout the x- and y-axes, respectively, are
and the polar moment of inertia about the origin of the mass in Ris
An example is given below.
Change of Variables in Double Integrals. Jacobian
Practical problems often require a change of the variables of integration in double integrals.
Recall from calculus that for a definite integral the formula for the change from xto uis
(5) .
Here we assume that is continuous and has a continuous derivative in some
interval such that and varies
between aand bwhen uvaries between and .
The formula for a change of variables in double integrals from x, yto u, vis (6) 冮R冮f(x, y) dx dy⫽ 冮R*冮f(x(u, v), y(u, v)) 2 0(x, y)
0(u, v) 2 du dv;
b a
x(a)⫽a, x(b)⫽b[or x(a)⫽b, x(b)⫽a] x(u) a⬉u⬉b
x⫽x(u)
冮abf(x) dx⫽ 冮abf(x(u)) dxdu du
I0⫽Ix⫹Iy⫽ 冮R冮(x2⫹y2)f(x, y) dx dy.
I0
Ix⫽ 冮R冮y2f(x, y) dx dy, Iy⫽ 冮R冮x2f(x, y) dx dy;
Iy
Ix
y⫽ 1
M冮R冮yf(x, y) dx dy;
x⫽ 1
M冮R冮xf(x, y) dx dy
x, y M⫽ 冮R冮f(x, y) dx dy;
⫽ f(x, y)
z
x y
R
f(x, y)
Fig. 231. Double integral as volume
that is, the integrand is expressed in terms of uand v, and dx dyis replaced by du dvtimes the absolute value of the Jacobian3
(7)
Here we assume the following. The functions
effecting the change are continuous and have continuous partial derivatives in some region in the uv-plane such that for every (u, v) in the corresponding point (x, y) lies in Rand, conversely, to every (x, y) in Rthere corresponds one and only one (u, v) in ; furthermore, the Jacobian Jis either positive throughout or negative throughout . For a proof, see Ref. [GenRef4] in App. 1.
E X A M P L E 1 Change of Variables in a Double Integral
Evaluate the following double integral over the square Rin Fig. 232.
Solution. The shape of R suggests the transformation Then The Jacobian is
Rcorresponds to the square Therefore,
冮R冮(x2⫹y2) dx dy⫽冮02冮0212 (u2⫹v2) 1 䊏
2 du dv⫽8 3. 0⬉u⬉2, 0⬉v⬉2.
J⫽0(x, y) 0(u, v)⫽ †
1 2 1 2
1 2
⫺12
† ⫽ ⫺1 2. y⫽12(u⫺v).
x⫽12(u⫹v), x⫹y⫽u, x⫺y⫽v.
冮R冮(x2⫹y2) dx dy
R*
R*
R*
R*
R*
x⫽x(u, v), y⫽y(u, v) J⫽ 0(x, y)
0(u, v)⫽4 0x 0u
0x 0v 0y 0u
0y 0v
4⫽ 0x 0u
0y 0v ⫺ 0x
0v 0y 0u.
3Named after the German mathematician CARL GUSTAV JACOB JACOBI (1804–1851), known for his contributions to elliptic functions, partial differential equations, and mechanics.
v = 0
v = 2 u =
2
u = 0 y
1 x
Fig. 232. Region Rin Example 1
Of particular practical interest are polar coordinatesrand , which can be introduced
by setting Then
and (8)
where is the region in the -plane corresponding to Rin the xy-plane.
E X A M P L E 2 Double Integrals in Polar Coordinates. Center of Gravity. Moments of Inertia
Let be the mass density in the region in Fig. 233. Find the total mass, the center of gravity, and the moments of inertia
Solution. We use the polar coordinates just defined and formula (8). This gives the total mass
The center of gravity has the coordinates
for reasons of symmetry.
The moments of inertia are
for reasons of symmetry,
Why are and less than ?
This is the end of our review on double integrals. These integrals will be needed in this chapter, beginning in the next section.
䊏
1
y 2
x
I0⫽Ix⫹Iy⫽p
8⫽0.3927.
Iy⫽ p 16
⫽ 冮0p>218 (1⫺cos 2u) du⫽1 8ap
2 ⫺0b⫽p
16⫽0.1963 Ix⫽ 冮R冮y2 dx dy⫽ 冮0p>2冮01r2 sin2 ur dr du⫽冮0p>214 sin2 udu
y⫽ 4 3p x⫽ 4
p冮0p>2冮01r cos u r dr du⫽ 4
p冮0p>213 cos u du⫽ 4
3p⫽0.4244 M⫽冮R冮dx dy⫽ 冮0p>2冮01r dr du⫽冮0p>212 du⫽p
4. Ix, Iy, I0.
f(x, y)⫽1
ru R*
冮R冮f(x, y) dx dy⫽ 冮R*冮f(r cos u, r sin u) r dr du J⫽ 0(x, y)
0(r, u)⫽2cos u sin u
⫺r sin u r cos u2⫽r x⫽r cos u, y⫽r sin u.
u
Fig. 233.
Example 2
y
1 x
1. Mean value theorem. Illustrate (2) with an example.
2–8 DOUBLE INTEGRALS
Describe the region of integration and evaluate.
2.
3.
4. Prob. 3, order reversed.
5.
6.
7. Prob. 6, order reversed.
8.
9–11 VOLUME
Find the volume of the given region in space.
9. The region beneath and above the rectangle with vertices (0, 0), (3, 0), (3, 2), (0, 2) in the xy-plane.
10. The first octant region bounded by the coordinate planes
and the surfaces Sketch it.
11. The region above the xy-plane and below the parabo- loid .
12–16 CENTER OF GRAVITY
Find the center of gravity of a mass of density in the given region R.
12.
13.
h y
b x R
1 2 h
b b
y
x R
f(x, y)⫽1
(x, y) z⫽1⫺(x2⫹y2)
y⫽1⫺x2, z⫽1⫺x2. z⫽4x2⫹9y2
冮0p>4冮0cos yx2 sin y dx dy
冮02冮0ysinh (x⫹y) dx dy 冮01冮xx2
(1⫺2xy) dy dx
冮03冮⫺yy(x2⫹y2) dx dy 冮02冮x2x(x⫹y)2 dy dx
14.
15.
16.
17–20 MOMENTS OF INERTIA
Find of a mass of density in the region Rin the figures, which the engineer is likely to need, along with other profiles listed in engineering handbooks.
17. Ras in Prob. 13.
18. Ras in Prob. 12.
19.
20. y
– x
– 0
h
b 2
b 2
a 2 a
2
y
– x –
–
b 2
b 2 h 2
h 2 a
2
a 2
f(x, y)⫽1 Ix, Iy, I0
y
x R
y
x R
r
y
x r1 r2 R
P R O B L E M S E T 1 0 . 3