Postgraduate Education of Board Members and R&D Investment—Evidence from China
6. Implication and Limitation
√2ρ +ˆ 1
√2φ + 3ˆˆ z , b =√
2 ˆφ + 2ˆz .
In spherical coordinates, the unit vectors have Cartesian representations ˆr = (1/√
2. After similar calculations for b, we have a = 3
6.3.5. Repeat the computations of Exercise 6.3.4 assuming the two vectors to be associated with the Cartesian coordinate point (1,1,1).
Solution:
At the location of the vectors, the cylindrical unit vectors are the same as in Exercise 6.3.4 so the answers for cylindrical coordinates are the same as
6.4. MULTIPLE INTEGRALS 85 found in that Exercise. In spherical coordinates, the unit vectors are ˆr = (1/√ spherical components of a and b are therefore
ar= 2√
We find (in spherical coordinates) the scalar product a· b = (2√
6.3.6. In spherical polar coordinates (r, θ, φ) two points P and Q are respectively at (1, π/4, π/3) and (3, 5π/8, 7π/6). Find the distance between P and Q.
Solution:
The computation is simplest if we convert P and Q to Cartesian coordinates.
P = ( sin(π/4) cos(π/3), sin(π/4) sin(π/3), cos(π/4) )
= (0.353553, 0.612372, 0.707107)
Q = ( 3 sin(5π/8) cos(7π/6), 3 sin(5π/8) sin(7π/6), 3 cos(5π/8) )
= (−2.400309, −1.385819, −1.148050) Now form [P− Q| = 3.875324.
6.4 MULTIPLE INTEGRALS
Exercises
6.4.1. Evaluate by changing the order of integration. Check your work using symbolic computation.
∫∞
Changing the order of integration,
∫ ∞ To check using maple, execute
> int(int((y-x)/y*exp(-y),y=x .. infinity), x=0 .. infinity);
mathematica will not evaluate this integral in its original form, but it can check the result after the order of the integrations has been interchanged.
6.4.2. Evaluate
Check this result using symbolic computation.
86 CHAPTER 6. MULTIDIMENSIONAL PROBLEMS Solution:
Reverse the order of integration, reaching
∫ ∞
The first of the above integrals evaluates to unity. An easy way to evaluate the second integral is to identify it as the real part of
∫ ∞
0
e(i−1)tdt = 1
1− i =1 + i 2 . Our overall result is 1
2 To check using maple, execute
> int(int(y^2*exp(-y^2)*sin(x*y),y=x .. infinity),
> x=0 .. infinity):
Using mathematica,
Integrate[Integrate[y^2*E^(-y^2)*Sin[x*y],
{y, x, Infinity}], {x, 0, Infinity}];
Both symbolic systems give the correct result, 1/4.
6.4.3. Evaluate
where S is a disk of radius c centered at the origin.
Solution: 6.4.4. Find the center of mass of a uniform solid hemisphere H of radius a.
Solution:
Put the flat side of the hemisphere H on the xy-plane with the sphere center at the origin and the curved surface at positive z. By symmetry the center of mass will be on the z-axis and given by the expression
(2πa3 3
)−1∫
H
z d3r .
6.4. MULTIPLE INTEGRALS 87 Perform the integration by summing over disks at each z, of thickness dz and volume π(a2− z2)dz. Each disk can be treated as a point mass at its center, so the center of mass will be on the z axis, given by
⟨z⟩ = 3
for a cylinder of height h and radius a with its center at the origin and its axis in the x-direction.
Solution:
Use cylindrical coordinates with the cylindrical z coordinate in the x-direction of the original Cartesian system. Then x2y2z2 becomes ρ4z2cos2φ sin2φ, and the volume element is ρ dρ dz dφ. The integral is
∫ h/2
The matrix product JK has elements that are chain-rule expressions:
JK =
6.4.7. Introduce a change of variables that makes this double integral elementary, and then evaluate it. Use symbolic computation to check your work.
∫ ∞ It is easiest to compute the Jacobian by evaluating
∂(s, t)
88 CHAPTER 6. MULTIDIMENSIONAL PROBLEMS and then take its reciprocal. Therefore,
I =
It is also possible to obtain this result by reversing the order of integration, performing the y integral, writing the resulting x integrand in the form f (x)e−x, then expanding f (x) in a power series and integrating in x term-by-term.
6.5 LINE AND SURFACE INTEGRALS
Exercises
6.5.1. Evaluate the line integral
∫
(2xy− y2) dx , where the integration is on the curve y = x3/2 from x = 0 to x = 4.
Solution:
Write entirely in terms of x:
∫ 4 6.5.2. Evaluate the line integral
∫
B· dr, where B = x ˆex− xy ˆey, for a straight-line path from (x, y) = (0, 0) to (x, y) = (2, 1).
Solution:
The integral can be written
∫
B· dr =
∫
(x dx− xy dy) .
Setting x = 2y in the second term and identifying the integration limits,
∫
6.5.3. A wire of uniform density lies along the portion of the parabola y = x2/2 that extends from (x, y) = (0, 0) to (x, y) = (2, 2). Find the position of its centroid.
Hint: Use symbolic computation to evaluate the necessary integrals.
Solution:
The integrals that are needed are L =
In maple, these integrals are
> L := int(sqrt(1+x^2), x = 0 .. 2):
> X := (1/L)*int(x*sqrt(1+x^2), x = 0 .. 2):
> Y := (1/L)*int((x^2/2)*sqrt(1+x^2), x = 0 .. 2):
> evalf(X), evalf(Y); 1.14725, 0.819960
6.5. LINE AND SURFACE INTEGRALS 89 In mathematica,
l = Integrate[Sqrt[1+x^2], {x,0,2} ];
xx = (1/l)*Integrate[x*Sqrt[1+x^2], {x,0,2} ];
yy = (1/l)*Integrate[x^2/2*Sqrt[1+x^2], {x,0,2} ];
{ N[xx], N[yy] } 1.14725, 0.81996
6.5.4. For the curved surface of a cylinder of radius r0with axis on the z axis and extending from z =−h to z = +h, evaluate ∫
6.5.5. For the curved cylindrical surface of Exercise 6.5.4, evaluate (a)
(a) Zero by symmetry;
∫
x2z2dA =
∫
y2z2dA.
(b) The xy term vanishes because it is an odd function of x (and also y). Thus, we compute
6.5.6. For the curved surface of the hemisphere of radius r0 and z≥ 0, evaluate
∫
6.5.7. What does Green’s theorem tell us (for an arbitrary area A) if we choose P (x, y) = x and Q(x, y) = 0? What is the general result if we apply the theorem to P (x, y) = 0, Q(x, y) = y?
90 CHAPTER 6. MULTIDIMENSIONAL PROBLEMS Solution:
If we set P (x, y) = x and Q(x, y) = 0, the integral over A becomes
∫
A
(1− 0) dA = A (the area enclosed).
If we take P (x, y) = 0 and Q(x, y) = y, the integral over A is−A. We therefore have the following line integrals for the area within a closed curve:
I
∂A
y dx =− I
∂A
x dy = A .
6.5.8. Verify a particular case of Green’s theorem by evaluation of the integrals on both sides of Eq. (6.46) when we choose ∂A to be the unit circle, P (x, y) = 0, and Q(x, y) = x3− y3.
Solution:
This case of Green’s theorem states I
(x2− y2) dx =
∫
A
2y dA = 0 .
The integral over A vanishes due to symmetry. Write the line integral in terms of the angle φ:
∫ 2π 0
(cos2φ− sin2φ)
(− sin φdφ) =
∫ 2π 0
(2 cos2φ− 1)
d cos φ = 0 . This zero result confirms Green’s theorem.
6.6 REARRANGEMENT OF DOUBLE SERIES
Exercises
6.6.1. Verify that the sums claimed to be equivalent in Example 6.6.1,
∑∞ p,q=0
xpyq p! q! and
∑∞ n=0
∑n p=0
xpyn−p p! (n− p)!, actually give identical results for x = 0.1, y =−0.3.
Solution:
In maple,
> x := 0.1: y := -0.3:
> evalf(sum(sum(x^p*y^q/p!/q!, p=0 .. infinity),
q=0 .. infinity)); 0.818731
> evalf(sum(sum(x^p*y^(n-p)/p!/(n-p)!,p=0 .. n),
n=0 .. infinity)); 0.818731
6.7. DIRAC DELTA FUNCTION 91 In mathematica,
x = 0.1; y = -0.3;
N[Sum[Sum[x^p*y^q/p!/q!, {p,0,Infinity} ],
{q,0,Infinity} ] ] 0.818731
N[Sum[Sum[x^p*y^(n-p)/p!/(n-p)!, {p,0,n} ],
{n,0,Infinity} ] ] 0.818731
6.6.2. Verify the equivalence of the following two double summations by comparing the coefficients of xptqfor all p and q when both are no larger than 6.
∑∞ n=0
∑n k=0
(−1)k(2n)!
22nk! n! (n− k)!(2x)n−ktn+k,
∑∞ n=0
[n/2]∑
k=0
(−1)k(2n− 2k)!
22n−2kk! (n− k)! (n − 2k)!(2x)n−2ktn. The notation [n/2] denotes the largest integer less than or equal to n/2.
Solution:
Expand, using symbolic computing, with finite limits that include all terms with both variables at powers no larger than 6. Then subtract one expansion from the other and check that the difference contains no terms with both x and t to powers≤ 6.
In maple, code that accomplishes this can be (not showing the output)
> S1 := expand(add(add((-1)^k*(2*m)!/2^(2*m)/k!/m!/(m-k)!
> *(2*x)^(m-k)*t^(m+k), k = 0 .. m), m = 0 .. 6));
> S2 := 0: for n from 0 to 6 do kk := floor(n/2);
> S2 := S2 + add((-1)^k*(2*n-2*k)!/2^(2*n-2*k)/k!/(n-k)!
> /(n-2*k)!*(2*x)^(n-2*k)*t^n, k = 0 .. kk) end do: S2;
> S1 - S2;
In mathematica, suitable code can be (suppressing the output)
S1 = Sum[Sum[(-1)^k*(2*m)!/2^(2*m)/k!/m!/(m-k)!*(2*x)^(m-k)
*t^(m+k), {k, 0, m}], {m, 0, 6}]
S2 = Sum[Sum[(-1)^k*(2*n-2*k)!/2^(2*n-2*k)/k!/(n-k)!
/(n-2*k)!*(2*x)^(n-2*k)*t^n,{k, 0, n/2}], {n, 0, 6}]
S1 - S2
6.7 DIRAC DELTA FUNCTION
Exercises
6.7.1. Show, that δ(ax) = δ(x)/|a|. Make a detailed argument justifying the absolute value signs that enclose a. Your argument should not depend upon the specific representation describing δ(x).
92 CHAPTER 6. MULTIDIMENSIONAL PROBLEMS Solution:
Let c be a finite positive number, and change the x integral below by writing it in terms of y = ax:
If a < 0 the integration is in the direction opposite to that defining a delta function; we can reverse the limits if we replace a by|a|.
6.7.2. Apply an integration by parts to establish Eq. (6.59).
Solution:
Letting c be a finite positive number, consider
∫ c
The generalized function δ(f (x)) will only be nonzero at the zeros of f ; if there is a zero at xi, the leading contribution to f in its neighborhood will be f′(xi)(x− xi), and the behavior of δ(f (x)) will be that of δ(x− xi)/|f′(xi)|, as indicated by Eq. (6.56). Similar contributions arise from all the zeros of f (x).
6.7.4. Justify Eq. (6.58).
Solution:
Near x = x1, δ[(x− x1)(x− x2)] approaches δ[a(x− x1)] with a = x1− x2, and therefore approaches δ(x−x1)/|x1−x2|. Near x = x2, we have δ(x−z2)/|x2−x1|.
These two contributions correspond to Eq. (6.58).
6.7.5. Show that the Heaviside and Dirac delta functions are related by u′(x− x0) = δ(x− x0) by for a general “well-behaved” f (x) and arbitrary values of a and b.
Solution:
Since u(x) is constant everywhere except at x = 0, we have u′(x) = 0 for all nonzero x. It is also the case that δ(x) is zero for all nonzero x, so what we need to prove reduces to
εlim→0 Integrating by parts, we reach
εlim→0 The first term within the limit approaches f (x0) and the second is zero, as is the third (the integral), which has a finite integrand and an infinitesimal interval of integration. The fourth term,−f(x0), cancels the first, showing the overall limit to be zero.