1002微微微甲甲甲08-12班班班期期期末末末考考考解解解答答答和和和評評評分分分標標標準準準 1. (12%) Evaluate
Z 1 0
Z 1−x 0
Z 1 y
sin(πz)
z(2 − z) dzdydx. [Hint: switch dzdydx into dxdydz.]
Solution:
We need to compute
Z 1 0
Z 1−x 0
Z 1 y
sin(πz)
z(2 − z)dzdydx.
The region of integration is bounded by [0, 1] × [0, 1] × [0, 1] ∩ {(x, y, z) ∈ R3|x + y ≤ 1, z ≥ y}. (4% If you derived the correct region or drawed a correct graph indicating the correct region)
Therefore, Z 1
0
Z 1−x 0
Z 1 y
f (x, y, z)dzdydx = Z 1
0
Z z 0
Z 1−y 0
f (x, y, z)dxdydz. (4%; you won’t get full points if you did not provide your reasons)
Hence
Z 1 0
Z 1−x 0
Z 1 y
sin(πz)
z(2 − z)dzdydx = Z 1
0
Z z 0
Z 1−y 0
sin(πz)
z(2 − z)dxdydz
= Z 1
0
sin(πz) z(2 − z)
Z z 0
(1 − y)dydz
= Z 1
0
sin(πz)
z(2 − z)(z − z2 2 )dz
= Z 1
0
1
2sin(πz)dz = − 1
2πcos(πz)
1
0
= 1
π. (4%)
2. (12%) Evaluate Z Z
R
e−4x2+12xy−10y2 dxdy where R is the region satisfying x ≥ 2y and y ≥ 0.
[Hint: rewrite 4x2− 12xy + 10y2 as u2+ v2by suitable changing of coordinates.]
Solution:
Let 4x2− 12xy + 10y2= u2+ v2 (2 points) Then let u = 2x − 3y and v = y then
(
x = u + 3v 2 y = v
(2 points)
The Jacobian is
∂(x, y)
∂(u, v) = 1
2 (2 points) Z Z
R
e−4x2+12xy−10y2dxdy
= Z ∞
0
Z u 0
e−(u2+v2)dvdu
= Z π4
0
Z ∞ 0
e−r21 2rdrdθ
= π
16 (6 points)
3. (12%) Evaluate Z 1
0
Z
√1−x2
0
Z
√
2−x2−y2
√
x2+y2
(x2+ y2+ z2)12 dzdydx.
Solution:
The domain is defined by
x2+ y2+ z2< 2, z2> x2+ y2, 0 < y <p
1 − x2, z > 0, x > 0. (4%) Change the variable by
x = r sin φ cos θ y = r sin φ sin θ
z = r cos φ And the domain is defined by 0 < r < 2, 0 < φ < π
4, 0 < θ < π
2. (4%) Z π4
0
Z π2
0
Z
√2
0
r3sin φ dr dθ dφ (2%)
= π
2 · (1 − 1
√2) ·
√ 24 4 =π
2 · (1 − 1
√2) (2%)
4. (12%) Find the work done by the force field
F = (2xyz + z2− 2y2+ 1) i + (x2z − 4xy + x) j + (x2y + 2xz + 2z) k in moving a particle along the curve C parameterized by r(t) = −2t i + sin−1t j + πt k, 0 ≤ t ≤ 1.
Solution:
Let φ(x, y, z) = x2yz + xz2− 2xy2+ x + z2, then
∇φ = (2xyz + z2− 2y2+ 1) i + (x2z − 4xy) j + (x2y + 2x + 2z) k F = ∇φ + x j (4%)
Z
C
F · dr = Z
C
(∇φ + x j) · dr = φ(x, y, z)
r(1)
r(0)
+ Z
C
xdy (4%)
φ(x, y, z)
r(1)
r(0)
= φ(x, y, z)
(−2,π2, π)
(0, 0, 0)
= 2π2− 2π2+ π2− 2 + π2= 2π2− 2 Z
C
xdy = Z 1
0
√−2t
1 − t2 = 2p 1 − t2
1
0
= −2 Z
C
F · dr = 2π2− 4 (4%)
5. (12%) Evaluate the line integral I
C
F · dr counterclockwise around the region bounded by 0 ≤ x ≤ 1 and x2≤ y ≤ 1, where F(x, y) = sin(x3) i + x2
1 + y2 j.
Solution:
By Green Theorem and Fubini Theorem
I
C
F · dr
= Z Z
R
∂
∂x x2 1 + y2 − ∂
∂ysin(x3)dA
= Z 1
0
Z 1 x2
2x
1 + y2dydx(6%)
= Z 1
0
Z √y 0
2x
1 + y2dxdy(3%)
= Z 1
0
x2 1 + y2
√y
x=0
dy
= Z 1
0
y 1 + y2dy
= 1
2ln(1 + y2)
1
0
= 1
2ln 2.(3%)
6. (12%) Evaluate the integral Z Z
S
F · ˆNdS, where
F = (x + y) i + (y + ez2) j + (2x sin y + 2z) k,
S is the surface of the region D inside the sphere x2+ y2+ z2= 4 and the cylinder x2+ y2= 1, N is the unit outward normal of S.ˆ
Solution:
By Divergence Theorem
Z Z
S
F · ˆNdS
= Z Z Z
D
divFdV
= Z Z Z
D
4dV = 4V (4%)
=4 Z 2π
0
Z 1 0
Z
√4−r2
−√ 4−r2
rdzdrdθ(6%)
=16π
3 (8 − 3√ 3).(2%)
7. (12%) Find the flux of F = yz i − xz j + (x2+ y2) k upward through the surface r(u, v) = eucos v i + eusin v j + u k, where 0 ≤ u ≤ 1, 0 ≤ v ≤ π.
Solution:
∂r
∂u = (eucosv, eusinv, 1)
∂r
∂v = (−eusinv, eucosv, 0)
∂r
∂u ×∂r
∂v = (−eucosv, −eusinv, e2u) (2 points)
flux = Z π
0
Z 1 0
(ueusinv, −ueucosv, e2u) · (−eucosv, −eusinv, e2u)dudv
= Z π
0
Z 1 0
e4ududv
= π
4(e4− 1) (10 points)
8. (12%) Evaluate I
C
F · dr around the curve
r(t) = cos t i + sin t j + (sin t + cos t) k, 0 ≤ t ≤ 2π, where F = (ex− y3) i + (ey+ x3) j + (ez+ x + y) k.
Solution:
Method 1: By Stokes’ theorem, I
C
F · dr = Z Z
R
∇ × F · Nds. (1 pt)
We also have C :
(x2+ y2= 1, z = x + y.
Let R :
(−x − y + z = 0,
x2+ y2≤ 1. Hence C is the boundary of R. Its parametric equation is R(x, y) = xi+yj+(x+y)k, x2+ y2≤ 1. N is the upper normal of R.
Nds = ∂R
∂x ×∂R
∂y dx dy = (−i − j + k)dx dy (3 pts)
∇ × F = det
i j k
∂
∂x
∂
∂y
∂
∂z ex− y3 ey+ x3 ez+ x + y
= i − j + (3x2+ 3y2)k (3 pts)
I
C
F · dr = Z Z
R
∇ × F · Nds
= Z Z
x2+y2≤1
i − j + (3x2+ 3y2)k · (−i − j + k)dx dy
= Z Z
x2+y2≤1
(3x2+ 3y2)dx dy
= Z 2π
0
Z 1 r=0
3r2r dr dθ
= 2π · 3 · r4 4
1
0
= 3π
2 . (5 pts) Method 2:
dr(t) = − sin t i + cos t j + (cos t − sin t)kdt (1 pts)
F(r(t)) · dr(t) =
(−ecos t+ sin3t) sin t + (esin t+ cos3t) cos t +(esin t+cos t+ cos t + sin t)(cos t − sin t)
dt (2 pts)
= ecos td cos t + sin4t dt + esin td sin t + cos4t dt
+esin t+cos td(sin t + cos t) + (cos2t − sin2t)dt (2 pts)
Z 2π t=0
F(r(t)) · dr(t) = ecos t+ esin t+ esin t+cos t2π
t=0+ Z 2π
t=0
sin4t + cos4t + cos2t − sin2t dt
= 0 + 2 Z 2π
t=0
sin4t dt + 0, since Z 2π
t=0
sinnt dt = Z 2π
t=0
cosnt dt
= 2
Z 2π 0
1 − cos 2t 2
2
dt
= Z 2π
t=0
1
2 − cos 2t +1
2cos22t dt
= π +1 8
Z 2π t=0
1 + cos 4t
2 d4t, since Z 2πn
0
cos s ds = 0
= π +π 2
= 3π
2 (7 pts)
9. (12%) Find the general solution of the differential equation
y00+ 3y0+ 2y = 2e−x+ 4e−xcos x.
Solution:
First, we consider the homogeneus equation y00+ 3y0+ 2y = 0 , try y = erx , we obtain the auxiliary equation r2+ 3r + 2 = 0, so (r + 1)(r + 2) = 0, thus we get r = −1, −2.
So the general solution to the homogeneous equation is yh= C1e−x+ C2e−2x, C1, C2 are constants
Since the differential equation is nonhomogeneus , the solution to this equation is given by y = yh+ yp, where yp is the particular solution to this equation .
To compute yp.
Method I
we need to solve the following two equations y00+ 3y0+ 2y = 2e−x (1)
y00+ 3y0+ 2y = 4e−xcos x (2)
For (1), since y = e−x is already a solution to this differertial equation, we try y = Axe−x, and we get A = 2, so the particular solution to (1) is given by yp1= 2xe−x
For (2), we try y = Be−xcos x + Ce−xsin x, and we get B = −2, C = 2, so the particular solution to (2) is given by yp2= −2e−xcos x + 2e−xsin x.
Since this is a linear differential equation, we have the particular solution to y00+ 3y0+ 2y = 2e−x+ 4e−xcos x , say yp, is the sum of the particular solution of (1) and (2), i.e., yp= yp1+ yp2= 2xe−x− 2e−xcos x + 2e−xsin x So, the solution to this equation is given by y = yh+ yp = C1e−x+ C2e−2x+ 2xe−x− 2e−xcos x + 2e−xsin x You can also try yp= Axe−x+ Be−xcos x + Ce−xsin x directly to solve A, B, C
Method II
The second method use variation of paramrters. For the homogeneus equation y00+ 3y0+ 2y = 0, from the above calculation, we have there are two linearly independent solutions , say y1 = e−x, y2= e−2x, and we search the particular solution of the form yp= u1(x)y1(x) + u2(x)y2(x).
For simplicity, we choose yp to satisfy the differential equation u01y1 + u02y2 = 0 , and by this assumption , for this yp need to satisfy this differential equation y00 + 3y0 + 2y = 2e−x + 4e−xcos x , we must have u01y10 + u02y02= 2e−x+ 4e−xcos x, so we have following two equations,
u01y1+ u02y2= u01e−x+ u02e−2x= 0 (3)
u01y10 + u02y02= −u01e−x− 2u02e−2x= 2e−x+ 4e−xcos x (4) (3) ∗ 2 + (4) → u01e−x= 2e−x+ 4e−xcos x → u01= 2 + 4 cos x ,so , u1= 2x + 4 sin x + C3 for som constant C3.
(3) + (4) → −u02e−2x= 2e−x+ 4e−xcos x → u02= −2ex− 4excos x .
Using integration by parts, we get Z
excos xdx = exsin x − Z
exsin xdx = exsin x + excos x − Z
excos xdx, so Z
excos xdx = 1
2(exsin x + excos x). Thus, u2= −2ex− 2(exsin x + excos x) + C4 , where C4 is a constant.
So, yp= u1(x)y1(x) + u2(x)y2(x) = u1(x)e−x+ u2(x)e−2x
= (2x + 4 sin x + C3)e−x+ (−2ex− 2(exsin x + excos x) + C4)e−2x
= 2xe−x− 2e−x+ 2e−xsin x − 2e−xcos x + C3e−x+ C4e−2x Thus the general solution is y = yh+ yp== C1e−x+ C2e−2x+ 2xe−x− 2e−x+ 2e−xsin x − 2e−xcos x + C3e−x+ C4e−2x
= C5e−x+ C6e−2x+ 2xe−x+ 2e−xsin x − 2e−xcos x. for some constant C5, C6.
10. (12%) Solve the differential equation
x3y000+ xy0− y = 0 in the interval x > 0.
Solution:
Let y = xr(2%) ,then we substitute it.
We have xr[(r − 1)r(r − 2) + r − 1] = 0.
=⇒ [r − 1]3= 0
=⇒ r = 1, 1, 1(7%)
Then by ”checking detail”(2%) the solution is y = c1x + c2x ln x + c3x(ln x)2(1%).
