ĐÁP ÁN ĐỀ THI TOÁN A3 Mã môn học: 1001113 Ngày thi: 13/01/2015 Nội dung
Câu
Điểm
Vẽ hình
0.25 2 −x 2
1
1.
I1 =
∫ dx ∫ f (x, y )dx −2
0.5
x
y
2
− 2−y
1
1
I1 =
2−y
∫ dy ∫ f (x, y )dx + ∫ dy ∫ −2
f (x , y )dx .
0.5
− 2−y
Vẽ hình
0.25
x = r cos ϕ y = r sin ϕ z = z 2π 2 4−r 2 2 3 x + y zdz = ∫ dϕ ∫ dr ∫ r zdz
I
4− x 2 +y 2
2.
I2 = 2
∫
∫ (
dxdy
2
x +y ≤4
2
x +y
4 −r 1 2 dr = r z 2 r
2
= 2π ∫
3
0
Đường
2
∫ r (16 − 8r )dr = 3
0
x + y 2 = 4x
π 2
K =
0
2
tròn
2.5 0.5
)
2
x = 2 + 2 cos t, y = 2 sin t, 1.
0
r
64π . 5
có
0.5
phương
trình
tham
số:
−π π ≤t ≤ . 2 2
0.5
∫ (2 + 2 cos t ) − (2 sin t ) 2dt = 8π + 32 2
2
0.5
π − 2
II
Đặt P = e x −y (1 + x + y ),Q = e x −y (1 − x − y ). Ta có Py′ = Qx′ = −e x −y (x + y ) , do đó tích phân đã cho không phụ thuộc vào đường đi. 2.
(−5,−5)
K =
∫
(
d e
(−1,−1)
1.
(−5,−5)
x −y
(x + y )) = ∫
(−1,−1)
(
)
d u (x, y )
2.5 0.5
0.5
= u (−5, −5) − u (−1, −1) = 10 + 2 = 12.
0.5
divF = x 2 + z 2 + z , rotF = −2yzi + (x − 2xz ) j + 0k
0.5
Gọi Ω là vật thể giới hạn bởi mặt cong S. Khi đó thông lượng của trường F qua S O −G Φ=
III
∫∫ xzdydz + yz dxdz + x zdxdy 2
2
=
∫∫∫ (x
2
)
+ z 2 + z dxdydz
0.5
Ω
S
2.
Tổng
2.0
Đổi biến sang hệ tọa độ trụ, ta được 1
Φ=
∫∫
x 2 +z 2 ≤1
dxdz
∫ (x
2
)
+ y + z dy = 2
x 2 +z 2
2π
1
1
∫ dϕ ∫ dr ∫ r dy (do ∫∫∫ zdxdydz = 0 ). 2
0
0
r
0.5
Ω
0.5 IV
1.
xy ′ − y y y y = tan ⇔ y ′ = + tan . x x x x
0.5
3.0
y dz , phương trình trở thành x = tan z . x dx tan z = 0 ⇔ z = n π ⇔ y = n πx , n ∈ ℕ là nghiệm của phương trình. tan z ≠ 0 phương trình trở thành: dz dx = ⇔ ln | sin z |= ln | x | + ln | C | hay sin z = Cx . Thay vào ta có tích tan z x phân tổng quát của phương trình đã cho là: y sin = Cx hay y = x arcsin (Cx ). x Nghiệm của phương trình thuần nhất: y = C 1 cos (2x ) + C 2 sin (2x ).
Đặt z =
y ′′ + 4y = 3 cos x yr = a cos x + b sin x . Thay vào ta tìm được yr = cos x .
Một
nghiệm
riêng
của
phương
1
2.
trình
có
0.5
0.5
0.5
dạng 0.5
1
Một nghiệm riêng của phương trình y ′′ + 4y = 4e −2x có dạng yr = ke −2x . Thay vào 2
1 ta tìm được yr = e −2x . 2 2 Nghiệm tổng quát của phương trình đã cho:
0.5
1 y = y + yr = C 1 cos (2x ) + C 2 sin (2x ) + cos x + e −2x . 2 Tổng điểm
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