Z Z R Z Z Z Z Z Z Z Z Z Z - Department of...
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1. Evaluate the double integral Z Z R 2xy dx dy and R is bounded by y = x, y =2 - x and y = 0. Answer: Z 1 0 Z 2-y y 2xy dx dy = Z 1 0 (4y - y 2 ) dy = 2 3 2. Evaluate the double integral Z Z R e -x 2 -y 2 dx dy where R = {(x, y):1 ≤ x 2 + y 2 ≤ 4}. Answer: Z 2π 0 Z 2 1 re -r 2 dr dθ = Z 2π 0 1 2 -e -4 + e -1 dθ = π(e -1 - e -4 ). 3. Evaluate the double integral Z Z R q 16 - x 2 - y 2 dx dy where R is inside x 2 + y 2 =9 and outside x 2 + y 2 = 4 and between y = x and y = -x with y ≥ 0. Answer: Z 3π/4 π/4 Z 3 2 r √ 16 - r 2 dr dθ = 1 3 Z 3π/4 π/4 (12 3/2 - 7 3/2 ) dθ = π 6 (12 3/2 - 7 3/2 ). 4. Evaluate the double integral Z Z R (8 + 3x - y 2 ) dx dy where R is bounded by y =2x - 1, y =2x + 5, y =1 - 3x and y =7 - 3x. Answer: u = y +3x and v = y - 2x implies x = 1 5 (u - v ) and y = 1 5 (2u +3v ). |J | = 1 5 1 5 Z 5 -1 Z 7 1 8+ 3 5 (u - v ) - 1 25 (2u +3v ) 2 du dv = 1 125 Z 5 -1 (1104-378v -54v 2 ) dv = - 36 25 . 5. Compute the volume of the solid bounded by x +2y + z = 8 and the coordinate planes, x = 0, y = 0, and z = 0. Answer: Z 4 0 Z 8-2y 0 (8-x-2y) dx dy = Z 4 0 64 - 32y - 1 2 (5 - 2y) 2 +4y 2 dy = 128 3 . 6. Compute the volume of the solid under z =6 - x 2 - y 2 and inside x 2 + y 2 = 1. Answer: Z 2π 0 Z 1 0 (6 - r 2 )r dr dθ = Z 2π 0 11 4 dθ = 11π 2 . 7. Find the surface area of that portion of the paraboloid z =1+ x 2 + y 2 that lies below the plane z = 5. Answer: Z Z S q 1+4x 2 +4y 2 dx dy = Z 2π 0 Z 2 0 r √ 1+4r 2 dr dθ = π 6 (17 3/2 - 1). 8. Determine the surface area of the portion of z =4y +3x 2 between y =2x, y =0 and x = 2.
Transcript of Z Z R Z Z Z Z Z Z Z Z Z Z - Department of...
1. Evaluate the double integral∫ ∫
R2xy dx dy and R is bounded by y = x, y = 2− x
and y = 0.
Answer:∫
1
0
∫
2−y
y2xy dx dy =
∫
1
0
(4y − y2) dy =2
3
2. Evaluate the double integral∫ ∫
Re−x2
−y2
dx dy where R = {(x, y) : 1 ≤ x2 + y2 ≤4}.Answer:
∫
2π
0
∫
2
1
re−r2
dr dθ =∫
2π
0
1
2
(
−e−4 + e−1)
dθ = π(e−1 − e−4).
3. Evaluate the double integral∫ ∫
R
√
16 − x2 − y2 dx dy where R is inside x2+y2 = 9
and outside x2 + y2 = 4 and between y = x and y = −x with y ≥ 0.
Answer:∫
3π/4
π/4
∫
3
2
r√
16 − r2 dr dθ =1
3
∫
3π/4
π/4
(123/2 − 73/2) dθ =π
6(123/2 − 73/2).
4. Evaluate the double integral∫ ∫
R(8 + 3x − y2) dx dy where R is bounded by
y = 2x − 1, y = 2x + 5, y = 1 − 3x and y = 7 − 3x.Answer: u = y + 3x and v = y − 2x implies x = 1
5(u − v) and y = 1
5(2u + 3v).
|J | = 1
5
1
5
∫
5
−1
∫
7
1
(
8 +3
5(u − v) − 1
25(2u + 3v)2
)
du dv =1
125
∫
5
−1
(1104−378v−54v2) dv =
−36
25.
5. Compute the volume of the solid bounded by x + 2y + z = 8 and the coordinateplanes, x = 0, y = 0, and z = 0.
Answer:∫
4
0
∫
8−2y
0
(8−x−2y) dx dy =∫
4
0
(
64 − 32y − 1
2(5 − 2y)2 + 4y2
)
dy =128
3.
6. Compute the volume of the solid under z = 6 − x2 − y2 and inside x2 + y2 = 1.
Answer:∫
2π
0
∫
1
0
(6 − r2)r dr dθ =∫
2π
0
(
11
4
)
dθ =11π
2.
7. Find the surface area of that portion of the paraboloid z = 1 + x2 + y2 that liesbelow the plane z = 5.
Answer:∫ ∫
S
√
1 + 4x2 + 4y2 dx dy =∫
2π
0
∫
2
0
r√
1 + 4r2 dr dθ =π
6(173/2 − 1).
8. Determine the surface area of the portion of z = 4y + 3x2 between y = 2x, y = 0and x = 2.
Answer:∫
2
0
∫
2x
0
√17 + 36x2 dy dx =
∫
2
0
2x√
17 + 36x2 dx =2
3
(
(161)3/2 − (17)3/2)
.
9. Determine the surface area of the surface defined by x = u, y = v + 2, z = 2uv for0 ≤ u ≤ 2 and 0 ≤ v ≤ 1.Answer: |〈1, 0, 2v〉×〈0, 1, 2u〉| =
√1 + 4u2 + 4v2. (Put integration on a calculator.)
∫
1
0
int20√
1 + 4u2 + 4v2 du dv = 5.2335.
10. Evaluate the triple integral∫ ∫ ∫
Q6xz2 dV where Q is the tetrahedron bounded
by −2x + y + z = 4 and the coordinate planes, x = 0, y = 0 and z = 0.
Answer:∫
0
−2
∫
2x+4
0
∫
4+2x−y
0
2xz3 dz dy dx =∫
0
−2
∫
2x+4
0
(128x + 192x2 − 96xy + 96x3 − 96x2y + 24xy2 + 16x4 − 24x3y + 12x2y2 −
2xy3) dy dx =∫
0
−2
(8x5 + 64x4 + 192x3 + 256x2 + 128x) dx = −256
15.
11. Evaluate the triple integral∫ ∫ ∫
Q(x − y) dV where Q is bounded by z = x2 + y2
and z = 4.
Answer:∫
2π
0
∫
2
0
∫
4
r2
(r2 cos θ − r2 sin θ) dz dr dθ = 0.
12. Convert to polar coordinates and evaluate the integral
∫
2
0
∫
√
4−x2
−
√
4−x2
x2 dx dy.
Answer:∫ π/2
−π/2
∫
2
0
r3 cos2 θ dr dθ =∫ π/2
−π/2
(2 + 2 cos (2θ)) dθ = 2π.
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