HG2M13 ________________________________________________________________________
Problem Sheet 2 1.
Evaluate the integrals in the following and sketch the regions of integration. (You might need to reverse the order of integration for some of the integrals.) i. iii. v. vii. ix.
2.
3
∫ ∫ (2 − y ) dydx 0
2
ii.
0
2 − x
1
∫ ∫ ( x
2
x
0
π
∫∫ 0
π
0
2
1
sin y
y
x
y 2
0
viii.
dydx
cos( x
2
iv. vi.
x sin y dydx
∫∫ ∫∫
) dydx
x
π
0
− y
3
) dxdy
x.
0
3
∫ ∫ ( x y − 2 xy ) dxdy −2
0
1
x 2
2
∫ ∫ ( x −1
− x
π
sin y
∫∫ 0
0
3
3
0
x
1
1
∫ ∫ ∫∫ 0
2
2
− y
) dydx
2 x dxdy
2 y 2 sin xy dydx
y
3
1 + x dxdy
In the following, sketch the region of integration and write an equivalent integral with the order of integration reversed. i.
3.
2
2
∫∫ 0
4 − 2 x 0
3 x 2 y dydx
ii.
1
∫∫ 0
y
y
3 dxdy
In the following, integrate the function f ( x, y ) over the given region D. i.
f ( x, y ) = y
ii. iii.
x y
= x, y =
f ( x, y ) =
; D is the region in the first quadrant bounded by the lines
2 x, x = 1 and x = 3 . 3 y
; D = {( x, y ) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2} 2 x + 1 f ( x, y ) = 2 xy − 3 x ; D is the triangular region with vertices (0,0 ), (2,2 ) and (0,4) .
iv.
f ( x, y ) =
x 1 + y
y
4.
= x
2
2
; D is the region in the first quadrant enclosed by
, y = 4 and x = 0 .
Find the volume of the solid whose base is the region in the xy - plane that is bounded by the parabola y = 4 − x 2 and the line y = 3 x , while the top of the solid is bounded by the plane z = x + 4 .
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HG2M13 ________________________________________________________________________
5.
In each of the following, evaluate i
∫∫ x y dA . 2
D
y (1,4)
D
(4,1)
(1,1)
x
0 y
y = x
2
y
=
2 x + 3
y
=
x
0
1 2
6.
Sketch the region bounded by the parabolas x = y 2 and x = 2 y − y 2 and find the region’s area.
7.
In the following, change the Cartesian integral into an equivalent polar integral and evaluate the polar integral. i. iii.
8.
1− x
1
∫ ∫ −1
0 2 − y
1
∫∫
Express
0
2
D
2
+ y
2
) dydx
ii.
∫∫ 0
2
4− y
2
x 2 + y 2
e
0
dxdy
2
y
∫∫
( x
x dxdy
f ( x, y ) dA as an iterated integral for each of the regions shown
below. i.
y
D
0
y
ii.
2
2
3 D
x 3
0
x
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HG2M13 ________________________________________________________________________
Answers 1.
i. -6
ii. 0 1
v. 2 −
π
2
vi.
2
ix. sin 1 1 2 − y 2
4
∫∫
2.
i.
3.
i. 4 ln 2
4.
0
x.
0
iii.
π
2 2 9
(2
iv. 4 5 viii. 9 − sin 9
)
2 −1
2
3
3
vii. 2
ii.
3 x y dxdy
ii.
−4
1
x
0
x 2
∫∫
3 dydx
iii. 8 3
π
2
iv.
1 2
(
)
17 − 1
625 12 711
5.
i.
ii. 103.2974
6.
13
7.
i.
8.
i.
∫ ∫ f (r cos θ, r sin θ) r drd θ
ii.
∫ ∫ f (r cos θ, r sin θ) r drd θ
20
1
ii.
π
4 2π
π
2
4
(e
4
)
−1
iii. 2 3
2
0
π
π
0
3
0
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