PP 32 : Double integral
1. Let R = [a, b] × [c, d] and f : R → R be defined by f (x, y) = p(x)q(y) where
! p : [a, b] →
!R
RR
Rb
Rd
and q : [c, d] → R are continuous. Show that f (x, y)dxdy =
p(x)dx
q(y)dy .
R
2. Let R = [0, π2 ] × [0, π2 ]. Evaluate
RR
3. Evaluate
RR
a
c
sin x cos ydxdy.
R
cos x3 dxdy where R is the region in R2 bounded by y = 3x2 , y = 0 and x = 1.
R
4. Let R be RR
the region lying below the curve y = cos x, − π2 ≤ x ≤
Evaluate
sin xdxdy.
π
2
and above the x-axis.
R
2
2
2
5. Let
RR R2 be the region in R bounded by the curves y = 2x and y = 1 + x . Evaluate
(2x + y)dxdy.
R
RR
6. Evaluate
x cos(y −
R
y3
3 )dxdy
where R = {(x, y) ∈ R2 : x2 + y 2 ≤ 1, x ≥ 0, y ≥ 0}.
7. Evaluate the following iterated integrals by interchanging the order of integration.
(a)
R1 R1
cos x2 dxdy.
0 y
(b)
R1 R1
0
(c)
√
3
ey dydx.
x
R1 R1
3
x3 ey dydx.
0 x2
(d)
R1 R1
0 y
8. Evaluate
1
dxdy.
1+x4
R1
(tan−1 πx − tan−1 x)dx.
0
9. Find the volume of the solid enclosed by the surfaces z = 6 − x2 − y 2 , z = 2x2 + y 2 − 1,
x = −1, x = 1 , y = −1 and y = 1.
10. Let D be the solid bounded by the surfaces y = x2 , y = 3x, z = 0 and z = x2 + y 2 . Find
the volume of D.
11. Led D be the solid bounded by the cylinder x2 + y 2 = 1 and the planes y + z = 1 and
z = 0. Find the volume of D.
12. Find the volume of the solid which is common to the cylinders x2 + y 2 = 1 and x2 + z 2 = 1.
See Problem 10 of PP 20.
Practice Problems 32: Hints/Solutions
1. Follows from the Fubini’s theorem.
π
2. By Fubuni’s theorem
π
R2
R2
sin x cos ydxdy = ( sin xdx)( cos ydy) = 1.
RR
0
R
RR
3. See Figure 1. By Fubuni’s theorem
0
R1 3x
R2
cos x3 dxdy =
cos x3 dydx =
0 0
R
R1
3x2 cos x3 dx =
0
sin 1.
π
RR
4. See Figure 2.
sin xdxdy =
− π2
R
RR
5. See Figure 3.
R2 cos
Rx
π
sin xdydx =
R
1
2
R1
sin2 x 2
2 |− π2
= 0.
R1 1+x
R 2 2
+ y)dxdy =
(2x + y)dydx.
(2x2
−1 2x2
√
6. See Figure 4.
sin x cos xdx =
− π2
0
R
RR
π
R2
3
x cos(y− y3 )dxdy
=
R1
2
1−y
R
0
0
3
x cos(y− y3 )dxdy
=
R1
0
√
1−y 2
y3
1 2
2 x cos(y− 3 )|0
=
2
(1 −
y 2 ) cos(y
−
0
7. (a)
R1 R1
y3
3 )dy
cos x2 dxdy =
0 y
(b)
(c)
√
cos tdt.
0
R1
cos x2 dydx =
=
R1 Ry
0
R1 R1
0 y
3
ey dxdy =
0 0
x
R1 R1
1
3
R1
0
=
1
dxdy
1+x4
8. Note that
R1
=
R1 R y
3
x3 ey dxdy =
0 0
R1 Rx
0 0
1
dydx
1+x4
=
(tan−1 πx − tan−1 x)dx =
1
2
R1
0
R1
0
1 2 y3
4 y e dy
du
1+u2
R1 πx
R
0 x
0
sin 1. See Figure 5.
eu du = 13 (e − 1). See Figure 6.
√
3
x3 ey dydx
1
2
x cos x2 dx =
0
2
3
ey dydx
x2
(d)
R1 Rx
R3
0 0
R1 R1
0
=
1
2
=
1
2
=
1
12 (e
− 1). See Figure 7.
tan−1 u|10 = π8 . See Figure 8.
1
dydx
1+y 2
=
R1 Ry
0
y
π
1
dxdy
1+y 2
+
Rπ R1
1
y
π
1
dxdy.
1+y 2
See
Figure 9.
9. Note that (6 − x2 − y 2 ) − (2x2 + y 2 − 1) ≥ 0 for all (x, y) ∈ [−1, 1] × [−1, 1]. The volume
R1 R1
of D =
(6 − x2 − y 2 ) − (2x2 + y 2 − 1)dydx.
−1 −1
10. Let R be the region in R2 bounded by the curves y = x2 and y = 3x. Then the volume of
RR
R3 R3x 2
D = (x2 + y 2 )dxdy =
(x + y 2 )dydx.
0 x2
R
11. Let R = {(x, y) ∈ R2 : x2 + y 2 ≤ 1}. ThenRRthe solid D lies RR
above the RR
region R and below
the graph z = 1 − y. The volume of D = (1 − y)dxdy = dxdy − ydxdy. Note that
RR
ydxdy =
R
R1
√
R
1−x
R 2
√
−1 − 1−x2
ydydx =
R1
R
R
0dx = 0. Therefore the required volume is the area of R
−1
which is π.
√
12. The solid
is enclosed by the cylinder x2 + y 2 = 1 and the surfaces z = − 1 − x2 and
√
z = 1 − x2 . Let R = {(x, y) ∈ R2 : x2 + y 2 ≤ 1}. The required volume is equal to
√
√
RR √
R1 1−x
R 2 √
( 1 − x2 − (− 1 − x2 )) =
2 1 − x2 = 16
3 .
R
√
−1 − 1−x2