MA22S1: Problem Set 8
Solutions
Due at the end of the tutorial, 2-4 December.
1. Using a double integral, find the volume under the surface
z = x2 cos y + x sin 2y ,
for the region given by R = {(x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤
π/2}.
Solution: We use the region to give us the limits and then
the volume is
ZZ
V =
(x2 cos y + x sin 2y)dA
R
Z
π/2
Z
2
dx(x2 cos y + x sin 2y)
0
0
2
Z π/2
1 3
1 2
=
dy x cos(y) + x sin(2y)
3
2
0
x=0
Z π/2
8 cos(y)
=
dy 2 sin(2y) +
3
0
π/2
8 sin(y)
= − cos(2y) +
3
y=0
8 sin(π/2)
8 sin(0)
= − cos(π) +
− − cos(0) +
3
3
14
8
.
= 1 + − [−1 + 0] =
3
3
=
dy
2. Find the volume below the surface
z = e2x ,
1
above the region R = {(x, y) : 0 ≤ x ≤ 1, x2 ≤ y ≤ x}.
Sketch R.
Solution: Let’s first sketch R to understand what region
we are integrating over.
1.0
0.8
0.6
x
0.4
x2
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
Figure 1: The region R = {(x, y) : 0 ≤ x ≤ 1, x2 ≤ y ≤ x}.
Note that since 0 ≤ x ≤ 1, x2 ≤ x. The only difference in
how we treat the integration in this question compared to
question 1 is that we must integrate over y first since the
limits of y depend on x. Of course, we could work out how
to exchange the limits by making the x limits depend on y,
but there is no reason to do that here when the limits have
been conveniently provided. We have
ZZ
Z 1
Z x
V =
e2x dA =
dx
dy e2x
Z
=
0
x2
0
R
1
x
dx y e2x y=x2 =
2
Z
0
1
dx (x − x2 ) e2x .
We will need to perform some integration by parts
Z
Z 1
1
1 1
1
1 1
1 2x 1
2x
dx e2x = x e2x x=0 − e2x x=0
dx x e = x e x=0 −
2
2
2
4
0
2 0
2
e
1
1
e
=
−0−
−
= (e2 + 1) ,
2
4
4
4
and
Z 1
Z
1 2 2x 1
1 1
dx x e = x e x=0 −
dx 2x e2x
2
2 0
0
e2
1
1 2
1
=
−0− 2
(e + 1) = (e2 − 1) ,
2
2
4
4
R1
where we made use of 0 dx x e2x again in the second integral. We now return to the volume, getting
Z 1
1
1
1
V =
dx (x − x2 ) e2x = (e2 + 1) − (e2 − 1) = .
4
4
2
0
2 2x
3. Prove that the integrals
Z 2 Z x2
0
and
0
Z 4Z
0
(x2 − y) dy dx ,
2
√
(x2 − y) dx dy ,
y
have the same value.
3
Solution: The first integral can be evaluated as
2
Z 2 Z x2
Z 2
2 x
y
dx
(x2 − y) dy dx =
x2 y −
2
0
0
0
0
Z 2
Z 2 4
4
x
x
=
x4 −
dx
dx =
2
0
0 2
2
32 16
x5 = =
=
.
10 0 10
5
The second integral can be evaluated as
2
Z 4 3
Z 4Z 2
x
2
dy
−
yx
(x
−
y)
dx
dy
=
√
√
3
y
0
0
y
3/2
Z 4 8
y
=
− 2y −
− y 3/2
dy
3
3
0
4
Z 4
2y 3/2
8y
4y 5/2
8
2
− 2y +
dy =
−y +
=
3
3
3
15 0
0
32
4(32) 160 − 240 + 128 48 16
=
− 16 +
=
=
=
.
3
15
15
15
5
This values clearly agree. Note that the second integral is
a bit harder. This shows how important it is to choose the
order of integration carefully.
4. Use a double integral to find the volume of the solid in the
first octant1 bounded by the surface
3x + y + 2z = 4 ,
which is a plane. Follow the steps outlined below:
(a) Sketch the solid and its projection onto the xy-plane.
1
The first octant is the region x, y, z ≥ 0.
4
(b) Use the projection to find the limits of integration.
(c) Perform the integration to find the volume.
Solution:
We will follow the suggested steps to work
through the problem.
(a) Note that we can rewrite the plane as
3
1
z = 2 − y − x.
2
2
The sketches of the solid and its projection onto the
xy-axes are
y
4
3
2
1
0
2.0
y
1.5
4
z 1.0
3
0.5
2
0.0
0.0
1
0.5
x
1.0
1.5
0.2
0.4
0.6
0.8
1.0
1.2
x
Figure 2: The solid and its projection.
(b) From the projection, we see that the lower limit of both
variables is 0. The upper limit of one will depend on the
other. Let’s choose to integrate one y first. Then the
upper limit will come from the equation of the surface
3x + y + 2z = 4 onto the xy-plane, i.e. when z = 0.
5
Then we have
3x + y + 2(0) = 4 ⇒ y = 4 − 3x ,
which means the y limits are 0 and 4−3x. What about
the limits in x? Well since the dependence on y has
been dealt with, we can now integrate x over its entire
range, which is 0 to 4/3. The 4/3 comes either from
the picture of the projection or setting both y and z to
0 to get 3x + 0 + 2(0) = 4 ⇒ x = 4/3. We set y = 0 to
find the upper limit of x since from the projection that
is where x is largest.
(c) We need to integrate over 2 − 12 y − 32 x. The volume is
ZZ 1
3
V =
2 − y − x dA
2
2
R
Z 4/3
Z 4−3x
3
1
=
dx
dy 2 − y − x
2
2
0
0
4−3x
Z 4/3
1
1
=
dx 2y − y 2 − xy
4
2
0
y=0
Z 4/3
1
1
=
dx 2(4 − 3x) − (4 − 3x)2 − x(4 − 3x) − 0
4
2
0
Z 4/3
9x2
=
dx 4 − 6x +
4
0
4/3
3x3
2
= 4x − 3x +
4 0
16
16 3 64
16 16 16 16
=
−3 +
−0=
−
+
=
.
3
9
4 27
3
3
9
9
6