Solutions #7
1. Evaluate
Z
x + y dA,
D
where D is the region in the first quadrant of the plane bounded by the lines y = x and
y = 3x and the hyperbola xy = 3.
Solution. It is not difficult to see that D is given by the inequalities
√
0 ≤ x ≤
3
x ≤ y ≤ ϕ(x),
where
Hence
R
D
(
3x 0 ≤ x ≤ 1
√
ϕ(x) = 3
1
≤
x
≤
3.
x
x + y dA =
=
=
=
=
R √3 R ϕ(x)
R01
R01
0
x
x + y dydx =
R 1 R 3x
x + y dydx +
R √3 R
3
x
1
x
3
R 0√3 x
1 2 y=3x
1 2 y= x
(xy + 2 y )|y=x dx + 1 (xy + 2√y )|y=x dx
R 3
+
(3 + 92 x12 ) − (x2
(3x2 + 92 x2 ) − (x2 + 12 x2 ) dx
1
√
91
1 3 x= 3
2x3 |x=1
)|x=1
x=0 + (3x
√ − 92 x1 − 2 1x √
2 − 0 + (3 3 − 2 √3 − 2 3 3) − (3 − 29 − 12 ) = 4.
x + y dydx
+ 21 x2 ) dx
2. Evaluate
π2
Z
0
Z
π
√
y
sin x
dx dy
x
sin x
Solution. Since the integral
dx cannot be expressed in terms of elementary functions,
x
we cannot evaluate it. Hence we will change the order of integration. First we write
Z
Z π2 Z π
sin x
sin x
dx
dy
=
dA,
√
x
x
D
0
y
R
where D is the region defined by
0 ≤ y ≤ π2
√
y ≤ x ≤ π.
The inequalities above imply that D is bounded by the lines y = 0, x = π and the parabola
y = x2 . Hence we can rewrite D as
0 ≤ x ≤ π
0 ≤ y ≤ x2 .
Combining the above we see that
Z π2 Z π
Z
Z π Z x2
sin x
sin x
sin x
dx
dy
=
dA
=
dy dx.
√
x
x
x
0
y
D
0
0
MATH 280: page 1 of 3
Solutions #7
MATH 280: page 2 of 3
The last integral is easy to evaluate:
Rπ
Rπ
R π sin x 2
Rπ
R π R x2 sin x
2
dy dx = 0 sinx x y|y=x
x dx = 0 x sin x dx = − 0 x d(cos x)
y=0 dx =
x
x
0
0 0
R
π
x=π
x=π
+ 0 cos x dx = −(x cos x)|x=π
= −(x cos x)|x=0
x=0 + sin x|x=0 = π.
3. Consider the iterated integral
Z 1Z
√
1−x2
Z √x2 +y2
dz dy dx.
0
0
x2 +y 2
(a) It calculates the volume of a solid. Identify the solid.
(b) Change the order of integration so that you first integrate with respect to x, then
with respect to y, then with respect to z.
Solution. (a) The solid W under consideration is given by the inequalities
0
≤ x ≤ √ 1
0
≤ y ≤ p 1 − x2
x2 + y 2 ≤ z ≤
x2 + y 2 .
p
The third inequality shows that W is above the cone z = x2 + y 2 and below the paraboloid
z = x2 + y 2 which intersect at the circle x2 + y 2 = 1 at level z = 1. The first two inequalities
show that W is above the quarter of the unit disk which lies in the first quadrant.
p Combining
the information above we conclude that W is the solid between the cone z = x2 + y 2 and
the paraboloid z = x2 + y 2 which lies in the first octant of R3 .
(b) The projection of W onto the z-axis is the interval [0, 1], hence W can be written as
0≤z ≤ 1
(x, y) ∈ Dz ,
where Dz denote the intersection of W with the horizontal
√ plane at level z. Clearly, Dz is
the part of the annulus between the circles of radii z and z which lies in the first quadrant.
Hence Dz can be described as
√
0
≤ y ≤ p z
ϕ(y, z) ≤ x ≤
z − y2,
where
(p
z2 − y2
ϕ(y, z) =
0
0≤y≤z
√
z ≤ y ≤ z.
In other words, W is expressed as
0
≤ z ≤
√1
0
≤ y ≤ p z
ϕ(x, y) ≤
≤
z − y2.
MATH 280: page 3 of 3
Finally,
R 1 R √1−x2 R √x2 +y2
0
0
x2 +y 2
Solutions #7
R 1 R √z R √z−y2
dx dy dz
ϕ(y,z)
0 0
√
R 1 R √z R √z−y2
R 1 R z R z−y2
dz dy dx.
= 0 0 √ 2 2 dx dy dz + 0 z 0
dz dy dx =
z −y