MA 51000 HOMEWORK ASSIGNMENT #9 SOLUTIONS

Problem 1. Sec. 5.1, pg. 326; prob. 5A lumberjack cuts out a wedge-shaped piece W of a cylindrical tree of radius r obtained by making two

saw cuts to the tree’s center, one horizontally and one at an angle θ. Compute the volume of the wedge Wusing Cavalieri’s principle.

Solution: Assume that the cylindrical tree of radius r can be described by the equation x2 + y2 = r2. LetPa denote the plane x = a. Then we have the cross-sectional area

A(x) = area of (W ∩ Px) =12b h =

12b2 tan θ =

12(r2 − x2

)tan θ

for −r ≤ x ≤ r. Cavalieri’s principle states that

volume of W =∫ r

−rA(x) dx =

∫ r

−r

12(r2 − x2

)tan θ dx =

12

(r2 x− x3

3

)tan θ

∣∣∣∣x=rx=−r

=23r3 tan θ.

Problem 2. Sec. 5.2, pg. 340; prob. 5Let f be continuous on [a, b] and g be continuous on [c, d]. Show that∫∫

R

[f(x) g(y)

]dx dy =

[∫ b

a

f(x) dx

][∫ d

c

g(y) dy

]where R = [a, b]× [c, d].

Solution: Define the function

G(y) =∫ b

a

f(x) g(y) dx = λ g(y) where λ =∫ b

a

f(x) dx.

Then we have the identity∫∫R

[f(x) g(y)

]dx dy =

∫ d

c

[∫ b

a

f(x) g(y) dx

]dy =

∫ d

c

G(y) dy = λ

[∫ d

c

g(y) dy

]

=

[∫ b

a

f(x) dx

][∫ d

c

g(y) dy

].

Problem 3. Sec. 5.2, pg. 340; prob. 111

2 MA 51000 HOMEWORK ASSIGNMENT #9 SOLUTIONS

Although Fubini’s theorem holds for most functions met in practice, one must still exercise some caution.This exercise gives a function for which it fails. By using a substitution involving the tangent function, showthat ∫ 1

0

∫ 1

0

x2 − y2

(x2 + y2)2dy dx =

π

4, yet

∫ 1

0

∫ 1

0

x2 − y2

(x2 + y2)2dx dy = −π

4.

Why does this not contradict Theorem 3 or 3’?

Solution: We have the (single) integrals∫ 1

0

x2 − y2

(x2 + y2)2dy =

y

x2 + y2

∣∣∣∣y=1

y=0

=1

1 + x2and

∫ 1

0

x2 − y2

(x2 + y2)2dx = − x

x2 + y2

∣∣∣∣x=1

x=0

= − 11 + y2

.

If we make the substitution

t = tan θ =⇒ dt

dθ= sec2 θ = 1 + t2 =⇒

∫ 1

0

11 + t2

dt =∫ π/4

0

dθ =π

4.

This gives the double integrals∫ 1

0

∫ 1

0

x2 − y2

(x2 + y2)2dy dx =

∫ 1

0

11 + x2

dx =π

4, yet

∫ 1

0

∫ 1

0

x2 − y2

(x2 + y2)2dx dy = −

∫ 1

0

11 + y2

dy = −π4.

This inequality does not contradict Theorem 3 because f(x, y) = (x2 − y2)/(x2 + y2)2 is not continuouson R = [0, 1] × [0, 1]. Indeed, f(x, y) is not defined when (x, y) = (0, 0). Similarly, this inequality does notcontradict Theorem 3’ because the integral

∫ 1

0f(x, y) dy (respectively

∫ 1

0f(x, y) dx) does not exist for all

x ∈ [0, 1]) (respectively y ∈ [0, 1]). Indeed, the integral does not exist for x = 0 (respectively, y = 0).

Problem 4. Sec. 5.3, pg. 348; prob. 3Use double integrals to compute the area of a circle of radius r.

Solution: Consider the interior of a circle of radius r:

D ={

(x, y) ∈ R2∣∣x2 + y2 ≤ r2

}.

This is a y-simple region because it is the collection of (x, y) ∈ R2 where −r ≤ x ≤ r and φ1(x) ≤ y ≤ φ2(x) interms of the continuous functions φ1, φ2 : [−r, r]→ R defined by φ1(x) = −

√r2 − x2 and φ2(x) =

√r2 − x2.

Hence we have the area

area of D =∫∫

D

dx dy =∫ r

−r

[∫ φ2(x)

φ1(x)

dy

]dx =

∫ r

−r2√r2 − x2 dx.

Upon substituting x = r sin θ we find the integral∫ r

−r2√r2 − x2 dx = r2

∫ π/2

−π/22 cos2 θ dθ = r2

[θ + cos θ sin θ

]∣∣∣∣θ=π/2θ=−π/2

= π r2.

Hence the area of a circle of radius r is π r2.

Problem 5. Sec. 5.3, pg. 348; prob. 12Set up the integral required to calculate the volume of a cone of base radius r and height h.

MA 51000 HOMEWORK ASSIGNMENT #9 SOLUTIONS 3

Solution: Let S denote the solid which is the interior of the region bounded by

z = 0 and z = h

[1−

√x2 + y2

r2

].

We compute the volume using Cavalieri’s principle. Let Pa denote the plane z = a. Then we have

A(z) = area of (S ∩ Pz) = area of circle x2 + y2 = r2(

1− z

h

)2

= π r2(

1− z

h

)2

, 0 ≤ z ≤ h.

Hence we have

volume of S =∫ h

0

A(z) dz = π r2∫ h

0

(1− z

h

)2

dz =13π r2 h.