Calculus 2: Volume of Solids of Revolution, Assignments of Calculus

Solutions to problems 1 and 2 from homework #7 in math 1230: calculus 2. The problems involve finding the volume of solids of revolution obtained by revolving the area under the curve y = 1/x^2 around the x-axis and y-axis, and determining the values of p for which the volumes are finite.

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Pre 2010

Uploaded on 07/29/2009

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Math 1230: Calculus 2
Homework #7 Due Monday, May 18
Read: Section 6.2 & 6.3
Problem 1 Take the area under the graph of y=1
xpfrom x= 1 to x=,
and revolve it around the x-axis. For which values of pis the volume of the
resulting solid finite?
Solution The volume of the slice at position xis
volume = π1
xp2
dx
so the total volume is
Z
1
π1
xp2
dx =πZ
1
1
x2pdx.
We know from class that this integral converges when 2p > 1 and that it
diverges otherwise. We conclude that the volume is finite when p > 1
2.2
Problem 2 Take the same region as in Problem 1 and revolve it around the
y-axis. For which values of pdoes that solid have finite volume?
Solution There are two ways to go about this one.
Method 1: Washers. The slice at position yhas volume
volume = πy1
p2
12dy
so the total volume is
Z1
0
πy1
p2
12dy =πZ1
0
y2
pdy π.
We know from Quiz 4, Problem 2 (in the future!) that this integral
converges when 2
p>1, and diverges otherwise. Thus the volume is
finite for p > 2.
pf2

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Math 1230: Calculus 2

Homework #7 Due Monday, May 18

Read: Section 6.2 & 6.

Problem 1 Take the area under the graph of y = (^) x^1 p from x = 1 to x = ∞, and revolve it around the x-axis. For which values of p is the volume of the resulting solid finite?

Solution The volume of the slice at position x is

volume = π

xp

dx

so the total volume is ∫ (^) ∞

1

π

xp

dx = π

1

x^2 p^

dx.

We know from class that this integral converges when 2p > 1 and that it diverges otherwise. We conclude that the volume is finite when p > 12. 2

Problem 2 Take the same region as in Problem 1 and revolve it around the y-axis. For which values of p does that solid have finite volume?

Solution There are two ways to go about this one.

  • Method 1: Washers. The slice at position y has volume

volume = π

y−^ p^1 )^2 − 12

dy

so the total volume is ∫ (^1)

0

π

y−^

(^1) p^ )^2 − 12

dy = π

0

y−^

(^2) p dy − π.

We know from Quiz 4, Problem 2 (in the future!) that this integral converges when − p^2 > −1, and diverges otherwise. Thus the volume is finite for p > 2.

  • Method 1: Shells. The volume of the shell at position x is

volume = 2πx

xp^

dx

so the total volume is ∫ (^) ∞

1

2 πx

xp^

dx = 2π

1

xp−^1

dx

and we know from class that this integral converges only when p− 1 > 1, which is to say that p > 2.

In either case we see that the volume is finite when p > 2 and infinite otherwise. 2