Math 142, Calculus II, Fall 2012
Written Homework Example 2
1. Write up the student study guide solution to 6.1.52.
A bowl is shaped like a hemisphere with diameter 30cm. A ball with diameter 10 cm is placed in the bowl
and water is poured into the bowl to a depth of h centimeters. Find the volume of water in the bowl.
First we will determine the volume of water in the bowl (without the ball) when the height of the water
is h cm for 0 ≤ h ≤ 15.
h
.
n
• Label the sides of the slabs xi = i∆x starting from the bottom of the lowest slab, and ending with
n∆x = h at the top of the top slab.
• Slice the water into n horizontal slabs of thickness ∆x =
• Consider the ith slab (the one labeled xi−1 on the
q bottom and xi on the top). The circle which is
p
2
2
the top face has radius ri = 15 − (15 − xi ) = 30xi − x2i as one can see by considering the right
triangle which the radius makes when the bowl is viewed from the side and using the Pythagorean
theorem (see the picture).
If the ith slab were a cylinder (instead of tapering inwards towards the bottom) its volume would be
(area of the face) × (thickness) = π(ri )2 ∆x = π(30xi − x2i )∆x.
So we can say that the volume of the ith slab is approximately π(30xi − x2i )∆x.
n
X
• Thus the volume of all the water is approximately
π(30xi − x2i )∆x.
i=1
• Since the error in our approximation appears to diminish as n gets large (indeed, the difference between
our slabs and actual cylinders seems to become very small as ∆x → 0), we take that the
volume = lim
n→∞
n
X
π(30xi − x2i )∆x
i=1
and of course this limit is equal to the usual integral
lim
n→∞
n
X
i=1
π(30xi −
x2i )∆x
Z
=
0
h
1
π(30x − x ) dx = π 15x − x3
3
2
2
h h3
2
π.
= 15h −
3
0
To compute the volume of the offending ball (in order to subtract it), we repeat this process with a
different radius. For 0 ≤ h ≤ 10
h
.
n
• Label the sides of the slabs xi = i∆x starting from the bottom of the lowest slab.
• Slice the ball into n horizontal slabs of thickness ∆x =
q
p
• Consider the ith slab. The circle which is the top face has radius ri = 52 − (5 − xi )2 = 10xi − x2i .
Note that it doesn’t matter whether h is less than or greater than 5 since we square (5 − xi ) in the
computation.
If the ith slab were a cylinder (instead of tapering inwards or outwards towards the bottom depending
if we are below or above the halfway mark on the ball) its volume would be
(area of the face) × (thickness) = π(ri )2 ∆x = π(10xi − x2i )∆x.
So we can say that the volume of the ith slab is approximately π(10xi − x2i )∆x.
• Thus the volume of the ball to height h is approximately
n
X
π(10xi − x2i )∆x.
i=1
• Since the error in our approximation appears to diminish as n gets large we take that the
volume = lim
n→∞
n
X
π(10xi − x2i )∆x.
i=1
This limit is equal to the usual integral
lim
n→∞
n
X
i=1
π(10xi −
x2i )∆x
Z
=
0
h
h 1 3 h3
2
π(10x − x ) dx = π 5x − x = 5h2 −
π, 0 ≤ h ≤ 10.
3
3
2
0
Soas long as the
height
of the 3water
is 0 ≤ h ≤ 10 (see picture), we compute that the volume of the water
3
h
h
is 15h2 −
π − 5h2 −
π = 10h2 π cubic cm. If the height of the water is 10 ≤ h ≤ 15 (again,
3
3
4
see picture), then the ball is covered completely and so we can simply subtract its volume, π(5)3 —it is
3
worth noting thatthe methodwe used above is one way to obtain the formula for the volume of a sphere.
4
h3
π − π53 cubic cm.
We are left with 15h2 −
3
3