# MATH 101 HOMEWORK 8 – SOLUTIONS ```MATH 101 HOMEWORK 8 – SOLUTIONS
1. A notch is cut to the centre of a cylindrical log of radius 20 cm. One plane face of the
notch is perpendicular to the axis of the log. The other one intersects it along a diameter
of the log at an angle of 60 degrees. (See Figure 7.20 in the textbook.) What volume of
wood was removed from the log by cutting the notch?
The cross-sections of the notch perpendicular to the diameter are right triangles, with
the angle at the diameter equal to 60 degrees. Consider a cross-section
at a distance x
√
2
2
from the√center of the log. Then one leg of the triangle is 20 − x , and the other is
tan(60◦ ) 202 − x2 . Thus the volume of the notch is
Z
20
−20
√ Z 20
1
3
◦
2
2
tan(60 )(20 − x )dx =
(400 − x2 )dx
2
2 −20
√
x3 20
3
16000
(400 − )
=
= √ ≈ 9237.6cm3 .
2
3 −20
3
2. Use Simpson’s Rule with n = 10 to find the approximate length of the curve y = x3
between x = 0 and x = 1. (You may use a calculator to find the numerical answer, but
you first have to write out the sum that you will be evaluating.)
The integral that we want to approximate is
Z
1
p
Z
1+
(3x2 )2 dx
=
0
1
p
1 + 9x4 dx.
0
Thus
p
p
p
p
1 √
1 + 4 1 + 9 &middot; 0.14 + 2 1 + 9 &middot; 0.24 + 4 1 + 9 &middot; 0.34 + 2 1 + 9 &middot; 0.44
30
p
p
p
p
p
√ +4 1 + 9 &middot; 0.54 + 2 1 + 9 &middot; 0.64 + 4 1 + 9 &middot; 0.74 + 2 1 + 9 &middot; 0.84 + 4 1 + 9 &middot; 0.94 + 2
S10 =
≈ 1.548.
3. Find the surface area of the torus obtained by rotating the circle (x − 2)2 + y 2 = 1
around the y axis.
1
On the top half of the circle, we have
y=
p
−2(x − 2)
−(x − 2)
1 − (x − 2)2 , y 0 = p
=p
,
2 1 − (x − 2)2
1 − (x − 2)2
so that
1 + (y 0 )2 = 1 +
(x − 2)2
1 − (x − 2)2 + (x − 2)2
1
=
=
.
2
2
1 − (x − 2)
1 − (x − 2)
1 − (x − 2)2
The surface area of the top part of the torus is
Z
2π
3
Z
p
0
2
x 1 + (y ) dx = 2π
1
1
3
x
p
dx = 2π
1 − (x − 2)2
Z
1
−1
u+2
√
du,
1 − u2
√
where we substituted x = u + 2. The function u/ 1 − u2 is odd, so the right side is equal
to
Z 1
1
du
√
= 4π
= 4π sin−1 u = 4π 2 .
2
−1
1−u
−1
By symmetry, the total area of the torus is twice as large, i.e. 8π 2 .
4. A circular cone standing upright on its base has base radius 10 cm and height 30
cm. It is made of material that compresses under pressure, so that its density at height x
(measured from the base) is 2 + 10−4 (30 − h)2 g/cm3 . Find the mass of the cone. (You
may use a calculator to find the numerical answer after you set up the integral.)
Let y = 30 − x denote the height measured from the top down, then the slice at x is a disc
of radius y/3 and has area π(y/3)2 . The density at y is 2 + 10−4 y 2 . Hence the mass is
Z
30
0
=
πy 2
y 5 30
π 2 3
(2 + 10−4 y 2 )dy =
y + 10−4
9
9 3
5 0
305 π2 3
30 + 10−4
= 2054π ≈ 6452.83g.
9 3
5
2
```