MATH 101 V2A
February 6th – Practice problems
Hints and Solutions
Volumes:
1. Calculate the volume of the solid obtained by rotating the region bounded by the x-axis and the curve
y = log(x) between x = 0 and x = 1 about the x-axis.
Hint: We want to calculate the volume of the solid obtained by rotating the shaded region about the
x-axis.
y
x
If we try to use the “washer” method, we’ll end up getting an ugly integral, so we’ll calculate the
volume using the method of cylindrical shells. If we use cylindrical shells, each shell will have to be
oriented so that the radius is in the y direction. Such a shell will have radius −y (since all of the points
on our curve have negative y values and the radius can’t be negative), “height” ey (the x-coordinate
of the point (x, y) on the curve y = log(x)) and thickness dy. So, the volume of each cylindrical shell
is Vshell = 2π(−y)ey dy. Therefore the volume of the solid is
Z 0
Vsolid = −2π
yey dy.
−∞
Now use integration by parts and L’Hospital’s rule to show that Vsolid = 2π.
2. Calculate the volume of a pyramid of height H and base an equilateral triangle with side L.
Hint: There are lots of ways you could do this, but one way would be to slice the
pyramid horizontally. If
√
3(H−y)2 L2
you do this, the volume of a triangular slice at height y will have volume Vslice =
. Therefore the
4H 2
volume of the pyramid is
√ 2Z H
3L
V =
(H 2 − 2Hy + y 2 ) dy.
4H 2 0
Answer: V =
√
3 2
12 L H.
Improper integrals: Determine whether each of the following integrals is convergent or divergent. Evaluate
those that are convergent.
Z ∞
2
1.
xe−x dx.
−∞
Solution: By definition, we have that
Z
∞
2
xe−x dx =
−∞
Z
0
lim
T1 →−∞
2
xex dx + lim
T2 →∞
T1
Z
T2
2
xex dx.
0
Now,
1 1 T12
1 x2 0
= lim
lim
e − e
= −∞.
T1 →−∞ 2
T1 →−∞ 2
2
T1
Z ∞
2
So, since the first integral is divergent,
xe−x dx is also divergent.
−∞
Z
∞
2.
1
log(x)
dx.
x
Solution: By definition we have that
Z
1
Z
∞
Therefore
1
Z
1
3.
−1
∞
log(x)
dx = lim
T →∞
x
Z
T
log(x)
dx
x
1
T 1
2
log(x) = lim
T →∞ 2
1
1
= lim log(T )2 = ∞.
T →∞ 2
log(x)
dx is divergent.
x
ex
dx.
ex − 1
Hint: Answer: This integral is also divergent!
Z
4.
0
1
1
dx. (Hint: Break the problem up into cases: p < 1, p = 1 and p > 1).
xp
Hint: Answer: The integral converges if p < 1 and diverges if p ≥ 1.
2