Exam 1 Solutions – MATH 230, Spring 2011
Instructor: Dr. Zachary Kilpatrick
1. (20pts) Evaluate the integral.
(a)
Z
3x − 4
dx =
x2 − 3x + 2
Z
B
A
+
dx
x−1 x−2
using partial fractions.
thus A(x − 2) + B(x − 1) = 3x − 4,
so A + B = 3 and −2A − B = −4, can be solved A = 1, B = 2, and we can evaluate
Z
1
2
+
dx = ln |x − 1| + 2 ln |x − 2| + C
x−1 x−2
(b)
Z
0
2
8
dx =
(x + 3)(x2 + 1)
Z
2
0
A
Bx + C
+ 2
dx
x+3
x +1
using partial fractions.
thus A(x2 + 1) + (Bx + C)(x + 3) = 8,
so A + B = 0, 3B + C = 0, A + 3C = 8, can be solved A = 4/5, B = −4/5, C = 12/5, and
we evaluate
2
Z
1
x−3
4
4 2 1
2
−1
ln |x + 3| − ln |x + 1| + 3 tan x
−
dx =
5 0 x + 3 x2 + 1
5
2
0
4
1
−1
−1
=
ln 5 − ln 3 − (ln 5 − ln 1) + 3 tan 2 − 3 tan 0
5
2
" √
#
4
5
=
ln
+ 3 tan−1 2
5
3
1
Determine whether the integral is convergent or divergent. If it is convergent, compute it.
(c)
Z
1
−1
1
dx =
x2
Z
0
−1
1
dx +
x2
Z
0
1
1
dx
x2
since 1/x2 blows up to ∞ at 0, we must evaluate the integrals using limits
1
Z 1
1
1
1
=
lim
lim+
−
dx
=
lim
− 1 = +∞
2
t→0
t→0+
x t t→0+ t
t x
t
Z t
1
1
1
lim−
= lim+ − − 1 = +∞
dx = lim− −
2
t→0
t→0
x −1 t→0
t
−1 x
thus
Z
1
−1
1
dx = +∞,
x2
so the integral diverges.
(d)
Z
1
0
√
1
Z 1
Z 1
√
2
1
2
2
2 3/2
3/2
3/2
x
dx = − lim+ t ln t −
x ln xdx = x ln x −
xdx,
3
3 0
x
3 t→0
3 0
0
and we can evaluate the limit using L’Hospital’s rule
lim+ t3/2 ln t = lim+
t→0
t→0
ln t
2
1/t
= lim+
= − lim+ t3/2 = 0
3/2
5/2
t→0 −(3/2)/t
1/t
3 t→0
so the integral becomes
Z
1
2 1√
4
4
xdx = − x3/2 0 = −
−
3 0
9
9
2
2. (20pts)
(a) Using the method of disks (washers), find the volume of the solid obtained by rotating
the region bounded by y = x3 and y = x1/3 about the x-axis. Start by sketching the region,
the solid, and a typical disk or washer.
the formula for the volume of a region rotated about an axis using the disc (or washer)
method is
Z b
V =π
ro (x)2 − ri (x)2 dx
a
to find the limits of the integral we note that we will integrate along the x direction and find
where the two curves cross: x3 = x1/3 ⇒ x = 0, 1.
the outer radius is ro (x) = x1/3 and inner radius ri (x) = x3 (shown below), so the integral
for the volume is
Z 1
2
2
V = π
x1/3 − x3 dx
0
0
1
2/3
x
2x5/2 x7
− x dx = π
−
5
7
6
1
0
16π
2 1
=π
=
−
5 7
35
1
0.8
y = x 1/3
0.6
y
= π
Z
ro
0.4
y = x3
0.2
0
0
ri
0.2
0.4
x
3
0.6
0.8
1
(b) Use the method of cylindrical shells to find the volume generated by rotating the region
√
bounded by x = y and x = y about the line y = 2. Start by sketching the region and a
typical shell.
the formula for the volume of a region rotated about an axis using the shell method is
Z b
V = 2π
r(y)h(y)dy
a
to find the limits of the integral, we note that we will integrate along the y direction and find
√
where the two curves cross: y = y ⇒ y = 0, 1.
the height of the shells will be the top curve (when looking from the y-axis) minus the bottom
√
axis so h = y − y, the radius is given by the distance of each shell from the center of the
volume at y = 2 given by r = 2 − y. thus the volume is
Z 1
Z 1
√
√
2 y − 2y − y 3/2 + y 2 dy
V = 2π
(2 − y)( y − y)dy = 2π
0
0
8π
4
2 1
= 2π
=
−1− +
3
5 3
15
2
y
1.5
1
r = 2−y
h=
√
y−y
0.5
0
0
0.2
0.4
x
4
0.6
0.8
1
3. (10pts) Find the arc length of the curve given by y = 16 x3 +
dy
1 2 1 −2
=
x − x
dx
2
2
2
2
1 2 1 −2
dy
x − x
= 1+
1+
dx
2
2
1 1
1
1 1
1
= 1 + x4 − + x−4 = x4 + + x−4
4
2 4
4
2 4
2
1 2 1 −2
=
x + x
2
2
so
L =
Z
4
s
dy
dx
2
dx
1+
s
2
Z 4 1 2 1 −2
=
x + x
dx
2
2
1
Z 4
1 2 1 −2
dx
x + x
=
2
2
1
4 21 3
87
1 3 1 −1
=
=
x − x
+
=
6
2
2
8
8
1
1
5
1
2x
when 1 ≤ x ≤ 4.
4. (10pts) Egyptian rioters filled a square-based pyramid (pictured) with water. Its height
is 10 m and all four edges of the base are 10 m each. Given water has density 1000 kg/m3
and gravity is 9.8 m/s2 , find the work required to pump the water out of the pyramid.
taking y = 0 as the bottom and y = 10 as the top, we need to evaluate the volume of water
and distance it has to be taken at each level y
volume at y: (10 − y)2 dy
distance to be lifted at y: (10 − y)
so the total work to pump all the water out is
Z 10
W = 9.8 · 1000
(10 − y)3 dy
0
if we then change variables u = 10 − y, we have
4 10
Z 10
u
3
= 9800(2500) = 2.45 × 107 J
9800
u du = 9800
4
0
0
6
5. (20pts)
(a) Show that x2 + y 2 + z 2 = 2x + 12y − 4z represents a sphere, and find its center and radius.
x2 − 2x + y 2 − 12y + z 2 + 4z = 0
completing the square
x2 − 2x + 1 + y 2 − 12y + 36 + z 2 + 4z + 4 = 1 + 36 + 4
(x − 1)2 + (y − 6)2 + (z + 2)2 = 41
so the equation represents a sphere with center (1, 6, −2) and radius
√
41
(b) Calculate the volume of the parallelepiped with the adjacent edges P Q, P R, P S where
P (−4, −1, 2), Q(−1, 1, 0), R(−2, 2, 3), S(−5, 3, 5).
P~Q = h3, 2, −2i
P~R = h2, 3, 1i
P~S = h−1, 4, 3i
we can then compute the volume of the parallelepiped using the triple product
3 2 −2 |P~Q · (P~R × P~S)| = 2 3 1 −1 4 3 = |3(9 − 4) − 2(6 + 1) − 2(8 + 3)| = |15 − 14 − 22| = 21
7
6. (20pts) (a) Compute the parametric equations for line through (1, −2, 3) and (−2, 1, 2).
~v = h−2 − 1, 1 − (−2), 2 − 3i = h−3, 3, −1i
using P (1, −2, 3), the parametric equations for the line are:
x = 1 − 3t,
y = −2 + 3t,
z =3−t
(b) Compute the parametric equations for the line through (−1, 6, −1) and (3, −1, 2).
~u = h3 − (−1), −1 − 6, 2 − (−1)i = h4, −7, 3i
using P (−1, 6, −1), the parametric equations for the line are:
x = −1 + 4s,
y = 6 − 7s,
z = −1 + 3s
(c) Determine whether the lines are parallel, skew, or intersecting. If they intersect, find the
point of intersection.
note that
−3
3
−1
6= 6=
,
4
7
3
so the lines are not parallel, but if we look for an intersection point by setting
1 − 3t = −1 + 4s
−2 + 3t = 6 − 7s
3 − t = −1 + 3s,
we find that t = −2 and s = 2 satisfies this system, so the lines intersect.
specifically, we find where they intersect by plugging t = −2 or s = 2 into either line’s
equation to find the point is (7, −8, 5)
8