MATH 221: PRACTICE PROBLEMS II
R
Problem 1: Evaluate the line integral C ydx + xdy, where C is the
parameterized path ~r(t) = (t2 , t3 ), 1 ≤ t ≤ 5.
Solution. On the path ~r(t), we have x(t) = t2 , y(t) = t3 , dx = 2tdt,
dy = 3t2 dt. Hence the line integral is
Z 5
Z 5
3
2
2
t (2tdt) + t (3t dt) =
5t4 dt = (55 − 1).
1
1
Problem 2. Consider the parabola C given by x = y 2 + 2.
(a) Write an integral for the arc length of the parabola between the
points (3, 1) and (6, −2).
(b) Give the parametric equation of the tangent line to the parabola
at the point (2, 0).
Solution. We parametrize C by putting y(t) = t and x(t) = t2 + 2 so
that the equation of the parabola holds for all values of t. This gives
dx/dt = 2t, dy/dt = 1.
(a) The points (3, 1), (6, −2) are reached at t = 1, −2 respectively. Arc
length is given by
s
Z 1√
Z 1 2 2
dx
dy
+
dt =
4t2 + 1 dt
dt
dt
−2
−2
Note that here we took the lower limit to be t = −2 and upper
limit to be t = 1, as arc length must be positive.
(b) The point (2, 0) is reached at t = 0 by the path ~r(t) = (t2 + 2)î + tĵ.
We first find the velocity vector at this point: ~v (t) = 2tî+ ĵ. Hence
the required tangent line is parallel to ~v (0) and passes through
the point (2, 0). This gives the parametric equation
~r(t) = (2, 0) + t~v (0) = 2î + tĵ
for the tangent line.
1
2
MATH 221: PRACTICE PROBLEMS II
Problem 3. Evaluate the integral
Z 1 Z √1−x2
√
− 1−x2
0
e−(x
2 +y 2 )
dydx.
Solution. The symmetry of the integrand and limits suggests that we
should use polar co-ordinates. If we sketch the region of integration,
we find that it is the right half of the interior of the unit circle x2 +y 2 =
1. This region can be described in polar co-ordinates by 0 ≤ r ≤ 1
and −π/2 ≤ θ ≤ π/2. However, as range of θ is [0, 2π], we should
divide this region into two parts - the top half with 0 ≤ θ ≤ π/2 and
≤ θ ≤ 2π. Hence the integral is
the bottom half, with 3π
2
Z π/2 Z 1
Z 2π Z 1
2
−r2
e rdrdθ +
e−r rdrdθ
0
0
3π/2
0
For the inner integral, we use the substitution u = r2 so that du =
2rdr, and u goes from 02 to 12 , giving
Z
1 1 −u
1
e du = (1 − e−1 ).
2 0
2
Hence,
Z π/2 Z 1
Z
π
1 − e−1 π/2
−r2
e rdrdθ =
dθ = (1 − e−1 ),
2
4
0
0
−π/2
and similarly,
Z
2π
3π/2
Z
1
2
e−r rdrdθ =
0
π
(1 − e−1 ).
4
This gives π(1 − e−1 )/2 as the value of the required integral.
Problem 4. Evaluate
Z
0
π
Z
y
π
sin x
dxdy
x
Solution. Sketching the region of integration shows that it is the interior of the triangle made by the points (0, 0), (π, 0) and (π, π). As this
integral does not seem easy to do as given, we reverse the order of
integration. For a fixed x, we see that y goes from 0 to x. Finally, x
goes from 0 to π. Hence the integral is
Z π
Z πZ x
sin x
sin x
dy dx =
xdx = (− cos x)|π0 = 2.
x
x
0
0
0