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The solutions to problem 1 to problem 8 of mathematics 2210 autumn 2006 practice midterm 3. The problems involve calculating integrals and expectations using various methods, including change of variables formula, probability density functions, and green's theorem.
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Mathematics 2210 Autumn 2006 Instructor: Mike Wills
Practice Midterm 3 Solutions
Problem #1 (5 points): Let R be the region in the plane given by
(1) {(x, y) ∈ R^2 | x = r cos θ, y = r sin θ,
π 4
≤ θ ≤
3 π 4
, 2 ≤ r ≤ 3 .}
Let f (x, y) = xy. Compute
R f dA. Solution Using the change of variables formula for polar coordinates, we com- pute as follows: ∫ ∫
R
f dA =
2
∫ 34 π
π 4
r cos θ r sin θ
rdθdr
2
rdr
∫ 34 π
π 4
cos θ sin θ
dθ
r^2 2
3 2
ln(sin θ)
34 π π 4
=
ln
− ln
R
f dA = 0
1
Let R = [0, 1] × [0, 1]. Define the probability density function f : R^2 → R via
(3) f (x) =
4 xy (x, y) ∈ R 0 (x, y) ∈/ R
for the random variables, X and Y. Find the expected values of X and Y.
Solution The expected value of X is given by (4)
E(X) =
R^2
xf (x)dx =
0
0
4 x^2 ydydx =
0
4 x^2
y^2 2
1 0
dx = 2
0
x^2 dx =
A virtually identical calculation shows that the expected value of Y is given by E(Y ) = 23.
Problem #3 (5 points):
Let E = {(x, y, z) ∈ R^3 | 0 ≤ z ≤ 1 , 0 ≤ y ≤ z, 0 ≤ x ≤ 3 y}. Compute
(5)
E
6 xzdV.
Solution We compute:
∫ ∫ ∫
E
6 xzdV = 6
0
∫ (^) z
0
∫ (^3) y
0
xzdxdydz
0
∫ (^) z
0
x^2 z
3 y 0
dydz
0
∫ (^) z
0
y^2 zdydz
0
y^3 z
z 0
dz
0
z^4 dz
9 z^5 5
1 0
Let f be defined on a smooth curve C in Rn. Define the following expression:
(10)
C
f ds.
During your explanation, it will be made clear what the term ‘smooth curve’ means. Your definition will involve a limit.
Solution Recall that C is smooth if there exists a C^1 function, r : [a, b] → Rn, such that the image of r is C, and ˙r is never zero. Let n ∈ N. Let ∆t = b−a n. For^ i^ = 0,^1 ,... , n^ define^ ti^ =^ a^ +^ i∆t. For^ i^ = 1,... , n, let^ t
∗ i ∈
[ti− 1 , ti] and let ∆si = |r(ti) − r(ti− 1 )|. Define Sn =
∑^ n
i=
f (r(t∗ i ))∆si.
Then,
C f ds^ is defined to be lim n→∞ Sn^ whenever this limit exists.
Problem #6 (5 points): Let D ⊂ Rn^ be an open set containing the smooth curve C. Let f : D → R be differentiable. Show that
(11)
C
∇f · dr = f (r(b)) − f (r(a)),
where r is a suitably chosen function on [a, b] which you will carefully define.
Solution Let r : [a, b] → Rn^ be a C^1 function whose derivative is never 0, and whose image is C. (Since C is smooth, such an r will exist, by definition.) Let us denote r = (x 1 , x 2 ,... , xn). By definition,
(12)
C
∇f · dr =
C
∑n
i=
∂f ∂xi
x ˙i =
∫ (^) b
a
Df Drdt,
where D denotes the derivative operator. By the chain rule, Df Dr = D(f ◦ r). Since f ◦ r is a real valued function of t, a real variable, we can denote D(f ◦ r) by (f ◦ r)′. The Fundamental Theorem of (single variable) Calculus now applies. We have:
(13)
∫ (^) b
a
Df Drdt =
∫ (^) b
a
(f ◦ r)′dt = (f ◦ r)
b a
= f (r(b)) − f (r(a)),
as desired.
Let C be the positively-oriented square in R^2 with vertices (0, 0), (1, 0), (1, 2), and (0, 2). Evaluate
(14)
C
eydx + 2xeydy.
Solution By Green’s Theorem, ∫
C
eydx + 2xeydy =
0
0
(2ey^ − ey)dydx
0
0
eydydx
= e^2 − 1.
Problem #8 (5 points):
Let D be an open set in R^3. Let F, G : D → R^3 be differentiable. Show that
(16) ∇ × (F + G) = (∇ × F) + (∇ × G).
Solution We start with the right hand side and show that we get the left hand side. Let us denote F = (P, Q, R) and G = (S, T, U ). Then,
(∇ × F) + (∇ × G) = (Ry − Qz , Pz − Rx, Qx − Py) + (Uy − Tz , Sz − Ux, Tx − Sy)
= ((R + U )y − (Q + T )z , (P + S)z − (R + U )x, (Q + T )x − (P + R)y) = ∇ × (P + S, Q + T, R + U ) = ∇ × (F + G),
as desired.
