Math 2263 Multivariable Calculus
Homework 16: 15.4 #18,34, 15.5 #28
July 7, 2011
15.4#18
Use a double integral to find the area of the region inside the cardioid r = 1 + cos θ and
outside the circle r = 3 cos θ.
Graphing the two curves, we find that they intersect in the first and fourth quadrants.
We can solve for the intersection points.
1 + cos θ = 3 cos θ
1 = 2 cos θ
1/2 = cos θ
θ = π/3, −π/3
Now we can set up the integral for area.
Z
2
π
π/3
1+cos θ
Z
Z
π/2
3 cos θ
Z
r dr dθ −
Z
π
2
π/3
0
π/3
π
2
Z
=
+
(3 cos θ)2 dθ
π/2
9 cos2 θ dθ
π/3
π/3
θ|ππ/3
π/2
π/3
(1 + 2 cos θ + cos θ) dθ −
=
Z
(1 + cos θ) dθ −
r dr dθ =
0
Z
2 sin θ|ππ/3
Z
π
+
π/2
√
1
(1 + cos 2θ) dθ −
2
Z
π/2
π/3
9
(1 + cos 2θ) dθ
2
2π
3
1
1
9
1
π/2
− 2(0 −
) + (θ + sin 2θ)|ππ/3 − (θ + sin 2θ)|π/3
3
2
2
2
2
2
√
2π
1
π
1
2π
9 π π
9
2π
=
+ 3 + (π − ) − (0 − sin ) − ( − ) − (0 − sin )
3
2
3
4
3
2 2
3
4
3
√
√
2π √
π π 1
3 9 π 9
3
π
=
+ 3+ − + ·
− · + ·
=
3
2
6 4 2
2 6 4 2
4
=
15.4#34
An agricultural sprinkler distributes water in a circular pattern of radius 100 ft. It supplies
water to a depth of e−r feet per hour at a distance of r feet from the sprinkler.
(a) If 0 < R ≤ 100, what is the total amount of water supplied per hour to the region inside
the circle of radius R centered at the sprinkler?
(b) Determine an expression for the average amount of water per hour per square foot supplied
to the region inside the circle of radius R.
(a)
Z Z
Z
2π
100
100
e−r r dr dθ = 2π
0
0
re−r dr
0
We use integration by parts with u = r and v = −e−r .
Z 100
−r 100
−r
= 2π − re |0 −
−e dr
0
−100
= 2π(−100e
100
Z
e−r dr
+
0
= −200πe−100 + 2π(−e−r |100
0 )
= −200πe−100 + 2π(−e−100 + 1)
= −202πe−100 + 2π
(b) The average amount of water per square foot will be the total amount of water
found in part (a), divided by the area of the region over which the water is
distributed.
1 − 101e−100
−202πe−100 + 2π
=
10000π
5000
15.5#28
(a) Verify that
if 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
otherwise
4xy
0
f (x, y) =
is a joint density function.
(b) If X and Y are random variables whose joint density function is the function f in part (a),
find (i) P (X ≥ 1/2) and (ii) P (X ≥ 1/2, Y ≤ 1/2).
(c) Find the expected values of X and Y .
R∞ R∞
(a) To be a joint density function, we need f (x, y) ≥ 0 and −∞ −∞ f (x, y) dx dy = 1.
The first is true by inspection. Confirm the second as follows:
Z ∞Z ∞
Z 1Z 1
f (x, y) dx dy =
4xy dx dy
−∞
−∞
Z
0
1
=
2x
2
Z
y|x=1
x=0
1
dy =
0
0
0
1
2y dy = y = 1
2
0
(b) (i)
1
Z
1
Z
Z
P (X ≥ 1/2) =
4xy dy dx =
1/2
Z
1
=
1/2
1
0
1/2
1
2
2x dx = x =1−
1/2
y=1
2xy dx
2
y=0
3
1
=
4
4
(ii)
Z
1
Z
P (x ≥ 1/2, y ≤ 1/2) =
1/2
Z
1
4xy dy dx =
1/2
Z
1
=
1/2
0
1/2
1
1
1 2 x dx = x 2
4
1/2
y=1/2
2xy dx
2
y=0
1
1
1 3
3
= (1 − ) = · =
4
4
4 4
16
(c) x-mean:
Z
1
Z
1
Z
1
Z
xf (x, y) dx dy =
0
0
=
0
1
Z
2
4x y dx dy =
0
Z
1
0
0
1
x=1
4 3 dy
x y
3
x=0
1
4
2
2 2 y dy = y =
3
3 0 3
y-mean:
Z
1
Z
1
Z
1
Z
yf (x, y dx dy =
0
0
0
0
1
4xy 2 dx dy =
2
3