Download Mathematics 205 HWK 22a Solutions Section 16.4 p759 Problem 2

Mathematics 205
HWK 22a Solutions
Section 16.4 p759
Problem 2, §16.4, p759.
For the region R (a half annulus) as shown, write
iterated integral in polar coordinates.
R
R
f dA as an
Solution. Since the function f isn’t specified, I will write f (P ) to indicate the f -value at the
point P of R, with f -values specified in terms of polar coordinates [r, θ] for P . For dA we use
rdrdθ. To determine the limits of integration, imagine shooting an r-arrow, so shooting radially
outward from the origin with θ fixed (see sketch). The arrow will enter the region R where r = 1
(on the inner semicircle) and it will leave where r = 2 (on the outer semicircle). Then θ will need
to vary from π2 (which gives the upper vertical edge of R) to 3π
2 (which gives the lower vertical
edge of R). Putting the pieces together, we have
Z
R
f dA =
Z
3π
2
π
2
Z
2
f (P ) r dr dθ.
1
Problem 4, §16.4, p759.
For the region R (a quarter disk) as shown, write
iterated integral in polar coordinates.
R
R
f dA as an
Solution. Here, too, I’ll write f (P ) to indicate the polar-coordinate version of the f -value at
the point P of R. For dA we use r dr dθ. To determine limits of integration, imagine shooting an
r-arrow for fixed θ. The arrow enters (is already in) the region R where r = 0 (at the origin). The
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Math 205 HWK 22a Solns continued
§16.4 p759
arrow leaves R when r = 0.5 (at the circular boundary arc of the region). Then θ must be allowed
to vary from 0 to π2 . This gives us
Z
R
Problem 7, §16.4, p759.
f dA =
Z
π
2
0
Z
0.5
f (P ) r dr dθ.
0
Sketch the region of integration for the integral
(with point P expressed in polar coordinates).
Z
π
3
π
6
Z
1
f (P )r dr dθ
0
Solution. The inner integral tells us that an r-arrow, shot radially outward for fixed θ enters
the region where r = 0 (so already in at the origin) and leaves the region where r = 1 (so when
the arrow hits the circle of radius 1 centered at the origin). Then the limits of integration on the
outer integral tell us that the arrows for π6 ≤ θ ≤ π3 are the ones that meet the region, or in other
words that the region lies between the rays θ = π6 and θ = π3 . The region of integration, R, say, is
a sector of a disk (like a piece of pie). Here’s a sketch of R, showing a typical r-arrow for fixed θ.
Problem 9, §16.4, p759. Sketch the region of integration for the integral
(with point P expressed in polar coordinates).
Z
π
4
0
Z
1
cos θ
f (P )r dr dθ
0
Solution. The inner integral tells us that an r-arrow, shot radially outward for fixed θ, enters
the region where r = 0 (so at the origin) and leaves the region when r = cos1 θ . Rewriting r = cos1 θ
as r cos θ = 1 and recognizing r cos θ as the rectangular coordinate x, we see that an r-arrow leaves
the region of integration when it hits the line x = 1, a vertical line one unit to the right of the
y-axis. The limits of integration on the outer integral tell us that θ is constrained between θ = 0
(so the ray that is the positive x-axis ) and θ = π4 (so the ray that is the first-quadrant portion of
the line y = x. So we end up with a triangular region, T , say, bounded by the rays θ = 0 and θ = π4
and by the vertical line r cos θ = 1 or x = 1. Here’s a sketch for T , showing a typical r-arrow.
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Math 205 HWK 22a Solns continued
§16.4 p759
R
Problem 12, §16.4, p759. Evaluate the integral
radius 2 centered at the origin.
R
sin(x2 + y 2 ) dA, where R is the disk of
Solution.
Step 1. Rewrite the integrand sin(x2 + y 2 ) as sin(r2 ).
Step 2. For dA, use the polar equivalent r dr dθ.
Step 3. Determine limits of integration. An r-arrow for fixed θ enters R (is already in R) where
r = 0 and leaves where r = 2. Since θ must give a full revolution, use the interval 0 ≤ θ ≤ 2π.
Step 4. Putting the pieces together and evaluating, we have
Z
2
2
sin(x + y ) dA =
R
=
Z
Z
2π
0
Z
2
r sin(r2 ) dr dθ
0
2π
0
·
¸2
1
2
− cos(r ) dθ
2
0
Z
1 2π
(1 − cos 4) dθ
2 0
= π(1 − cos 4)
=
Problem 13, §16.4, p759. Evaluate the integral
region between the circles of radius 1 and radius 2.
R
R
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(x2 − y 2 ) dA, where R is the first-quadrant
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Math 205 HWK 22a Solns continued
§16.4 p759
Step 1. In polar coordinates, the integrand x2 −y 2 becomes r2 cos2 θ −r2 sin2 θ or r2 (cos2 θ −sin2 θ.
A little later we’ll see that it will be helpful to use a double-angle formula and rewrite things once
again: x2 − y 2 = r2 (cos2 θ − sin2 θ) = r2 cos(2θ).
Step 2. For dA use its polar equivalent dA = r dr dθ.
Step 3. Determine the limits of integration. For fixed θ, an r-arrow shot radially outward from
the origin will enter R where r = 1 and leave R where r = 2. The values for θ must vary over the
interval 0 ≤ θ ≤ π2 .
Step 4. Putting the pieces together and evaluating, we have
Z
Z πZ 2
2
2
2
(x − y ( dA =
r2 (cos2 θ − sin2 θ)r dr dθ
R
0
=
Z
1
π
2
0
=
Z
2
r3 cos 2θ dr dθ
1
π
2
0
Z
Z
·
r4
cos 2θ
4
π
2
¸2
dθ
1
15
cos 2θ dθ
4
0
·
¸ π2
15
=
sin 2θ
8
0
=
= 0.
It might have been better to integrate in the opposite order. This is easy to do in this problem, since
the limits of integration are constant, and it will more readily enable us to exploit the symmetry
of the eventual integrand relative to the midpoint of the interval of integration. Here’s what that
calculation would look like.
Z
Z π2 Z 2
2
2
(x − y ( dA =
r2 (cos2 θ − sin2 θ)r dr dθ
R
0
1
!
Z ÃZ π
2
2
=
cos 2θ dθ
1
=
Z
r3 dr
0
2
0 dr
1
= 0.
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Math 205 HWK 22a Solns continued
§16.4 p759
Problem 15, §16.4, p759. Consider the integral
Z
0
3
Z
1
f (x, y) dy dx.
x
3
(a) Sketch the region R over which the integration is being performed.
(b) Rewrite the integral with the order of integration reversed.
(c) Rewrite the integral in polar coordinates.
Solution. (a) The inner integral tells us that a vertical arrow (a y-arrow for fixed x) enters R
where y = x3 (or x = 3y) and leaves R where y = 1. According to the outer limits of integration,
x then varies over the interval 0 ≤ x ≤ 3. Sketch the lines y = x3 , y = 1, x = 0, observing that
the line y = x3 meets the line y = 1 where x = 3. The region R is a triangle, shown below with a
typical vertical arrow.
This time we use horizontal arrows (or x-arrows, for fixed y), as shown in the sketch that follows.
Such an arrow enters R when x = 0 and leaves R when x = 3y. Then y must vary between y = 0
and y = 1. This gives us the new iterated integral
Z
0
1
Z
3y
f (x, y) dx dy.
0
(c) We will need to replace f (x, y) by f (r cos θ, r sin θ) and use rdrdθ for dA. To find the new
limits of integration, shoot an r-arrow radially outward from the origin.
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Math 205 HWK 22a Solns continued
§16.4 p759
Such an arrow (corresponding to fixed θ) enters R when r = 0 and leaves when it hits the line
y = 1, which we can rewrite as r sin θ = 1 or r = sin1 θ . Thus we’ll use 0 ≤ r ≤ sin1 θ as limits of
integration for the inner integral. It’s easy to see from the sketch that the upper limit of integration
for θ will be π2 , but what about the lower limit? The smallest θ would be where the ray for that θ
coincides with the first-quadrant portion of the line 3x = y, i.e. the ray emanating from the origin
and passing through the point (3, 1). Using a little trigonometry and the definition of the inverse
tangent function, we can write this angle as arctan 13 . Thus the polar-coordinate version of our
integral is
Z
π
2
arctan
1
3
Z
1
sin θ
f ( r cos θ, r sin θ) r dr dθ.
0
Problem 17, §16.4, p759. Convert the integral
R √6 R x
0
−x
dy dx to polar coordinates and evaluate.
Solution. The current integrand is the constant function 1, which is also 1 in polar coordinates.
The rectangular dy dx will be replaced by its polar equivalent r dr dθ. To find the new limits of
integration, we need to identify and sketch the region of integration, T , say. The current inner
integral tells us that a vertical arrow enters T when y = −x and leaves T when
√ y = x. The outer
limits tell us that the region lies between the vertical lines√x = 0 and x = 6. The region T is
triangular, and is bounded by the lines y = x, y = −x, x = 6.
To find limits of integration for the polar-coordinate integral, then, imagine shooting an r-arrow,
radially outward from the origin, for fixed θ.
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Math 205 HWK 22a Solns continued
§16.4 p759
The √
arrow is already in √
the triangle where r = 0, and it leaves
√ when the arrow hits the line
x = 6. Rewriting x = 6 in polar
coordinates as r cos θ = 6 and solving for r, we see that
√
the r-arrow leaves T where r = cos6θ . The range of values for θ will be − π4 ≤ θ ≤ π4 . Putting the
pieces together and evaluating, we have
Z
0
√
6
Z
x
dy dx =
Z
π
4
−π
4
−x
=
=
Z
π
4
−π
4
π
4
Z
−π
4
=
Z
π
4
Z
√
6
cos θ
r dr dθ
0
·
·
2
r
2
√
¸ cos6θ
0
3
cos2 θ
¸
dθ
dθ
3 sec2 θ dθ
−π
4
π
= [3 tan θ]−4 π
4
=3+3
=6
Note, however, that there is a much better way to evaluate
√ this√integral! This particular integral
just gives the area of the triangular region T , which is 12 ( 6)(2 6) = 6.
Problem
21, §16.4, p759. Find p
the volume of an ice-cream cone bounded by the hemisphere
p
z = 8 − x2 − y 2 and the cone z = x2 + y 2 .
Solution.
Here’s a sketch of p
the surfaces that bound the specified solid. The top rounded
boundary surface comes from z = 8 − x2 − y 2 , p
which is the top half of the sphere x2 +y 2 +z 2 = 8.
The conical boundary surface comes from z = x2 + y 2 , which meets the yz-plane in the lines
z = ±y and meets the xz-plane in the lines z = ±x. The solid itself is bounded above by the sphere
and belowpby the cone.pThe points on the circle where the two boundary surfaces meet satisfy the
equation x2 + y 2 = 8 − x2 − y 2 , which can be rewritten as x2 + y 2 = 4. These points all lie
on a circle of radius 2 centered on the z-axis and on the plane z = 2 parallel to the xy-plane.
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Math 205 HWK 22a Solns continued
§16.4 p759
Let D be the shadow in the xy-plane that is cast by the disk x2 + y 2 = 4, z = 2.
p
At
each
point
in
the
shadow
region
D,
the
height
function
for
our
solid
will
be
8 − x2 − y 2 −
p
2
2
x + y . If we integrate the height function over the shadow or base region D, we will have the
volume. In other words, the volume we want can be represented by
Z
D
p
p
( 8 − x2 − y 2 − x2 + y 2 ) dA.
Polar coordinates, however, will
√ be much easier to use here than rectangular! In polar coordinates,
the height function becomes 8 − r2 − r and dA will appear as its polar equivalent r dr dθ. Since
D is just a disk of radius 2 centered at the origin, the limits of integration will be 0 ≤ r ≤ 2,
0 ≤ θ ≤ 2π. Thus the volume we want is
Z
2π
0
Z
0
2
Z
p
( 8 − r2 − r) r dr dθ =
2π
0
Z
2
(r
0
p
8 − r2 − r2 ) dr dθ
·
¸2
1
r3
2 32
=
− (8 − r ) −
dθ
3
3 0
0
Z 2π
1 √
8 1 √
=
(− 4 4 − + 8 8 + 0) dθ
3
3 3
0
Z 2π
√
16
=
( ( 2 − 1)
3
0
32π √
=
( 2 − 1).
3
Z
2π
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