Assignment 7 - Solutions Math 209 { Fall 2008
Assignment 7 - Solutions
Math 209 ? Fall 2008
1. (Sec. 15.4, exercise 8.) Use polar coordinates to evaluate the double integral
(x + y) dA,
R
where R is the region that lies to the left of the y-axis between the circles x2 + y2 = 1 and x2 + y2 = 4.
Solution: This region R can be described in polar coordinates as the set of all points (r, ) with 1 r 2 and /2 3/2. Then
3/2 2
3/2 2
(x + y) dA =
(r cos + r sin ) r dr d =
r2(cos + sin ) dr d
R
/2 1
/2 1
3/2
=
(cos + sin ) d
/2
2
r2 dr
1
= sin - cos |3/2/2
r3 2
14
= ??? = -
31
3
2. (Sec. 15.4, exercise 16.) Use a double integral to find the area of the region enclosed by the curve r = 4 + 3 cos .
Solution: This region, call it D, can be described in polar coordinates as consisting of all points (r, ) with 0 r 4 + 3 cos and 0 2. Then
Area(D) =
2 4+3 cos
2 r2 r=4+3 cos
dA =
r dr d =
d
D
00
0
2 r=0
1 =
2
(4
+ 3 cos )2 d
=
1
2
(16 + 24 cos + 9 cos2 ) d
20
20
1 2
1 + cos(2)
=
16 + 24 cos + 9 ?
d
20
2
1
99
2 41
= 16 + 24 sin + + sin(2) = .
2
24
0
2
3. (Sec. 15.4, exercise 22.) Use polar coordinates to find he volume of the solid inside the sphere x2 + y2 + z2 = 16 and outside the cylinder x2 + y2 = 4.
Solution: The sphere x2 + y2 + z2 = 16 intersects the xy-plane along the circle with equation x2 + y2 = 16. Since the solid is symmetric about the xy-plane, we may compute
its total volume as twice the volume of the part that lies above the xy-plane, and this latter is the solid that lies below the graph of z = 16 - x2 - y2 and above the annular region D = {(x, y) | 4 x2 + y2 16}. Hence, changing polar coordinates,
Vol = 2
2 4
16 - x2 - y2 dA = 2
16 - r2 r dr d
D
02
2
4
= 2 d
16 - r2 r dr
0
2
= 2 |20
1 - (16
-
r2)3/2
4
= ? ? ? = 32 3.
3
2
4. (Sec. 15.4, part of exercise 36.) Use polar coordinates to evaluate the double integral
e-(x2+y2) dA,
D(a)
where D(a) = {(x, y) | x2 + y2 a2} is the disk of radius a centered at the origin.
Solution:
2 a
e-(x2+y2) dA, =
e-r2 r dr d = 2
D(a)
00
- 1 e-r2 a
2
0
= (1 - e-a2).
5. (Sec. 15.5, exercise 2.) Electric charge is distributed over the disk x2 + y2 4 so that the charge density at (x, y) is (x, y) = x + y + x2 + y2 (in coulombs per square meter).
Find the total charge on the disk.
Solution: Call Q the total charge on the disk. We have
Q= = = = =
(x, y) dA = (x + y + x2 + y2) dA
D
D
2 2
2 2
(r cos + r sin + r2) r dr d =
(r2(cos + sin ) + r3) dr d
00
00
2 2
2 2
r2(cos + sin ) dr d +
r3 dr d
00
00
2
(cos + sin ) d
2
2
r2 dr + 2 r3 dr
0
0
0
sin - cos |20
r3 2
r4 2
+ 2
= ? ? ? = 8 Coulombs.
30
40
6. (Sec. 15.5, exercise 10.) Find the mass and center of mass of the lamina that occupies the regionD bounded by the parabolas y = x2 and x = y2 and has density function (x, y) = x.
Sx2olutyion:xWaenmd a0ydexscrib1e
D as a region of type I: it is the (draw a picture!). Then, calling
set of all points (x, m the mass of the
y) with lamina,
we have
m=
1
(x, y) dA =
x x dy dx =
1
x( x
-
x2)
dx
=
1
(x - x5/2) dx =
3 .
D
0 x2
0
0
14
Next let's compute the (first) moments of the lamina:
Mx =
1 0
x y x dy dx =
x2
1
x
?
1 (x
-
x4) dx
=
1
0
2
2
1
(x3/2 - x9/2) dx =
6
0
55
and
Mx =
1 0
x x x dy dx =
x2
1
x x( x
-
x2)
dx
=
0
1
(x2
-
x7/2) dx
=
1 .
0
9
Finally,
the
center
of
mass
of
the
lamina
is
(x,
y)
=
(
My m
,
Mx m
)
=
(
14 27
,
28 55
).
7. (Sec. 15.5, exercise 12.) A lamina occupies the part of the disk x2 + y2 1 in the first quadrant. Find its center of mass if the density at any point is proportional to the square of its distance from the origin.
Solution: The region of integration (part of the disk x2 + y2 1 in the first quadrant) is
described
easily
in
polar
coordinates
as
the
set
of
all
(r, )
with
0
r
1
and
0
2
.
Also, the density is (x, y) = k( x2 + y2)2 = k(x2 + y2) = kr2, where k is a constant of
proportionality. Then the mass m of the lamina is
m=
/2
1
kr2 r dr d =
/2
1
kr3 dr d
=
k ;
0
0
0
0
8
and its first moments are
Mx = My =
y ? (x, y) dx dy =
/2
1 kr4 sin dr d = k
/2
k
sin d = ,
region
0
0
50
5
x ? (x, y) dx dy =
/2
1 kr4 cos dr d = k
/2
k
cos d = .
region
0
0
50
5
Finally,
the
center
of
mass
is
(x,
y)
=
(
8 5
,
8 5
).
8. (Sec. 15.5, exercise 20.) Consider a square fan blade with sides of length 2 and the lower left corner placed at the origin. If the density of the blade is (x, y) = 1 + 0.1x, is it more difficult to rotate the blade about the x-axis or about the y-axis?
Solution: We can determine in which direction rotation will be more difficult by comparing the moments of inertia about the x-axis and about the y-axis. Calling D the region of the xy-plane occupied by the fan blade, we have
Ix = =
y2(x, y) dA =
D
x2 x=2 x + 0.1 ?
2 x=0
22
y2(1 + 0.1x) dy dx =
00
y3 y=2
8
= (2 + 0.2) ?
3 y=0
3
2
(1 + 0.1x) dx
0
5.87,
2
y2 dy
0
and similarly
Iy = =
x2(x, y) dA =
D
x3
x4 x=2
+ 0.1 ?
3
4 x=0
22
x2(1 + 0.1x) dy dx =
00
2
(x2 + 0.1x3) dx
0
y|yy==20
=
8 + 0.4
3
? (2) 6.13.
2
dy
0
Since Iy > Ix, more force is required to rotate the blade about the y-axis; in other words, it is easier to rotate it about the x-axis.
9. (Sec. 15.6, exercises 4 and 8.) Evaluate the iterated integrals
1 2x y
(a)
2xyz dz dy dx
0x 0
x xz
(b)
x2 sin y dy dz dx
0
00
Solution: For (a) we have
1 2x y
1 2x
1 2x
2xyz dz dy dx =
xyz2 z=y dy dx =
xy3 dy dx
z=0
0x 0
0x
0x
=
1 xy4 y=2x dx =
1 15 x5 dx = 5
0
4 y=x
04
8
while for (b) we have
x xz
x
x2 sin y dy dz dx =
-x2 cos y y=xz dz dx y=0
0
00
0 0 x
=
x2 - x2 cos(xz) dz dx
0 0
=
x2z
- x sin(xz)
z=x z=0
dx
0
=
(x3
-
x sin(x2)) dx
=
2
-
1.
0
4
10. (Sec. 15.6, exercise 10.) Evaluate the tripple integral E = {(x, y, z) | 0 x 1, 0 y x, x z 2x}.
E yz cos(x5) dV, where
Solution: Set this triple integral as an iterated integral:
yz cos(x5) dV =
1
x
2x
yz cos(x5) dz dy dx =
1
x yz2 cos(x5) z=2x dy dx
E
00x
00 2
z=x
1 =
1
x 3x2y cos(x5) dy dx = 1
1 3 x2y2 cos(x5) y=x dx
20 0
20 2
y=0
3 =
1 x4 cos(x5) dx = 3
1 sin(x5)
1
=
3
sin(1).
40
45
0 20
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- table of useful integrals
- techniques of integration whitman college
- table of basic integrals basic forms
- contiune on 16 7 triple integrals
- double integrals stankova
- assignment 7 solutions math 209 fall 2008
- trig substitution
- integrals in cylindrical spherical coordinates sect 15
- residue calculus part ii
- math 104 improper integrals with solutions
Related searches
- 7 grade math problems and answers
- grade 7 summer math packet
- grade 7 printable math worksheets
- 7 grade math word problems
- 7 grade math practice
- grade 7 ncert math textbook pdf
- infinitely many solutions math definition
- 7 grade math percentage worksheets
- 7 grade math games
- 7 grade math printable worksheets
- second grade math assignment worksheet
- 7 grade math lesson