a. Let B be the region between the upper concentric hemispheres of radii a and b centered at the origin and situated in the first octant, where 0 < a < b. Consider F a function defined on B whose form in spherical coordinates ( ρ , θ , φ ) is F ( x , y , z ) = f ( ρ ) cos φ . Show that if g ( a ) = g ( b ) = 0 and ∫ a b h ( h ρ ) d ρ = 0 . then ∭ B F ( x , y , z ) d V = π 2 4 [ a h ( a ) − b h ( b ) ] where g is an antiderivative of f and h is an antiderivative of g. b. Use the previous result to show that ∭ B z cos x 2 + y 2 + z 2 x 2 + y 2 + z 2 d V = 3 π 2 . where B is the region between the upper concentric hemispheres of radii π and 2 π centered at the origin and situated in the first octant.
a. Let B be the region between the upper concentric hemispheres of radii a and b centered at the origin and situated in the first octant, where 0 < a < b. Consider F a function defined on B whose form in spherical coordinates ( ρ , θ , φ ) is F ( x , y , z ) = f ( ρ ) cos φ . Show that if g ( a ) = g ( b ) = 0 and ∫ a b h ( h ρ ) d ρ = 0 . then ∭ B F ( x , y , z ) d V = π 2 4 [ a h ( a ) − b h ( b ) ] where g is an antiderivative of f and h is an antiderivative of g. b. Use the previous result to show that ∭ B z cos x 2 + y 2 + z 2 x 2 + y 2 + z 2 d V = 3 π 2 . where B is the region between the upper concentric hemispheres of radii π and 2 π centered at the origin and situated in the first octant.
a. Let B be the region between the upper concentric hemispheres of radii a and b centered at the origin and situated in the first octant, where 0 < a < b. Consider F a function defined on B whose form in spherical coordinates (
ρ
,
θ
,
φ
) is
F
(
x
,
y
,
z
)
=
f
(
ρ
)
cos
φ
. Show that if
g
(
a
)
=
g
(
b
)
=
0
and
∫
a
b
h
(
h
ρ
)
d
ρ
=
0
. then
∭
B
F
(
x
,
y
,
z
)
d
V
=
π
2
4
[
a
h
(
a
)
−
b
h
(
b
)
]
where g is an antiderivative of f and h is an antiderivative of g.
b. Use the previous result to show that
∭
B
z
cos
x
2
+
y
2
+
z
2
x
2
+
y
2
+
z
2
d
V
=
3
π
2
. where B is the region between the upper concentric hemispheres of radii
π
and 2
π
centered at the origin and situated in the first octant.
A: Tan Latitude / Tan P
A = Tan 04° 30'/ Tan 77° 50.3'
A= 0.016960 803 S CA named opposite to latitude,
except when hour angle between 090° and 270°)
B: Tan Declination | Sin P
B Tan 052° 42.1'/ Sin 77° 50.3'
B = 1.34 2905601 SCB is alway named same as
declination)
C = A + B = 1.35 9866404 S CC correction, A+/- B:
if A and B have same name - add, If
different name- subtract)
=
Tan Azimuth 1/Ccx cos Latitude)
Tan Azimuth = 0.737640253
Azimuth
=
S 36.4° E CAzimuth takes combined
name of C correction and Hour Angle - If LHA
is between 0° and 180°, it is named "west", if
LHA is between 180° and 360° it is named "east"
True Azimuth= 143.6°
Compass Azimuth = 145.0°
Compass Error = 1.4° West
Variation 4.0 East
Deviation: 5.4 West
A: Tan Latitude / Tan P
A = Tan 04° 30'/ Tan 77° 50.3'
A= 0.016960 803 S CA named opposite to latitude,
except when hour angle between 090° and 270°)
B: Tan Declination | Sin P
B Tan 052° 42.1'/ Sin 77° 50.3'
B = 1.34 2905601 SCB is alway named same as
declination)
C = A + B = 1.35 9866404 S CC correction, A+/- B:
if A and B have same name - add, If
different name- subtract)
=
Tan Azimuth 1/Ccx cos Latitude)
Tan Azimuth = 0.737640253
Azimuth
=
S 36.4° E CAzimuth takes combined
name of C correction and Hour Angle - If LHA
is between 0° and 180°, it is named "west", if
LHA is between 180° and 360° it is named "east"
True Azimuth= 143.6°
Compass Azimuth = 145.0°
Compass Error = 1.4° West
Variation 4.0 East
Deviation: 5.4 West
Direction: Strictly write in 4 bond paper, because my activity
sheet is have 4 spaces. This is actually for maritime.
industry course, but I think geometry can do this.
use nautical almanac.
Sample Calculation (Amplitude- Sun):
On 07th May 2006 at Sunset, a vesel in position 10°00'N
0 10°00' W observed the sun bearing 288° by compass. Find
the
compass error.
LMT Sunset
07d
18h
13m
(+)00d
00h
40 м
LIT:
UTC Sunset:
07d
18h
53 m
added - since
longitude is
westerly
Declination Co7d 18h): N016° 55.5'
d(0.7):
(+)
00-6
N016 56.1'
Declination Sun:
Sin Amplitude Sin Declination (Los Latitude
- Sin 016° 56.1'/Cos 10°00'
= 0.295780189
Amplitude = WI. 2N (The prefix of amplitude is
named easterly if body is rising.
and westerly of body is setting.
The suffix is named came as
declination.)
True Bearing: 287.20
Compass Bearing
288.0°
Compass Error: 0.8' West
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