√x MATH SOLV
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![6:40 ✔✔✔
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The scaler field F is given as F = x²z+y²z
Step 1: Introduction
As per the question, we are asked to compute the surface integral of a scalar field F = x²z+y² z over a
hemisphere defined by x² + y² + ² = 36 with z > 0.
We will accomplish this by first parametrizing the surface using spherical coordinates, substituting these
parameters into the scalar field F, and then performing the double integration over the parameter domain.
The term F d.S in the double integral refers to the flux of a vector field through surface S.
First, we need to parameterize the hemisphere.
Step 2: Finding the surface integral
We can use spherical coordinates for this.
& ASK
To calculate the integral FdS of a scalar field F over a surface S, we will parametrize the surface S,
plug this parameterization into F, and then integrate over the parameter space.
The spherical coordinates are related to Cartesian coordinates by
x = p sin cos
y=psin osin
z = p cos o
The surface area element dS in spherical coordinates is given by p² sin o do do.
Splitting the integral gives:
J.F
The 0-integral just gives a factor of 2πT.
In spherical coordinates, the hemisphere x² + y² + ² = 36, z ≥ 0 is given by p = 6,0 ≤ ≤ π/2, and
0 ≤ 0 < 2TT.
Substituting the expressions for x, y, z into the scalar field F gives
F = (6 sin cos 0)2 (6 cos p) + (6 sin o sin 0)2(6 cos ) = 216 sin² cos
Now we can substitute these values into the surface integral:
π/2
[[Fds = [" ["
FdS=
Therefore, the surface integral is
Solution
T/2
|||
Vo))
LTE1
FdS = 216-6²
The -integral can e calculated by using the substitution u = sind, du = cos o do:
= [² ²³ du = [ + ][ -= =
S
T/2
5.6² 1²h 40 6*¹² sin³ cos o do
216 sin² p cos 6²sin o do de
sin³ o cos ó do =
VO) 4G+ 35%
LTE2
+
J FdS216-62.2πT.
√x MATH SOLV
1
4
O
= 3888T
Therefore the value of the given scaler
surface integral is : 38881
11](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fed5788d2-a0b4-4eb1-998c-a4248e13736e%2F2ec9e714-31a3-40f0-a038-0e290bbd32fa%2Fwr5cy9n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:6:40 ✔✔✔
bartleby.com/questior
Q SEARCH
The scaler field F is given as F = x²z+y²z
Step 1: Introduction
As per the question, we are asked to compute the surface integral of a scalar field F = x²z+y² z over a
hemisphere defined by x² + y² + ² = 36 with z > 0.
We will accomplish this by first parametrizing the surface using spherical coordinates, substituting these
parameters into the scalar field F, and then performing the double integration over the parameter domain.
The term F d.S in the double integral refers to the flux of a vector field through surface S.
First, we need to parameterize the hemisphere.
Step 2: Finding the surface integral
We can use spherical coordinates for this.
& ASK
To calculate the integral FdS of a scalar field F over a surface S, we will parametrize the surface S,
plug this parameterization into F, and then integrate over the parameter space.
The spherical coordinates are related to Cartesian coordinates by
x = p sin cos
y=psin osin
z = p cos o
The surface area element dS in spherical coordinates is given by p² sin o do do.
Splitting the integral gives:
J.F
The 0-integral just gives a factor of 2πT.
In spherical coordinates, the hemisphere x² + y² + ² = 36, z ≥ 0 is given by p = 6,0 ≤ ≤ π/2, and
0 ≤ 0 < 2TT.
Substituting the expressions for x, y, z into the scalar field F gives
F = (6 sin cos 0)2 (6 cos p) + (6 sin o sin 0)2(6 cos ) = 216 sin² cos
Now we can substitute these values into the surface integral:
π/2
[[Fds = [" ["
FdS=
Therefore, the surface integral is
Solution
T/2
|||
Vo))
LTE1
FdS = 216-6²
The -integral can e calculated by using the substitution u = sind, du = cos o do:
= [² ²³ du = [ + ][ -= =
S
T/2
5.6² 1²h 40 6*¹² sin³ cos o do
216 sin² p cos 6²sin o do de
sin³ o cos ó do =
VO) 4G+ 35%
LTE2
+
J FdS216-62.2πT.
√x MATH SOLV
1
4
O
= 3888T
Therefore the value of the given scaler
surface integral is : 38881
11
Transcribed Image Text:Find [[ F·dS, F = x²z+ y²z,S is the hemisphere x² + y² +z² = 36,z≥0.
(a) 1296л (b) 324л (c) 64л (d) 32л
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