A student has evaluated this integral as follows. The unit normal to S is the vector N = xi+yj+zk. The angle y between γ N and k is y = 0 in spherical coordinates, so that the area element Also in spherical coordinates, is 1 COS Y = 1 COS sphere. Therefore √ √x² + y²dS V 22. 2π = = x² + y² = sin on the x² + y² -do de 0 COS Y •2π sin = do de Cos = 2π [ – In(cos )] - = 2π ( - In(cos(7/4)) + In(cos 0)) = 2π(— In(√√2/2) - In(1)) = -2π In(√√2/2). = =-
A student has evaluated this integral as follows. The unit normal to S is the vector N = xi+yj+zk. The angle y between γ N and k is y = 0 in spherical coordinates, so that the area element Also in spherical coordinates, is 1 COS Y = 1 COS sphere. Therefore √ √x² + y²dS V 22. 2π = = x² + y² = sin on the x² + y² -do de 0 COS Y •2π sin = do de Cos = 2π [ – In(cos )] - = 2π ( - In(cos(7/4)) + In(cos 0)) = 2π(— In(√√2/2) - In(1)) = -2π In(√√2/2). = =-
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter8: Applications Of Trigonometry
Section8.4: The Dot Product
Problem 31E
Related questions
Question
Let S be the part of the sphere x^2 +y^2 +z^2 = 1 that lies above the cone z = sqrt(x^2 + y^2).
Consider the following solution, where is the mistake?
![SL
x² + y²ds.
A student has evaluated this integral as follows.
The unit normal to S is the vector N = xi+yj+zk. The angle y between
N and k is Y
=
1
=
is
COS Y
Cos
sphere. Therefore
1
> in spherical coordinates, so that the area element
Also in spherical coordinates, v
sino on the
x² + y²
=
2πT
π/4
+ y²
x² + y²dS
=
do do
COS Y
2πT
sin o
=
do de
0
COS O
π/4
= 2π [ – In(cos ¢)]™*
0
= 2π ( − In(cos(π/4)) + In(cos 0))
= 2π(− ln(√√2/2) – ln(1)) = −2π ln(√√2/2).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb8d7f700-77ae-4ba4-b5a1-ce4a4719e247%2Fa53bf830-ab23-4735-953f-ab77736f3777%2Fa8yiz56_processed.png&w=3840&q=75)
Transcribed Image Text:SL
x² + y²ds.
A student has evaluated this integral as follows.
The unit normal to S is the vector N = xi+yj+zk. The angle y between
N and k is Y
=
1
=
is
COS Y
Cos
sphere. Therefore
1
> in spherical coordinates, so that the area element
Also in spherical coordinates, v
sino on the
x² + y²
=
2πT
π/4
+ y²
x² + y²dS
=
do do
COS Y
2πT
sin o
=
do de
0
COS O
π/4
= 2π [ – In(cos ¢)]™*
0
= 2π ( − In(cos(π/4)) + In(cos 0))
= 2π(− ln(√√2/2) – ln(1)) = −2π ln(√√2/2).
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