Step 1: Solution-(1) Given: For R = [0,1] × [0,1] 2 I = √ √(xy)² cos(x³)dxdy R Our aim is to evaluate the given integral Step 2: Calculation: I = S√(xy)² cos(x³)dxdy R I = S₁₂ f₁ x² y² cos(x³) dxdy I = (f²x² cos(x³ ) d x) (f₁ y²dy) = Let x³ 3 t⇒ 3x² dx = dt f₁² fox² cos(x³) dx = cos(t)dt fox² cos(x³)dx = (sint) | So x² cos(x3)dx=(sin1 - sin0) Sox2 cos(x³)dx x² cos(x³)dx = (sin1 - 0) sin1 = 3 So y²dy = (5) 1 0 So y² dy = -0 = 1/3 Hence we get: I 3 = (sinl) (½³) = (sinl) 9
Step 1: Solution-(1) Given: For R = [0,1] × [0,1] 2 I = √ √(xy)² cos(x³)dxdy R Our aim is to evaluate the given integral Step 2: Calculation: I = S√(xy)² cos(x³)dxdy R I = S₁₂ f₁ x² y² cos(x³) dxdy I = (f²x² cos(x³ ) d x) (f₁ y²dy) = Let x³ 3 t⇒ 3x² dx = dt f₁² fox² cos(x³) dx = cos(t)dt fox² cos(x³)dx = (sint) | So x² cos(x3)dx=(sin1 - sin0) Sox2 cos(x³)dx x² cos(x³)dx = (sin1 - 0) sin1 = 3 So y²dy = (5) 1 0 So y² dy = -0 = 1/3 Hence we get: I 3 = (sinl) (½³) = (sinl) 9
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Where the 1/3 come from pls explain?
![Step 1: Solution-(1)
Given: For R
=
[0,1] × [0,1]
2
I = √ √(xy)² cos(x³)dxdy
R
Our aim is to evaluate the given integral
Step 2: Calculation:
I =
S√(xy)² cos(x³)dxdy
R
I = S₁₂ f₁ x² y² cos(x³) dxdy
I = (f²x² cos(x³ ) d x) (f₁ y²dy)
=
Let x³
3
t⇒ 3x² dx
= dt
f₁²
fox² cos(x³) dx = cos(t)dt
fox² cos(x³)dx =
(sint) |
So x² cos(x3)dx=(sin1 - sin0)
Sox2 cos(x³)dx
x² cos(x³)dx
=
(sin1 - 0)
sin1
=
3
So y²dy = (5)
1
0
So y² dy = -0 = 1/3
Hence we get:
I
3
= (sinl) (½³) = (sinl)
9](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e53cf2c-4bf5-4ccd-bd0e-fb990a97eda5%2F4acf1b3e-42c2-4928-b2af-5b94b2d3ff63%2Ftzlo22_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Step 1: Solution-(1)
Given: For R
=
[0,1] × [0,1]
2
I = √ √(xy)² cos(x³)dxdy
R
Our aim is to evaluate the given integral
Step 2: Calculation:
I =
S√(xy)² cos(x³)dxdy
R
I = S₁₂ f₁ x² y² cos(x³) dxdy
I = (f²x² cos(x³ ) d x) (f₁ y²dy)
=
Let x³
3
t⇒ 3x² dx
= dt
f₁²
fox² cos(x³) dx = cos(t)dt
fox² cos(x³)dx =
(sint) |
So x² cos(x3)dx=(sin1 - sin0)
Sox2 cos(x³)dx
x² cos(x³)dx
=
(sin1 - 0)
sin1
=
3
So y²dy = (5)
1
0
So y² dy = -0 = 1/3
Hence we get:
I
3
= (sinl) (½³) = (sinl)
9
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