Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section: Chapter Questions
Problem 30RE
Related questions
Question
100%
(Urgent) please answer step 8, thank you
![Consider the following equations.
f(y) =
у
√ 36- y²
V
g(y) = 0
y = 3
Sketch the region bounded by the graphs of the functions. Find the area of the region.
Step 1
Write the original algebraic functions:
f(y) =
Y✔
36-y
and g(y) = 0✔
2✔
0.
y
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F32c5117c-6cba-4a4e-a2e4-0e889cc62fe2%2F96a28bf2-18fa-498e-8f38-99dc7393c788%2F9kytjmf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the following equations.
f(y) =
у
√ 36- y²
V
g(y) = 0
y = 3
Sketch the region bounded by the graphs of the functions. Find the area of the region.
Step 1
Write the original algebraic functions:
f(y) =
Y✔
36-y
and g(y) = 0✔
2✔
0.
y
2
![Step 5
Assume g₁(y) = 36-y2. Therefore, g₁'(y) = -2✔
F₁(x) = √/y Therefore, f₁(g₁(y)) =
Define f₁(y)
Step 6
Multiply and divide the given integrand by-2.
-2y
66²²2² 7²/²0
= 72²71/6² (36 - y²) - ¹1/12
2 √ 36-y2 dy =
Hence F(
Step 7
F(y) = 2√
2V
=
11
Step 8
= = 2¹ [F(9₁₂ (Y))]
whereF(g₁(y)) is the antiderivative off (g₁(y)).
Therefore,
2✔
y
Therefore, F(g₁(y)) = 2V 36✔✔
-1
The antiderivative of f₁(y) is
¹15 | √ √ dy = 2√ y✔
2V
36 y
[ (36-²)-¹/22 -2 yoy]
dy
2
-2y
-1
2
dy ====2²2-11
16² = 2²² √ 36-y²
= 6-
36 - Y
0.
32
1
2
-2 y.
✓
[ f(9₁(v) 92 (v) dy
3
2
y
2
-2✔
√36-1² 100
y](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F32c5117c-6cba-4a4e-a2e4-0e889cc62fe2%2F96a28bf2-18fa-498e-8f38-99dc7393c788%2F10q2zj3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Step 5
Assume g₁(y) = 36-y2. Therefore, g₁'(y) = -2✔
F₁(x) = √/y Therefore, f₁(g₁(y)) =
Define f₁(y)
Step 6
Multiply and divide the given integrand by-2.
-2y
66²²2² 7²/²0
= 72²71/6² (36 - y²) - ¹1/12
2 √ 36-y2 dy =
Hence F(
Step 7
F(y) = 2√
2V
=
11
Step 8
= = 2¹ [F(9₁₂ (Y))]
whereF(g₁(y)) is the antiderivative off (g₁(y)).
Therefore,
2✔
y
Therefore, F(g₁(y)) = 2V 36✔✔
-1
The antiderivative of f₁(y) is
¹15 | √ √ dy = 2√ y✔
2V
36 y
[ (36-²)-¹/22 -2 yoy]
dy
2
-2y
-1
2
dy ====2²2-11
16² = 2²² √ 36-y²
= 6-
36 - Y
0.
32
1
2
-2 y.
✓
[ f(9₁(v) 92 (v) dy
3
2
y
2
-2✔
√36-1² 100
y
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