Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
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answer step 8 please
Transcribed Image Text:Consider the following equations.
f(y) =
g(y) = 0
y = 2
Sketch the region bounded by the graphs of the functions. Find the area of the region.
Step 1
Write the original algebraic functions:
Step 3
f(y) =
and g(y)
3.0
Step 2
Sketch the region bounded by the graphs of the functions.
y
2.5
2.0
1.5
1.0
0.5
y
3.0
y
√36-
2.5
2.0
1.5
1.0
0.5
V
36-
0.2
0.4
0.2 0.4 0.6
⇒ y =
0.6 0.8 1.0 1.2
Set f(y) equal to g(y) and solve for y.
y
V 36-
0✔ 0
✔
0
0.8 1.0
1.2
X
X
2.5
2.0
1.5
1.0
0.5
y
3.0
y
3.0
2.5
2.0
1.5
1.0
0.5
0.2 0.4
0.2 0.4
0.6 0.8 1.0 1.2
0.6 0.8 1.0 1.2
X
X
![Step 3
Set f(y) equal to g(y) and solve for y.
y
Step 4
A =
Define f₁(y) =
Step 6
36
Because g(y)< f(y) for all y in the interval [0, 2], the area of the region bounded by the graphs of f(y) =
S
Step 7
2✔
⇒ y =
Step 8
= ²√36-1²-0
1
Hence F(y) = 2V
Therefore,
0✔
Y✔
✓
Step 5
Assume g₁(y) = 36 - y². Therefore, g₁'(y) = -2 ✓
0
Therefore, F(g₁(y)) = 2V
0
[f(y) - g(y)] dy
Multiply and divide the given integrand by -2.
-2y
√√² = ²²√36²
[²
dy.
dy =
The antiderivative of f₁(y) is
is √ √ v d
Therefore, f₁(9₁(y)) = √
-=[Fe₂OND]²
whereF(g₁(y)) is the antiderivative off₁ (9₁(y)).
0 dy
dy= 2V
36
√²2=²²√36²-√² dy = 72² 11
= 6-
36
(36- -²-¹/2✓
12✔
1
0
(9 (v) [(v) dy
36
-2 y dy
y
V 36 y
29(y) = 0, and y = 2 is:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F32c5117c-6cba-4a4e-a2e4-0e889cc62fe2%2F8ca0dd05-bbbc-4d3e-ad25-60426f14fcfb%2F49d5k37_processed.png&w=3840&q=75)
Transcribed Image Text:Step 3
Set f(y) equal to g(y) and solve for y.
y
Step 4
A =
Define f₁(y) =
Step 6
36
Because g(y)< f(y) for all y in the interval [0, 2], the area of the region bounded by the graphs of f(y) =
S
Step 7
2✔
⇒ y =
Step 8
= ²√36-1²-0
1
Hence F(y) = 2V
Therefore,
0✔
Y✔
✓
Step 5
Assume g₁(y) = 36 - y². Therefore, g₁'(y) = -2 ✓
0
Therefore, F(g₁(y)) = 2V
0
[f(y) - g(y)] dy
Multiply and divide the given integrand by -2.
-2y
√√² = ²²√36²
[²
dy.
dy =
The antiderivative of f₁(y) is
is √ √ v d
Therefore, f₁(9₁(y)) = √
-=[Fe₂OND]²
whereF(g₁(y)) is the antiderivative off₁ (9₁(y)).
0 dy
dy= 2V
36
√²2=²²√36²-√² dy = 72² 11
= 6-
36
(36- -²-¹/2✓
12✔
1
0
(9 (v) [(v) dy
36
-2 y dy
y
V 36 y
29(y) = 0, and y = 2 is:
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