EXAMPLE 3 Evaluate [√x² + zdv, where E is the region bounded by the paraboloid y = x² + 2² and the plane y = 1. SOLUTION The solid E is shown in the top figure. If we regard it as a type 1 region, then we need to consider its projection D₁ onto the xy-plane, which is the parabolic region in the middle figure. (The trace of y=x² + 2² in the plane z = 0 is the parabola y = .) From y = x² + 2² we obtain z = + , so the lower boundary surface of E is z = -√y-x and the upper boundary surface is z = x x . Therefore the description of E as a type 1 region is x E = {(x, y, z) | -1 sx1, x² sys1, -√y-x² ≤z≤ √y-x²) and so we obtain Mar2 [[[√x √x² + Z²dv= LIV √x² + z dz dy dx 1-2 Although this expression is correct, it is extremely difficult to evaluate. So let's instead consider E as a type 3 region. As such, its projection D3 onto the xz-plane is the disk x² + y² ≤ 1 shown in the bottom figure. Then the left boundary of E is the paraboloid y=x+² and the boundary is the plane y = 1, so taking u₁(x, z) = x² + 2² and u₂(x, z) = 1, we have [[[√x √x² + z²dv- INY √x² + zdy]dA dA X Although this integral could be written as [E -x²² - 2²³²) √x²³² + z²dz dx it's easier to convert to polar coordinates in the xz-plane: x = rcos(8), z = rsin(8). This gives [[[√x √x² + z²dV= (1-x²-2²) √x² +2²)dA - SS ₁₁. -6²6 - 1. de C x = 2π - 4x 15 ✓ 14 dr de dr X
EXAMPLE 3 Evaluate [√x² + zdv, where E is the region bounded by the paraboloid y = x² + 2² and the plane y = 1. SOLUTION The solid E is shown in the top figure. If we regard it as a type 1 region, then we need to consider its projection D₁ onto the xy-plane, which is the parabolic region in the middle figure. (The trace of y=x² + 2² in the plane z = 0 is the parabola y = .) From y = x² + 2² we obtain z = + , so the lower boundary surface of E is z = -√y-x and the upper boundary surface is z = x x . Therefore the description of E as a type 1 region is x E = {(x, y, z) | -1 sx1, x² sys1, -√y-x² ≤z≤ √y-x²) and so we obtain Mar2 [[[√x √x² + Z²dv= LIV √x² + z dz dy dx 1-2 Although this expression is correct, it is extremely difficult to evaluate. So let's instead consider E as a type 3 region. As such, its projection D3 onto the xz-plane is the disk x² + y² ≤ 1 shown in the bottom figure. Then the left boundary of E is the paraboloid y=x+² and the boundary is the plane y = 1, so taking u₁(x, z) = x² + 2² and u₂(x, z) = 1, we have [[[√x √x² + z²dv- INY √x² + zdy]dA dA X Although this integral could be written as [E -x²² - 2²³²) √x²³² + z²dz dx it's easier to convert to polar coordinates in the xz-plane: x = rcos(8), z = rsin(8). This gives [[[√x √x² + z²dV= (1-x²-2²) √x² +2²)dA - SS ₁₁. -6²6 - 1. de C x = 2π - 4x 15 ✓ 14 dr de dr X
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
![EXAMPLE 3
Evaluate JJSE √x² + zdv, where E is the region bounded by the paraboloid y = x² + 2² and the plane y = 1.
SOLUTION The solid E is shown in the top figure. If we regard it as a type 1 region, then we need to consider its projection D₁ onto the xy-plane, which is the parabolic region
in the middle figure. (The trace of y = x² + z² in the plane z = 0 is the parabola y =
.) From y = x² + z² we obtain z = +
, so the lower boundary surface of Eis z = -√y - x² and the upper boundary surface is z =
x
x
. Therefore the description of E as a type 1 region is
x
E = {(x, y, z) | -1 ≤ x1, x² ≤ y ≤ 1, -√y - x² ≤z≤ √y-x²}
and so we obtain
]] [√x + ZOV =¸
-L
/y-x²
√x² + 2-dz dy dx
Although this expression is correct, it is extremely difficult to evaluate. So let's instead consider E as a type 3 region. As such, its projection D3 onto the xz-plane is the disk x² +
y² ≤ 1 shown in the bottom figure. Then the left boundary of E is the paraboloid y = x² + z² and the boundary is the plane y = 1, so taking u₁(x, z) = x² + 2² and u₂(x, z) = 1, we
have
√x² + z-dy]dA
]] [√x² + ZOV =¸
-IN₂²
- SS[
dA
X
Although this integral could be written as
√1-2²
IN
(1-x²-2²) √x² + z²dz dx
√1-22
it's easier to convert to polar coordinates in the xz-plane: x = rcos(8), z = rsin(0). This gives
[[[
√x² + z²dV=
(1-x²-2²) √x² + z²)dA
v = [[ ₁₁.
-LLL
- ™ de
X
= 2π
4₁
15
1
dr de
dr
X](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feace2bdd-ed98-480f-bf7e-07f2184cf474%2F22fb7d55-82c0-4d0d-9a32-b9bce8fe9a70%2Fvn8a81_processed.png&w=3840&q=75)
Transcribed Image Text:EXAMPLE 3
Evaluate JJSE √x² + zdv, where E is the region bounded by the paraboloid y = x² + 2² and the plane y = 1.
SOLUTION The solid E is shown in the top figure. If we regard it as a type 1 region, then we need to consider its projection D₁ onto the xy-plane, which is the parabolic region
in the middle figure. (The trace of y = x² + z² in the plane z = 0 is the parabola y =
.) From y = x² + z² we obtain z = +
, so the lower boundary surface of Eis z = -√y - x² and the upper boundary surface is z =
x
x
. Therefore the description of E as a type 1 region is
x
E = {(x, y, z) | -1 ≤ x1, x² ≤ y ≤ 1, -√y - x² ≤z≤ √y-x²}
and so we obtain
]] [√x + ZOV =¸
-L
/y-x²
√x² + 2-dz dy dx
Although this expression is correct, it is extremely difficult to evaluate. So let's instead consider E as a type 3 region. As such, its projection D3 onto the xz-plane is the disk x² +
y² ≤ 1 shown in the bottom figure. Then the left boundary of E is the paraboloid y = x² + z² and the boundary is the plane y = 1, so taking u₁(x, z) = x² + 2² and u₂(x, z) = 1, we
have
√x² + z-dy]dA
]] [√x² + ZOV =¸
-IN₂²
- SS[
dA
X
Although this integral could be written as
√1-2²
IN
(1-x²-2²) √x² + z²dz dx
√1-22
it's easier to convert to polar coordinates in the xz-plane: x = rcos(8), z = rsin(0). This gives
[[[
√x² + z²dV=
(1-x²-2²) √x² + z²)dA
v = [[ ₁₁.
-LLL
- ™ de
X
= 2π
4₁
15
1
dr de
dr
X
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