EXAMPLE 3 Evaluate fle Vx + z'dv, where E is the region bounded by the paraboloid y = x + z? and the plane y = 9. 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 D, onto the xy-plane, which is the parabolic region in the middle figure. (The trace of y = O is the parabola y = + z? in the plane z = .) From y = x + z we obtain z = + , so the lower boundary surface of E is z = -Vy - x and the upper boundary surface is z = Therefore the description of E as a type 1 region is E = {(x, y, z) | -3 s x 3, x² s y s 9, -vy - x² s zs Vy - x³} and so we obtain Vx + z*dV = Vx + zdz dy dx correct, it is extremely difficult to evaluate. So let's instead consider E as such, its projection Dz onto the xz-plane is the disk x + y s 9 shown in the bottom figure. Then the left boundary of E is Although this expression type 3 region. As the paraboloid y = x2 + z? and the boundary is the plane y = 9, so taking u,(x, z) = x2 + z? and p2(x, z) = 9, we have Vx + zidV = - SSC dA Although this integral could be written as (9 - x - z?) vx + z*dz dx it's easier to convert to polar coordinates in the xz-plane: x = rcos(e), z = rsin(e). This gives Vx + z*dV = (9 - x² - z) vx + z*)dA dr de dA dr
EXAMPLE 3 Evaluate fle Vx + z'dv, where E is the region bounded by the paraboloid y = x + z? and the plane y = 9. 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 D, onto the xy-plane, which is the parabolic region in the middle figure. (The trace of y = O is the parabola y = + z? in the plane z = .) From y = x + z we obtain z = + , so the lower boundary surface of E is z = -Vy - x and the upper boundary surface is z = Therefore the description of E as a type 1 region is E = {(x, y, z) | -3 s x 3, x² s y s 9, -vy - x² s zs Vy - x³} and so we obtain Vx + z*dV = Vx + zdz dy dx correct, it is extremely difficult to evaluate. So let's instead consider E as such, its projection Dz onto the xz-plane is the disk x + y s 9 shown in the bottom figure. Then the left boundary of E is Although this expression type 3 region. As the paraboloid y = x2 + z? and the boundary is the plane y = 9, so taking u,(x, z) = x2 + z? and p2(x, z) = 9, we have Vx + zidV = - SSC dA Although this integral could be written as (9 - x - z?) vx + z*dz dx it's easier to convert to polar coordinates in the xz-plane: x = rcos(e), z = rsin(e). This gives Vx + z*dV = (9 - x² - z) vx + z*)dA dr de dA dr
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 fSSE Vx² + z?dV, where E is the region bounded by the paraboloid y = x2 + z? and the plane y = 9.
y = x²+z?
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
E
projection D, onto the xy-plane, which is the parabolic region in the middle figure. (The trace of y = x2 + z² in the plane z =
the parabola y =
.) From y = x² + z? we obtain z = +
, so the lower boundary surface of E is z =
and the upper boundary surface is z =
. Therefore the description of E as a type 1 region is
E = {(x, y, z) | -3 < x 3, x² < y s 9, -Vy - x² < z s Vy - x²}
y
and so we obtain
Vx2 + z?dV =
Vx2 + z?dz dy dx
Vy-z2
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 x? + y? < 9 shown in the bottom figure. Then the left boundary of E is
the paraboloid y = x² + z? and the boundary is the plane y = 9, so taking (x, z) = x² + z? and µ2(x, z) = 9, we have
Vx2 + z?dV =
Vx? + z?dy]dA
dA
Although this integral could be written as
/9-z2
(9 - x2 - z?) Vx² + z?dz dx
it's easier to
polar coordinates in the xz-plane: x = rcos(0), z = rsin(0). This gives
Vx2 + z?dV =
(9 - x2 - z?) vx? + z²]dA
-27
-LI
dr de
70
2T
de
dr
=
– 21|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F67afbb67-017a-4ebc-bbcb-7bf247bdda00%2Fdbde1cd1-c476-4027-b515-c7facb593a70%2Furxe3nl_processed.png&w=3840&q=75)
Transcribed Image Text:EXAMPLE 3
Evaluate fSSE Vx² + z?dV, where E is the region bounded by the paraboloid y = x2 + z? and the plane y = 9.
y = x²+z?
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
E
projection D, onto the xy-plane, which is the parabolic region in the middle figure. (The trace of y = x2 + z² in the plane z =
the parabola y =
.) From y = x² + z? we obtain z = +
, so the lower boundary surface of E is z =
and the upper boundary surface is z =
. Therefore the description of E as a type 1 region is
E = {(x, y, z) | -3 < x 3, x² < y s 9, -Vy - x² < z s Vy - x²}
y
and so we obtain
Vx2 + z?dV =
Vx2 + z?dz dy dx
Vy-z2
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 x? + y? < 9 shown in the bottom figure. Then the left boundary of E is
the paraboloid y = x² + z? and the boundary is the plane y = 9, so taking (x, z) = x² + z? and µ2(x, z) = 9, we have
Vx2 + z?dV =
Vx? + z?dy]dA
dA
Although this integral could be written as
/9-z2
(9 - x2 - z?) Vx² + z?dz dx
it's easier to
polar coordinates in the xz-plane: x = rcos(0), z = rsin(0). This gives
Vx2 + z?dV =
(9 - x2 - z?) vx? + z²]dA
-27
-LI
dr de
70
2T
de
dr
=
– 21|
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