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EXAMPLE 3 Evaluate fE Vx + z'dv, where E is the region bounded by the paraboloid y = x + z and the plane y = 4. 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 = x + z in the plane z = SOLUTION %3D O is the parabola y =

EXAMPLE 3 Evaluate fE Vx + z'dv, where E is the region bounded by the paraboloid y = x + z and the plane y = 4. 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 = x + z in the plane z = SOLUTION %3D O is the parabola y =

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 12T
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no. 3 2.6
EXAMPLE 3 Evaluate [[ſE Vx + z*dV, where E is the region bounded by the paraboloid y = x2 + z² and the plane y = 4. y=+ 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 = x + z' in the plane z = O is the parabola y = ) From v = 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) | -28x2, x? sy s 4, -vy xszs vy - x} and so we obtain Vx + zdv = Vx + z*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 D, onto the xz-plane is the disk x + y s 1 shown in the bottom figure. Then the left boundary of E is the paraboloid y = x' + z? and the boundary is the plane y- 4, so taking u,(x, z) = x2 + z? and u,(x, z) = 4, we have Vx? + z*dV = dA Although this integral could be written as (4 - x - z') vx + z'dz dx /4-r it's easier to convert to polar coordinates in the xz-plane: x = rcos(8), z = rsin(8). This gives
Transcribed Image Text:EXAMPLE 3 Evaluate [[ſE Vx + z*dV, where E is the region bounded by the paraboloid y = x2 + z² and the plane y = 4. y=+ 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 = x + z' in the plane z = O is the parabola y = ) From v = 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) | -28x2, x? sy s 4, -vy xszs vy - x} and so we obtain Vx + zdv = Vx + z*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 D, onto the xz-plane is the disk x + y s 1 shown in the bottom figure. Then the left boundary of E is the paraboloid y = x' + z? and the boundary is the plane y- 4, so taking u,(x, z) = x2 + z? and u,(x, z) = 4, we have Vx? + z*dV = dA Although this integral could be written as (4 - x - z') vx + z'dz dx /4-r it's easier to convert to polar coordinates in the xz-plane: x = rcos(8), z = rsin(8). This gives
Although this integral could be written as (4 - x² - z²) Vx² + z*dz_dx it's easier to convert to polar coordinates in the xz-plane: x = rcos(0), z = rsin(@). This gives Vx² + z²dV_ | đi dên dr op
Transcribed Image Text:Although this integral could be written as (4 - x² - z²) Vx² + z*dz_dx it's easier to convert to polar coordinates in the xz-plane: x = rcos(0), z = rsin(@). This gives Vx² + z²dV_ | đi dên dr op
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