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EXAMPLE 3 Evaluate fle Vx + z-ov, where E is the region bounded by the paraboloid y - x +z 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 D, onto the xy-plane, which is the parabolic region in the middle figure. (The trace of y - x + 2 in the plane z-0 is the parabola y= .) 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) | -1 sx1, x sys 1, -vy - xszs vy- x"> and so we obtain +zdz 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 + ys1 shown in the bottom figure. Then the left boundary of E is the paraboloid y= x + z and the boundary is the plane y - 1, so taking u(x, z) - x + z and u2(x, z) - 1, we have dA Although this integral could be written as (1 - x - z) v + z'dz dx it's easier to convert to polar coordinates in the xz-plane: x - rcos(8), z - rsin(9). This gives v + zov dr de dr

EXAMPLE 3 Evaluate fle Vx + z-ov, where E is the region bounded by the paraboloid y - x +z 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 D, onto the xy-plane, which is the parabolic region in the middle figure. (The trace of y - x + 2 in the plane z-0 is the parabola y= .) 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) | -1 sx1, x sys 1, -vy - xszs vy- x"> and so we obtain +zdz 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 + ys1 shown in the bottom figure. Then the left boundary of E is the paraboloid y= x + z and the boundary is the plane y - 1, so taking u(x, z) - x + z and u2(x, z) - 1, we have dA Although this integral could be written as (1 - x - z) v + z'dz dx it's easier to convert to polar coordinates in the xz-plane: x - rcos(8), z - rsin(9). This gives v + zov dr de dr

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section: Chapter Questions
Problem 18T
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EXAMPLE 3 Evaluate ff; vx +z-dv, where E is the region bounded by the paraboloid y = x + z 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 D, onto the xy-plane, which is the parabolic region in the middle figure. (The trace of y = x + z? in the plane z = 0 is the parabola y = .) 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) | -1s x1, x² s y s 1, -vy - x szs vy - x*} and so we obtain Vx + z'dz dy dx Although this expression is correct, it is extremely difficult to evaluate. So let's instead consider as a type 3 region. As such, its projection D3 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 = 1, so taking p(x, z) = x² + z and P2(x, z) = 1, we have vx² + z-dV = vt + z-dy]dA dA Although this integral could be written as (1 - x - z?) v + z*dz dx it's easier to convert to polar coordinates in the xz-plane: x = rcos(0), z = rsin(e). This gives Vx + z-dV = dr de dr 27
Transcribed Image Text:EXAMPLE 3 Evaluate ff; vx +z-dv, where E is the region bounded by the paraboloid y = x + z 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 D, onto the xy-plane, which is the parabolic region in the middle figure. (The trace of y = x + z? in the plane z = 0 is the parabola y = .) 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) | -1s x1, x² s y s 1, -vy - x szs vy - x*} and so we obtain Vx + z'dz dy dx Although this expression is correct, it is extremely difficult to evaluate. So let's instead consider as a type 3 region. As such, its projection D3 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 = 1, so taking p(x, z) = x² + z and P2(x, z) = 1, we have vx² + z-dV = vt + z-dy]dA dA Although this integral could be written as (1 - x - z?) v + z*dz dx it's easier to convert to polar coordinates in the xz-plane: x = rcos(0), z = rsin(e). This gives Vx + z-dV = dr de dr 27
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