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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

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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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|
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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