Identify the surfaces with cylindrical equations z = 4 r2 and 2r2 + z2 = 1.
Answer:
Not sure if I did it right… §15.8: Triple Integrals in Cylindrical Coordinates Outcome A: Convert an equation from rectangular coordinates to cylindrical coordinates, and vice versa. The cylindrical coordinate system describes a point (x,y,z) in rectangular space in terms of the triple (r,θ,z) where r and θ are the polar coordinates of the projection (x,y,0) and z is the (signed) distance from the projection (x,y,0) to (x,y,z). The transformation from cylindrical coordinates to rectangular coordinates is
x = rcosθ, y = rsinθ, z = z,
and the inverse transformation is r2 = x2 + y2, tanθ = y/x, z = z.
We can transformation equations in x, y, and z to equations in r, θ, and z, and vice versa. Examples. (a) The hyperboloid of two sheets −x2 −y2 + z2 = 1 becomes −r2 + z2 = 1 or z2 = 1 + r2. (b) The shifted paraboloid x2 + y2 −z = 2y becomes r2 −z = 2rsinθ or z = r2 −rsinθ = r(r−sinθ). © The equation 2r2 + z2 = 1 becomes the ellipsoid 2(x2 + y2) + z2 = 1.
(d) The equation r = 5 becomes the cylinder x2 + y2 = 25.
Outcome B: Describe a solid in cylindrical coordinates. Cylindrical coordinates are useful in problems where there is a symmetry about the z-axis. Examples. (a) A solid lies between the cylinder x2+y2 = 1 and the sphere x2+y2+z2 = 4. In cylindrical coordinates, these equations are r2 = 1, r2 + z2 = 4.
These surfaces intersect along the two curves given by r2 = 1,z2 = 3.
The cylindrical coordinate description of the solid is {(r,θ,z) : 0 ≤ r ≤ 1,0 ≤ θ ≤ 2π,− √4−r2 ≤ z ≤√4−r2. Here is a picture of this solid.
(b) What is the solid {(r,θ,z) : 0 ≤ θ ≤ π/2,0 ≤ r ≤ 2,r ≤ z ≤ 2}? The solid lies between z = r and z = 2 over the first quadrant 0 ≤ θ ≤ π/2. The surfaces z = r and z = 2 intersect along the curve r = 2, z = 2. The surface z = r is the cone z =px2 + y2, and so the solid lies above this cone, below the horizontal plane z = 2, all over that part of the disk of radius 2 in the first quadrant in the xy-plane. Here is a picture of this solid.
Outcome C: Evaluate a triple integral by converting it to cylindrical coordinates. Conversion of a triple integral to cylindrical coordinates MAY simplify the iterated integration. Let f(x,y,z) be continuous on the solid E = {(x,y,z) : (x,y) ∈ D,u1(x,y) ≤ z ≤ u2(x,y)} where D = {(x,y) : α ≤ θ ≤ β,h1(θ) ≤ r ≤ h2(θ)}.
The triple integral of f in cylindrical coordinates is ZZZE f dV =ZZD"Z u2(x,y) u1(x,y) f(x,y,z) dz#dA =Z β α Z h2(θ) h1(θ) Z u2(r cosθ,r sinθ) u1(r cosθ,r sinθ) f(rcosθ,rsinθ,z) rdrdθ.
Example. Let E be the solid enclosed by the two planes z = 0, z = x + y + 5 and between the two cylinders x2 + y2 = 4, x2 + y2 = 9. The bottom of this solid is z = 0, and the top of this surface is given by z = x + y + 5, or in cylindrical coordinates,
z = rcosθ + rsinθ + 5 = r(cosθ + sinθ) + 5.
The “sides” of this solid are the cylinders, r2 = 4 and r2 = 9. A cylindrical coordinate description of this solid is E = {(r,θ,z) : 2 ≤ r ≤ 3,0 ≤ θ ≤ 2π,0 ≤ z ≤ 5 + r(cosθ + sinθ)}. Here is a picture of this solid.
The triple integral of f(x,y,z) = x over E is Z 2π 0 Z 3 2 Z 5+r(cosθ+sinθ) 0 r2 cosθ dzdrdθ =Z 2π 0 Z 3 2 zr2 cosθz=5+r(cosθ+sinθ) z=0 drdθ =Z 2π 0 Z 3 2 (5 + r(cosθ + sinθ)r2 cosθ drdθ =Z 2π 0 Z 3 2 5r2 cosθ + r3(cos2 θ + cosθsinθ)drdθ =Z 2π 0 5r3 3 cosθ + r4 4 (cos2 θ + cosθsinθ)r=3 r=2 dθ =Z 2π 0 5 327−8cosθ + 81−16 4 cos2 θ + cosθsinθdθ =Z 2π 0 95 3 cosθ + 65 4 (cos2 θ + cosθsinθdθ =Z 2π 0 95 3 cosθ + 65 41 + cos2θ 2 + cosθsinθdθ =95sinθ 3 + 65 4θ 2 + sin2θ 4
