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Surface of revolution: equation, points, area

Short answer

Rotating the line from (2,0,0) to (4,0,6) about the Z-axis yields P(s,phi)=((2+2s)cos phi, (2+2s)sin phi, 6s). Points: (0,3,3); (-4,0,6); (1.768,1.768,1.5). Area = pi(2+4)*sqrt(40) = 119.2 square units.

Describe the surface of revolution. A straight line from (2, 0, 0) to (4, 0, 6) in the XZ plane is revolved about the Z axis. Write the parametric equation of the surface, find the points for (s, phi) = (0.5, 90 deg), (1, 180 deg) and (0.25, 45 deg), and calculate the curved surface area.

A surface of revolution is formed by rotating a plane curve (the generatrix or profile) about an axis in its plane. Every point of the curve moves on a circle (parallel); each position of the curve is a meridian. For a profile r=f(s)r=f(s), z=g(s)z=g(s) about the Z axis:

P(s,ϕ)=(f(s)cos⁡ϕ, f(s)sin⁡ϕ, g(s)),0≤ϕ≤2π\mathbf P(s,\phi)=\big(f(s)\cos\phi,\ f(s)\sin\phi,\ g(s)\big),\qquad 0\le\phi\le2\pi

Generatrix

The line is r=2+2sr=2+2s, z=6sz=6s for 0≤s≤10\le s\le1.

P(s,ϕ)=((2+2s)cos⁡ϕ, (2+2s)sin⁡ϕ, 6s)\mathbf P(s,\phi)=\big((2+2s)\cos\phi,\ (2+2s)\sin\phi,\ 6s\big)

The surface is the side of a frustum of a cone with bottom radius 2, top radius 4 and height 6.

Points

ssϕ\phir=2+2sr=2+2sP\mathbf P
0.590 deg3(0, 3, 3)
1180 deg4(-4, 0, 6)
0.2545 deg2.5(1.768, 1.768, 1.5)

Curved surface area

Slant length l=(4−2)2+62=40=6.325l=\sqrt{(4-2)^2+6^2}=\sqrt{40}=6.325.

A=π (r1+r2) l=π(2+4)(6.325)=119.2 square unitsA=\pi\,(r_1+r_2)\,l=\pi(2+4)(6.325)=119.2\ \text{square units}

Answer: P(s, phi) = ((2+2s)cos phi, (2+2s)sin phi, 6s); points (0,3,3), (-4,0,6), (1.768,1.768,1.5); area = 119.2 square units.

✓ Answer checked by Claude Opus 5.5 (high effort)

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