I’m trying to calculate the outside surface area to paint a water tank that’s a right cylinder with a hemisphere on top. The cylinder has radius 3 m and height 8 m; the hemisphere has the same radius 3 m. The bottom of the cylinder is open (not painted).
I’m confused about which circular areas to include or exclude. Do I subtract the circle where the hemisphere meets the cylinder? I keep second-guessing whether that shared circle is visible or counted twice.
My (probably wrong) attempt: I treated the cylinder as if it had both top and bottom and the hemisphere as a full sphere, so I did 2πr(h + r) + 4πr^2. Plugging in r = 3, h = 8, I got 2π·3·(8+3) + 4π·9 = 66π + 36π = 102π m². This feels off because I’m almost certainly counting hidden surfaces.
What’s the correct way to set up the surface area expression here? Which surfaces exactly should be included, and should that shared circular face be subtracted entirely?
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3 Responses
Ohhh, nice shape! A cylinder wearing a little hemispherical hat. This is exactly the kind of geometry mash-up that makes my pattern-loving brain light up.
What counts as “outside surface” here?
– The curved side of the cylinder (that’s the rectangle you’d get if you unrolled it).
– The curved surface of the hemisphere (half of a sphere’s surface).
– Not the bottom (it’s open).
– Not the flat circular interface where the hemisphere meets the cylinder. That circle is just a seam/edge; it’s not an exposed flat face you can paint.
So the pieces to add are:
– Cylinder lateral area: 2πrh
– Hemisphere curved area: 2πr^2
Why no subtraction for the shared circle?
– Tempting thought: “Should I subtract πr^2 for the circle where they meet?” But the 2πr^2 formula for a hemisphere already excludes the flat base circle, and 2πrh only covers the cylinder’s side. Neither of these formulas includes a flat circular disk to begin with, so there’s nothing to subtract. The circle at the seam has zero area-it’s a 1D boundary, not a 2D face.
Putting it together for r = 3 m, h = 8 m:
– Cylinder side: 2πrh = 2π·3·8 = 48π
– Hemisphere: 2πr^2 = 2π·9 = 18π
– Total painted area = 48π + 18π = 66π m² ≈ 207.3 m²
Where the common “double-count” trap happens
– If you treat the cylinder as having a top and bottom and the hemisphere as a full sphere (like you tried), you get 2πr(h + r) + 4πr^2. Then it feels natural to subtract the shared circle once (−πr^2) and subtract the open bottom (−πr^2). That gives 2πrh + 4πr^2, which still overshoots because the full sphere wasn’t the right starting point for a hemisphere. I always get tangled there if I go that route.
A quick, simple check example
– Take an even simpler tank: radius 1 m, height 2 m, with a hemisphere on top and an open bottom.
– Our “just add the curved bits” rule says:
– Cylinder side: 2πrh = 2π·1·2 = 4π
– Hemisphere: 2πr^2 = 2π·1 = 2π
– Total = 6π m²
– Sanity checks:
– If the height were 0 (just a hemisphere cap), you’d get 2πr^2. For r = 1, that’s 2π, which matches the known area of a hemisphere’s curved surface. Nice!
Bottom line
– Include: cylinder’s curved side 2πrh and hemisphere’s curved surface 2πr^2.
– Exclude: the open bottom and the circular seam.
– For your numbers: total painted area = 66π m².
I always want to subtract the “mystery circle” where the dome kisses the cylinder, so my first wobbly instinct is A = 2πrh + 2πr^2 − πr^2 (eep!). But that’s me over-subtracting: that shared circle is just a rim with no paintable area-no top disk to remove, since the hemisphere covers it, and the bottom is open, so no bottom disk either. For the outside only, you paint two curved bits: the cylinder’s curved side (2πrh) plus the hemisphere’s curved surface (half a sphere, so 2πr^2). So the total is A = 2πrh + 2πr^2. Worked example with r = 3 m and h = 8 m: A = 2π·3·8 + 2π·3^2 = 48π + 18π = 66π m². The circle where they meet isn’t counted at all, and nothing is double-counted if you just stick to the curved surfaces. If you want a refresher on these pieces, Khan Academy has a nice walkthrough of cylinder and sphere surface areas: https://www.khanacademy.org/math/geometry/hs-geo-solids/hs-geo-surface-area/a/surface-area-of-a-cylinder
Count only the exposed curved bits: cylinder’s lateral area 2πrh plus the hemisphere’s curved area 2πr²-the bottom is open and the shared circle is hidden inside, so don’t include it (and there’s nothing to subtract if you never added it). With r=3, h=8 that’s 2π·3·8 + 2π·3² = 48π + 18π = 66π m², which I’m pretty sure is right.