Help me with my hw

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mathworld.wolfram.com/CylindricalSegment.html
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inb4 mobilefag

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V = pi * R^2 * h * cos(pi/6) = pi * 9 * 8 * sqrt{3}/2 = 36*pi*sqrt{3} = 195.890

A * cos(pi/6) = pi * R^2 = 9 pi = A * sqrt{3}/2 -> A = 18 pi/sqrt{3} = 97.945

>A * cos(pi/6) = pi * R^2 = 9 pi = A * sqrt{3}/2 -> A = 18 pi/sqrt{3} = 97.945

pi/3, damn

A * cos(pi/3) = pi * R^2 = 9 pi = A * 1/2 -> A = 18 pi = 56.548

it said that one was wrong for the cylinder but that's the last attempt i have

(int from ln2 to ln3 of e^(3x)) - (int from ln2 to ln3 of e^(x)) = 1/3(3^3-2^3) - (3-2) = 1/3(27-8)-1=19/3-1=16/3

This is wrong. The only way I can see getting this done is through Fubini's theorem.

isn't it formula 22 though ?
mathworld.wolfram.com/CylindricalSegment.html

I found it doing scalar product, but maybe I am misled...

fuck im a brainlet

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>surface area of a cylindrical segment

bruh

>horizontal cross-section

ok, I read wrong.
I cut it vertically at some random spot

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Let me get my IUTT notes out...

Shiiet, the first one gets my juices flowing. The rest is brainlet tier, but the first one is something one should really know and it's not even that hard, you just have to either find a good parametrization or brute-force it through

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shear is irrelevant to volume, only to surface area. A sheared cylinder will have the same volume as a non-sheared one with same height.

So the volume is pi * r^2 * h * cos(theta)

I forgot to add

brainlets

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Shiiiet, nigga. That's right ain't it. Studying highly theoretical shit makes you forget basic shit I suppose.

I don't have any more attempts on the slanted cylinder one lmao pls get the others

Why the fuck wouldn't it be just pi*r^2*h tho

Look at the image. h isn't the height of a straight cylinder.

keep watching that shit hard and realize how much of a brainlet are you

so it was right with theta=pi/6 instead of pi/3

A = 18 pi/sqrt{3} = 97.945

the more you keep spamming questions, the less likely it is for anyone to answer

it is, i had a brainlet moment too and assumed h was the length along the axis

If you straighten that cylinder h will be too short and the volume will be too small. You want the hypotenuse of the imaginary triangle with the two smaller sides being h and the horizontal bit between h and the bottom-right of the cylinder.