What happens if you spam Science & Math

Do you get the ban hammer? Do people stop coming to this board?
Do you receive SCCIIIEEEEEEEEEENCNCCCE?

No mods in dis.4chan. Empirically proven many times by individual posters.

Did you READ the POAST

Again we want the area of the yellow section.  Assume first that h<R, so the water level is below the horizontal line through the center of the circle.  The area of the yellow region is equal to half the area of the circle (i.e. \pi R^2) minus the area of the pink and blue areas.

The angle 90-x is equal to \cos^{-1} \frac{R-h}{R}, so the angle x is equal to \frac{\pi}{2} - \cos^{-1} \frac{R-h}{R} = \sin^{-1} \frac{R-h}{R}.  This shows that the area of the two pink sectors is equal to

2 \cdot \frac{\sin^{-1} \frac{R-h}{R}}{2\pi} * \pi R^2 =  R^2 \sin^{-1} \frac{R-h}{R}.

By the formula for the area of triangles, the combined area of the two blue triangles is

2 \left[\frac{1}{2} (R-h)\sqrt{R^2-(R-h)^2}\right] = (R-h)\sqrt{R^2-(R-h)^2}

Subtracting these two areas from half the area of the circle, and multiplying by L, we find the exact same formula for the volume as we did in problem 1.

Now if h > R and the water level is above midway, the picture is a little different, and you're going to be *adding* an area which is above the midline to the area of the bottom half of the circle.  It's almost completely similar to what I just did, and I'll let you work it out.

Again we want the area of the yellow section.  Assume first that h<R, so the water level is below the horizontal line through the center of the circle.  The area of the yellow region is equal to half the area of the circle (i.e. \pi R^2) minus the area of the pink and blue areas.

The angle 90-x is equal to \cos^{-1} \frac{R-h}{R}, so the angle x is equal to \frac{\pi}{2} - \cos^{-1} \frac{R-h}{R} = \sin^{-1} \frac{R-h}{R}.  This shows that the area of the two pink sectors is equal to

2 \left[\cdot \frac{\sin^{-1} \frac{R-h}{R}}{2\pi} \cdot \pi R^2\right] =  R^2 \sin^{-1} \frac{R-h}{R}.

By the formula for the area of triangles, the combined area of the two blue triangles is

2 \left[\frac{1}{2} (R-h)\sqrt{R^2-(R-h)^2}\right] = (R-h)\sqrt{R^2-(R-h)^2}

Subtracting these two areas from half the area of the circle, and multiplying by L, we find the exact same formula for the volume as we did in problem 1.

Now if h > R and the water level is above midway, the picture is a little different, and you're going to be *adding* an area which is above the midline to the area of the bottom half of the circle.  It's almost completely similar to what I just did, and I'll let you work it out.

Again we want the area of the yellow section.  Assume first that h<R, so the water level is below the horizontal line through the center of the circle.  The area of the yellow region is equal to half the area of the circle (i.e. \pi R^2) minus the area of the pink and blue areas.

The angle 90-x is equal to \cos^{-1} \frac{R-h}{R}, so the angle x is equal to \frac{\pi}{2} - \cos^{-1} \frac{R-h}{R} = \sin^{-1} \frac{R-h}{R}.  This shows that the area of the two pink sectors is equal to

2 \left[\cdot \frac{\sin^{-1} \frac{R-h}{R}}{2\pi} \cdot \pi R^2\right] =  R^2 \sin^{-1} \frac{R-h}{R}.

By the formula for the area of triangles, the combined area of the two blue triangles is

2 \left[\frac{1}{2} (R-h)\sqrt{R^2-(R-h)^2}\right] = (R-h)\sqrt{R^2-(R-h)^2}

Subtracting these two areas from half the area of the circle, and multiplying by L, we find the exact same formula for the volume as we did in problem 1.

Now if h > R and the water level is above midway, the picture is a little different, and you're going to be *adding* an area which is above the midline to the area of the bottom half of the circle.  It's almost completely similar to what I just did, and I'll let you work it out.

V(h) = 2000000 \pi-4000000 \sin^{-1} \left(1-\frac{h}{100}\right) - (40000-400h) \sqrt{200h-h^2}

V(h) = \frac{1}{231}\left[2000000 \pi-4000000 \sin^{-1} \left(1-\frac{h}{100}\right) - (40000-400h) \sqrt{200h-h^2}\right]

V(h) = \frac{1}{231}\left[2000000 \pi-4000000 \sin^{-1} \left(1-\frac{h}{100}\right) - (40000-400h) \sqrt{200h-h^2}\right]
= 2755.93-17316.02 \sin^{-1} \left(1-\frac{h}{100}\right) - (173.16-1.73h)\sqrt{200h-h^2}

V(h) = \frac{1}{231}\left[2000000 \pi-4000000 \sin^{-1} \left(1-\frac{h}{100}\right) - (40000-400h) \sqrt{200h-h^2}\right]
= 27199.94-17316.02 \sin^{-1} \left(1-\frac{h}{100}\right) - (173.16-1.73h)\sqrt{200h-h^2}