A bowl shaped like a hemisphere with radius 60 cm is partially filled with water. The depth of the water is 10cm in the center of the bowl. How many liters of water are in the bowl?
My somewhat silly attempt to solve this was the following: 1/6th of the bowl is filled with water (I think), and the volume of a hemisphere is $(2/3)\pi r^3$, the amount of liters of water in the bowl must be 1/6th of that: $(1/9 )\pi r^3$ which not surprisingly gave the wrong answer. I know I'm supposed to solve this using integration; I just don't know where to begin.
The red circled triangle has sides $R$ (the radius of the sphere), $r$ (the radius of the water surface) and $R-h$ (where $h$ is the water height). Pythagoras tells you that $R^2 = r^2 + (R-h)^2.$
Best Answer
Your bowl can be modeled in 2-D by the equation $x^2+y^2=60^2$. We will make that in terms of $y$ to be $y=\sqrt{3600-x^2}$. We know the bounds are $50, 60$ because your shape is $10$ cm deep. In order to calculate the volume sweeped out by the curve between $50, 60$, we use the integral $$\pi\int_{50}^{60}\sqrt{3600-x^2}^2 dx$$ where the $\pi \sqrt{3600-x^2}^2 dx$ is the volume of a tiny cylindrical slice of your shape, and the integral is to add up those infinitelay many infinitely thin cylindrical slices into the shape the water forms in the bowl.
I think you can calculate the integral on your own.