The problem involves finding the volume of water in a hemispherical bowl with radius "a" filled to a depth "h". The discussion centers around setting up the integral using the disk method to calculate this volume.
Participants are actively engaging with the problem, exploring different aspects of the setup and seeking clarification on the equations involved. Some guidance has been offered regarding sketching the graph and understanding the geometry, but there is still confusion about specific equations and dimensions.
There is a mention of the depth "h" being measured from the bottom of the bowl, and participants are trying to clarify the relationship between "h" and the radius "a". The discussion reflects uncertainty about the correct mathematical representation of the bowl's shape.
Start by sketching a graph of the function that represents the bowl. I recommend sketching the lower half of the graph of x2 + y2 = a2 to make your computations a little simpler.htoor9 said:Homework Statement
A hemispherical bowl of radius "a" contains water to a depth "h"
Fine the volume of water in the bowl.
Homework Equations
Pi * Integral(R(x))^2
basically the disk method
The Attempt at a Solution
The answer is (Pi*h2*(3a-h))/3...i just have no idea how to set up the integral to get that. Thanks guys
Mark44 said:Start by sketching a graph of the function that represents the bowl. I recommend sketching the lower half of the graph of x2 + y2 = a2 to make your computations a little simpler.
Draw a line across the bowl representing a level of h units. Since h is measured from the bottom of the bowl, if you have located the bowl as I suggested, the water level crosses the y-axis at (0, h - a). For example, if the bowl's radius is 6'' and the water is 2" deep, the water level on the y-axis is at (0, 2 - 6) = (0, -4).
Draw a typical "slice" of the water. What is its volume? That's what you will use for your integrand.
Mark44 said:The bottom point of the bowl is at (0, -a).