Volume of Revolution: Finding % Filled in Bowl

In summary, the equation for the water volume of a hemisphere is x^2 + y^2 = 81y=f(x). To find the water volume when the bowl is filled to a certain height, integrate the equation between x=-9 and x=-4.5.
  • #1
dan greig
41
0
I have a question about finding the volume of a hemisphere, I've got that bit sorted but the next bit asks,

if the bowl is partially filled what percentage of the bowl is filled.

I think i understand the method but i need to find the x value when the bowl is filled 4.5cm of a hemisphere of radius 9cm
 
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  • #2
Have you rotated about the x-axis or y-axis?

~H
 
  • #3
You have the equation of the shape (outline) of the vessel, you have the height (plane) it needs to be filled to...

ie. integrate the outline^2 between 0 and this height and times by pi.

Of course - you want the height:redface: but from what you write... this is 4.5cm ?!?
 
  • #4
about the y axis.

the shape is a circle - x^2 + y^2

is this what you mean by outline?

the equation i have been given is, x^2 + y^2 = 81
 
  • #5
y=f(x)

then [tex]\pi\int_0^{4.5}f(x)^2dx[/tex] should be your answer.

from my immediate thought...
 
  • #6
No, I don't think you have the limits of integration right. Since the water will "fill" from the bottom of the circle up, the intgration should be from x= 9 to x= 4.5 (or, more correctly, -9 to -4.5 so the water doesn't spill out!). Also the problem asked for the percentage that was filled. If the water is 1/2 up the side of the bowl, what percentage is filled?
 
  • #7
if i use -9 and -4.5 as the limits of integraton, what is the function of x -
f(x) i use?

as i have been given the equation for a circle it has confused me a little.
 
  • #8
Úsing my above equation with...

[tex]f(x)=(81-(x-9)^2)^{1/2}[/tex] as the outline should work.

ie. [tex]\pi\int_0^{4.5}(18x-x^2)dx[/tex]

Giving a numerical value of the water filling 31% of the hemisphere - which seems about right...
 
Last edited:

1. How do you calculate the volume of revolution for a bowl?

To calculate the volume of revolution for a bowl, you will need to use the formula V = π∫(R(x))^2 dx, where R(x) is the radius of the bowl at a given point and the integral is taken over the height of the bowl. This will give you the total volume of the bowl.

2. Can you explain the concept of "finding % filled" in a bowl?

"Finding % filled" in a bowl refers to determining the percentage of the bowl that is occupied by a given substance. In the context of volume of revolution, it involves calculating the volume of the substance and dividing it by the total volume of the bowl, then multiplying by 100 to get a percentage.

3. How does the shape of the bowl affect the volume of revolution calculation?

The shape of the bowl is a crucial factor in the volume of revolution calculation. A bowl with a wider base and narrower top will have a larger volume compared to a bowl with a narrower base and wider top, even if they have the same height. This is because the radius of the bowl changes as you move up or down its height.

4. What are some real-world applications of calculating the volume of revolution for a bowl?

Calculating the volume of revolution for a bowl has many real-world applications. It can be used in engineering to design containers or vessels with specific capacities. It is also useful in the food industry to determine the amount of product that can be held in a bowl before spilling over. Additionally, it can be used in physics to calculate the volume of a curved object or in architecture to design curved structures.

5. Are there any limitations to using the volume of revolution method for calculating the filled percentage in a bowl?

There are a few limitations to using the volume of revolution method for calculating the filled percentage in a bowl. Firstly, it assumes that the bowl is a perfect geometric shape and does not take into account any irregularities or imperfections. Secondly, it does not account for any changes in volume due to the substance being compressed or expanding. Lastly, it may not be suitable for very complex bowl shapes, as the mathematical calculations may become too complicated.

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