Volume of revolution (1 Viewer)

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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
 

Hootenanny

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Have you rotated about the x-axis or y-axis?

~H
 

J77

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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 ?!?
 
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
 

J77

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y=f(x)

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

from my immediate thought...
 

HallsofIvy

Science Advisor
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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?
 
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.
 

J77

1,061
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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:

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