# Volume of revolution (1 Viewer)

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#### dan greig

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

Staff Emeritus
Gold Member
Have you rotated about the x-axis or y-axis?

~H

#### J77

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 but from what you write... this is 4.5cm ?!?

#### dan greig

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

y=f(x)

then $$\pi\int_0^{4.5}f(x)^2dx$$ should be your answer.

from my immediate thought...

#### HallsofIvy

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?

#### dan greig

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

Úsing my above equation with...

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

ie. $$\pi\int_0^{4.5}(18x-x^2)dx$$

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

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