Cereal in a hemisphirical bowl with a radius of three inches.

[cnnagbo]

There exists a hemisphirical bowl with a radius of three inches. What depth must cereal be poured to until the volume of the cereal in the bowl is half of the volume of a full bowl of cereal?

 

[mmm4444bot]

Hello cnnagbo. You didn't mention why you're working this exercise; so I guessed. You're a calculus student, and you've been instructed to use calculus to complete the exercise. (This is why your thread moved from the Advanced Math board to the Calculus board.)

If I've guessed wrongly, please provide some context.

:idea: It helps volunteer tutors, when students explain why they're stuck or show what they've already done.

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[sinx]

There exists a hemisphirical bowl with a radius of three inches. What depth must cereal be poured to until the volume of the cereal in the bowl is half of the volume of a full bowl of cereal?

how would you calculate the volume of a hemisphere, as a summation of circular areas?

 

[cnnagbo]

Hello cnnagbo. You didn't mention why you're working this exercise; so I guessed. You're a calculus student, and you've been instructed to use calculus to complete the exercise. (This is why your thread moved from the Advanced Math board to the Calculus board.)

If I've guessed wrongly, please provide some context.

:idea: It helps volunteer tutors, when students explain why they're stuck or show what they've already done.

These suggestions are mentioned in the forum guidelines; please, take a look. You can start with this summary. Thank you! :cool:

Volume of hemisphere is (2/3) * pi * r ^ 3 = 18 * pi in.^3 if r = 3 in.

Half of the above volume is 9 * pi in.^3 = (2/3) * pi * r ^ 3 gives r = 2.382 in., which is I think is incorrect.

 

[cnnagbo]

how would you calculate the volume of a hemisphere, as a summation of circular areas?

Using a definite integral of 0 to 1.5 and integrating the function (2/3) * pi * r ^ 3

 

[sinx]

Volume of hemisphere is (2/3) * pi * r ^ 3 = 18 * pi in.^3 if r = 3 in.

Half of the above volume is 9 * pi in.^3 = (2/3) * pi * r ^ 3 gives r = 2.382 in., which is I think is incorrect.

draw a bowl with milk in it.
the surface of the milk has area of pi*r²
[where r=3sin theta.] [theta measured from vertical = 0 to horizontal = pi/2, vertical again is pi, etc]
diff Volume of the milk is area x diff depth of milk.
integrate to get the general eqn for volume.
when set Volume eqn= 1/2 of the total hemisphere Volume, that will give you theta for 1/2 volume.

let me know the detailed soln.
the general eqn for volume should generate 2/3*pi*r³ when you integrate from 0 to pi/2, the whole hemisphere.
I did not get through it, but i think this is the way to go.

 

[sinx]

There exists a hemisphirical bowl with a radius of three inches. What depth must cereal be poured to until the volume of the cereal in the bowl is half of the volume of a full bowl of cereal?

instead of using Rsin(theta) as the radius of the top of the milk; try this.
use pathagorean theorem, then the radius of the top of milk=(R_m)²=r²-z² ; where z=depth, and r=radius of bowl.
then, Area of milk=A_m=pi*(r²-z²) and dVolume=A_mdz, where z goes from 0 to z.
then integrate dVolume.
you should get Volume=pi*R²(z-1/3z³)
1/2 volume of hemisphere =1/3(pi)R³
set them equal and solve for z (depth)

with R=3, you should get z=sqrt3.
let me know if i'm right.