Volume of bottle from cross section

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March 17th 2011, 06:53 AM
mathguy80

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Volume of bottle from cross section

Hi,

Another geometry problem!

Attachment 21170

The diagram shows the vertical cross-section of a bottle of air refresher. The three curved parts of the diagram are circular arcs of radius 4 cm, with center O. AB is the diameter of circular base, CD is the diameter of the horizontal circular cross-section through O, and EF is the diameter of the cylindrical neck of the bottle. AF and BE are perpendicular to each other.

(a) Find AB.

(b) Hence find Volume of the bottle.

For (a) the right triangle gives AB = 5.66. But for (b) there seems to be so much happening here, not sure how to proceed.

Thanks for your help.
March 18th 2011, 05:41 AM
earboth

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Quote:

Originally Posted by mathguy80

Hi,

Another geometry problem!

Attachment 21170

The diagram shows the vertical cross-section of a bottle of air refresher. The three curved parts of the diagram are circular arcs of radius 4 cm, with center O. AB is the diameter of circular base, CD is the diameter of the horizontal circular cross-section through O, and EF is the diameter of the cylindrical neck of the bottle. AF and BE are perpendicular to each other.

(a) Find AB.

(b) Hence find Volume of the bottle.

For (a) the right triangle gives AB = 5.66. But for (b) there seems to be so much happening here, not sure how to proceed.

Thanks for your help.

1. All given points are placed on a circle around O with r = 4 cm.

2. $|\overline{AB}| =|\overline{EF}|$

$x = 2 \sqrt{2}$ and the length of the orange distance has the same length.

3. The bottle is composed of

- a spherical layer with r = 4 and the thickness x;
- a cylinder with r = x and h = x
- a spherical segment with r = 4 and h = r - x
March 18th 2011, 08:00 AM
mathguy80

Quote:

3. The bottle is composed of

- a spherical layer with r = 4 and the thickness x;
- a cylinder with r = x and h = x
- a spherical segment with r = 4 and h = r - x

Damn, you made it so straightforward! So the bottle can be rearranged as 1/2 sphere with radius 4 cm and one cylinder with radius and height 2.82. So volume is 205.1 cubic cm. That checks out!

Very cool, @earboth. Thanks.
March 18th 2011, 09:07 AM
Soroban

Hello, mathguy80!

Another approach . . .

Quote:

Code:

* * * * X * E o---------------o F .| * * |. | * * | . | * * | . C o-*-------o-------*-o D * * O * * 4* *4 * * * * A o---------------o B . Y . . . .

The diagram shows the vertical cross-section of a bottle of air refresher.
The three curved parts of the diagram are circular arcs of radius 4 cm, with center O.
AB is the diameter of circular base,
CD is the diameter of the horizontal circular cross-section through O,
and EF is the diameter of the cylindrical neck of the bottle.
AF and BE are perpendicular to each other. . ABFE is a square!

(a) Find AB.

You are right . . . $AB \:=\:4\sqrt{2}$

Quote:

(b) Hence find Volume of the bottle.

Take the spherical cap at $\,X$ and move it to the South Pole $\,Y.$

Then the top solid is a cylinder with radius $2\sqrt{2}$ and height $2\sqrt{2}$

. . Its volume is: . $V_1 \:=\:\pi r^2h \:=\:\pi(2\sqrt{2})^2(2\sqrt{2}) \:=\:16\sqrt{2}\pi$

The bottom solid is a hemisphere of radius 4.

. . Its volume is: . $V_2 \:=\:\frac{1}{2} \times \frac{4}{3}\pi(4^3) \:=\:\dfrac{128}{3}\pi$

The total volume is: . $16\sqrt{2}\pi + \dfrac{128}{3}\pi \;=\;\dfrac{16(8 + 3\sqrt{2})}{3}\pi$
March 25th 2011, 03:03 AM
mathguy80

Hey Soroban,

Sorry I missed this reply! Your solution work pretty well too. I didn't make the connection that the bottle is sphere-like. Thanks for your help.