# Volume of bottle from cross section

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• Mar 17th 2011, 06:53 AM
mathguy80
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.
• Mar 18th 2011, 05:41 AM
earboth
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. $\displaystyle |\overline{AB}| =|\overline{EF}|$

$\displaystyle x = 2 \sqrt{2}$ 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
• Mar 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.
• Mar 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 . . . $\displaystyle AB \:=\:4\sqrt{2}$

Quote:

(b) Hence find Volume of the bottle.

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

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

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

The bottom solid is a hemisphere of radius 4.

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

The total volume is: .$\displaystyle 16\sqrt{2}\pi + \dfrac{128}{3}\pi \;=\;\dfrac{16(8 + 3\sqrt{2})}{3}\pi$

• Mar 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.