My Math Forum [ASK] Water Pipe

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July 4th, 2018, 11:04 PM   #1
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Joined: Nov 2010
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Math Focus: Trigonometry

A water pipe whose diameter is 84 cm and length is 2,4 m can contain rain water with the water height 68 cm like in the attached picture.

Determine:
a. The surface area which gets contact with the water
b. The volume of the water (in liters)

So... How do I do? What is the simple way to determine the area of a... truncated circle? (Don't know what the proper term is.)
Attached Images
 Water Pipe.JPG (12.3 KB, 0 views)

Last edited by skipjack; July 5th, 2018 at 09:52 AM.

 July 5th, 2018, 03:42 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Looks like a pretty standard Calculus II problem. Start by drawing a cross section of the pipe on a coordinate system- a circle with center at (0, 0) and diameter 84, so radius 42, has equation $\displaystyle x^2+ y^2= 42^2$. The bottom of that circle is at y= -42, the top is at y= 42. The surface of the water is a horizontal line at y= 68- 42= 26. At every y, a horizontal line intersects the circle at $\displaystyle x= \pm \sqrt{42^2- y^2}$ so has length $\displaystyle 2\sqrt{42^2- y^2}$. Taking a 'thickness' of $\displaystyle \Delta y$, the area of that thin horizontal strip is $\displaystyle 2\sqrt{42^2- y^2}\Delta y$. Adding those gives the Riemann sum, $\displaystyle \sigma 2\sqrt{42^2- y^2}\Delta y$, that, in the limit, is the integral for the area, $\displaystyle 2\int_{-42}^{26}\sqrt{42^2- y^2}dy$.
 July 5th, 2018, 10:10 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 A "truncated circle" is called a circular segment, and has a standard area formula that can be used for part b.
 July 5th, 2018, 04:59 PM #4 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry Looks like 9 graders wouldn't be able to solve this after all...

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