July 5th, 2018, 12:04 AM  #1 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry  [ASK] Water Pipe
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.) Last edited by skipjack; July 5th, 2018 at 10:52 AM. 
July 5th, 2018, 04:42 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
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, 11:10 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,877 Thanks: 1834 
A "truncated circle" is called a circular segment, and has a standard area formula that can be used for part b.

July 5th, 2018, 05:59 PM  #4 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry 
Looks like 9 graders wouldn't be able to solve this after all...


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