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Thread: [ASK] Water Pipe

  1. #1
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    [ASK] Water Pipe

    A water pipe whose diameter is 84 cm dan length is 2,4 m can contain rain water with the water height 68 cm like in the attached picture.
    [ASK] Water Pipe-water-pipe.jpg
    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? (dunno what the proper term is)
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  2. #2
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    Re: [ASK] Water Pipe

    Pretty basic Calculus II. Draw a cross-section of the pipe on a coordinate system: a circle with center at (0, 0) and radius 42. Its equation is $\displaystyle x^2+ y^2= 42^2$. The bottom of the circle is at y= -42 and the top is at y= 42. Draw a line at height y= -42+ 68= 26 representing the surface of the water. A line across that circle at any given y has length $\displaystyle 2\sqrt{42^2- y^2}$. The cross section area from the bottom of the pipe, y= -42, to y= 26 is given by $\displaystyle 2 \int_{-42}^{26} \sqrt{42^2- y^2}dy$.

    The volume of water in the pipe is that area times the length of the pipe.
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    Re: [ASK] Water Pipe

    Is there any other approach without using calculus or trigonometry? This is supposed to be for 9 graders.
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    Senior Member Vinod's Avatar
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    Re: [ASK] Water Pipe

    Quote Originally Posted by Monoxdifly View Post
    Is there any other approach without using calculus or trigonometry? This is supposed to be for 9 graders.
    Hello,
    So the volume of water is $2.4m(882\pi+1764\arcsin(\frac{13}{21})+208\sqrt{1 7}) i-e 115.3443 metre$
    Last edited by Vinod; Jul 5th 2018 at 08:55 PM.
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    Re: [ASK] Water Pipe

    I think no need calculus, but requires simple trigonometry
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    Re: [ASK] Water Pipe

    Okay, whether it is calculus or trigonometry, it is incomprehensible for a 9th grader.
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    Re: [ASK] Water Pipe

    Quote Originally Posted by Monoxdifly View Post
    A water pipe whose diameter is 84 cm dan length is 2,4 m can contain rain water
    with the water height 68 cm like in the attached picture.
    Click image for larger version. 

Name:	Water Pipe.JPG 
Views:	4 
Size:	12.3 KB 
ID:	38828
    Determine:
    a. The surface area which gets contact with the water
    b. The volume of the water (in liters)
    Part b. is easy: dump the water from the pipe into a measuring glass !!
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  8. #8
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    Re: [ASK] Water Pipe

    [ASK] Water Pipe-clipboard.jpg

    The area above half full is seen to be the sum of 2 equal sized sectors and 2 equal sized triangles.

    The area of each triangle is given by

    $A_T=\dfrac{b h}{2} = \dfrac{\sqrt{r^2-h^2}\cdot h}{2}$

    The sector angle is given by

    $\theta = \arcsin\left(\dfrac{h}{r}\right)$

    The area of each sector is thus

    $A_S = \pi r^2 \cdot \dfrac{\arcsin\left(\dfrac{h}{r}\right)}{2\pi} = r^2 \dfrac{\arcsin\left(\dfrac{h}{r}\right)}{2}$

    combining these we get the area of the blue portion is

    $A_{total} = \dfrac{\pi r^2}{2} + 2(A_T+A_S) = \dfrac{\pi r^2}{2} + \sqrt{r^2-h^2}\cdot h+ r^2 \arcsin\left(\dfrac{h}{r}\right)$
    Thanks from topsquark
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