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.

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)

2. ## 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.

3. ## Re: [ASK] Water Pipe

Is there any other approach without using calculus or trigonometry? This is supposed to be for 9 graders.

4. ## Re: [ASK] Water Pipe

Originally Posted by Monoxdifly
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 $2.4m(882\pi+1764\arcsin(\frac{13}{21})+208\sqrt{1 7}) i-e 115.3443 metre$

5. ## Re: [ASK] Water Pipe

I think no need calculus, but requires simple trigonometry

6. ## Re: [ASK] Water Pipe

Okay, whether it is calculus or trigonometry, it is incomprehensible for a 9th grader.

7. ## Re: [ASK] Water Pipe

Originally Posted by Monoxdifly
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.

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 !!

8. ## Re: [ASK] Water Pipe

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

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