"A cylindrical glass of radius r and height L is filled with water and then tilted until the water remaining in the glass exactly covers its base.
"A cylindrical glass of radius r and height L is filled with water and then tilted until the water remaining in the glass exactly covers its base.
Determine a way to "slice" the water into parallel cross-sections that are trapezoids and then set up a definite integral for the volume of the water."
Can you please provide a solution to this problem and ? and if possible a diagram of your cross sections and computations ? Thanks
View attachment 1389
Per wjm's suggestion, I have been looking at this.
The diagram has no axes pictured, so I am going to place the y axis vertical.
From the top where the water just touches the base.
Now, as wjm stated, make the slices parallel to this y axis.
The area of a trapezoid is \(\displaystyle \frac{1}{2}h(a+b)\). Where a and b are the top and bottom base lengths.
This leaves dx to be the thickness of each slice.
So, the volume of each slice will be the area of the trapezoid times the thickness.
What may be initially difficult to envision is that the sum of the lengths of the trapezoids bases is equal to
L. And, if we place a longitudinal axis through the cylinder (maybe call it z), then the height of the
trapezoid is a distance y on either side of it. Giving the height of the trapezoid equal to 2y.
This means the area is \(\displaystyle \frac{1}{2}2y\cdot L=yL\)
So, the volume of each trapezoid would be \(\displaystyle yLdx\)
You are correct about \(\displaystyle y^{2}=r^{2}-x^{2}\).
So, we end up with \(\displaystyle \int_{-r}^{r}\frac{1}{2}L\cdot 2\sqrt{r^{2}-x^{2}}dx\)
\(\displaystyle =\frac{{\pi}r^{2}L}{2}\).
Which is the result we would expect to arrive at because half of the cylinder is full.
I will try to make a diagram of what I am trying to explain. See the trapezoid in the center?.