Integral using height to find volume

Question

How do you find the volume of a "pit" which is circular in horizontal cross-section, and parabolic in vertical cross-section using height by "sticking". "Sticking" is when we insert a dipstick through an opening over the deepest part of the buried container until the dipstick touches the bottom, then we pull the stick out and read the liquid level showing on the stick. The specifications of the "pit" are: 40 feet deep at the center and 40 feet across at the top.

Answer 1

If you know three points of a parabola, you the equation of that parabola as well. We know the points (0,0) which is the center, (20, 40) which is the right "side" when looking at our pit from the side, and (-20, 40) which is the other side of the pit, at the top. The method is really quite cool. The general equation for a parabola is

$$ax^2 + bx + c = y$$

so we can input our values for x and y, thus giving us a system of equations in three variables. Another thing to realize here is that this quadratic equation is even and consequently consists of only even terms ($x^2$, $x^0$). Therefore, $b$ has to be 0.

$a(0)^2 + b(0)+c=0$

$a(20)^2+b(20)+c=40$

$a(-20)^2+b(20)+c=40$

Obviously, from the first equation, c=0, so let's update our stuff.

$400a+20b=40$

$400a+20b=40$

Remembering that b=0, we have

$400a = 40$

$a=1/10$

Great! So our quadratic equation is $y=\frac{x^2}{10}$. Now, we just have to integrate. The inverse of the function is $y = \sqrt{10x}$. Although that is only half of the function, we don't need the other half for our purposes.

Think about the area of a circle. For any value x, the area of the circle at that particular point is going to be $\pi r^2$. But the radius is just our height, or y value. Therefore, the area of the circle at any point is going to be $\pi [f(x)]^2$.

As a result, our integral becomes:

$$\int^{40}_{0}\pi \Big(\sqrt{10x}\Big)^2 dx$$

$$\pi \int^{40}_{0} 10x \space dx$$

$${5 \pi x^2}{}$$

will give us the volume between 0 and 40 if we put in 40. Therefor, we have

$${5 \pi 40^2}{}$$

Therefore, the volume is $8000 \pi$.

Answer 2

Hint:

fix the bottom apex of the pit to be (0,0). $y = bx^2$ Put x = 20 and y = 40 and find $b = \frac{1}{10}$

radius $= x$, height $= 40-y$, you have the parabola $y = \frac{1}{10}x^2$

Volume $$ V= 2\pi\int_{0}^{20}x\left(40-\frac{1}{10}.x^2\right)dx = 8000\pi$$

Or using ring method, $$V=\pi \int_{0}^{40} (\sqrt{10y})^2dy$$

Either way you get $V = 8000\pi$

Go ahead and find the volume V

Answer 3

First write down an equation relating the depth of the pit $z$ and the distance from the center $r$.

Mouse-over the empty boxes to reveal each step.

A parabola such that $z = 0$ when $r = \pm 20$ can be written: $z = z_0 (r - 20)(r + 20)$. The condition that $z = 40$ when $r = 0$ gives $z_0 = -\frac{1}{10}$. Therefore, $z = \frac{1}{10} \bigl( 400 - r^2 \bigr)$.

Slicing the volume of the pit into horizontal circular disks, express the area of each disk in terms of the depth.

$A = \pi r^2 = \pi \bigl( 400 - 10z \bigr)$

Integrate to find the total volume.

$V = \displaystyle \int_0^{40} \!\!\! A(z) \, dz = \pi \int_0^{40} \!\! \bigl( 400 - 10z \bigr) \, dz = \pi \bigl( 400z - 5z^2 \bigr) \bigg|_0^{40} = 8000\pi$ ft$^3\!$.