# Thread: Integration applied to Physics problem

1. ## Integration applied to Physics problem

Ok, here's the problem: "A gas station stores its gasoline in a tank under the ground. The tank is a cylinder lying horizontally on its side. (iIn other words, the tank is not standing vertically on one of its flat ends.) If the radius of the cylinder is 1 meter, its length is 5 meters, and it stop is 4 meters under the ground, find the total amount of work needed to pump the gasoline out of the tank. (The density of gasoline is 673kg per cubic meter; use g=9.8m/s^2"

I've got an equation for total work that says that Total Work is equal to 32977* the integral (from 0 to 2) of w(6-h)dh with "w" being the width of the cylinder and "h" being the height.

My issue is that I can't figure out a formula by which to convert w to h; only by doing that can I solve the problem. So does anyone have any suggestions? Any help is appreciated.

2. I would set it up like this:

It is 4 feet from the ground to the top of the tank. Therefore, it is 5 feet to the center of the tank.

The distance from the center to the top of the tank we can call x. So, the

height of the gas at x level has width w(x).

$673(5)\int_{-1}^{1}(5-x)(2\sqrt{1-x^{2}})dx$

3. Originally Posted by galactus
I would set it up like this:

It is 4 feet from the ground to the top of the tank. Therefore, it is 5 feet to the center of the tank.

The distance from the center to the top of the tank we can call x. So, the

height of the gas at x level has width w(x).

$673(5)\int_{-1}^{1}(5-x)(2\sqrt{1-x^{2}})dx$
Just to help, for that integral, it may not be the easiest if you dont see it

distributing we get

$\int\bigg[10\sqrt{1-x^2}-2x\sqrt{1-x^2}\bigg]dx$

Lets do the seperately, easiest first

for the second one

$\int{-2x\sqrt{1-x^2}dx}$

either seeing we have the derivative of the quantity, or making th sub

$u=1-x^2$

$\frac{2}{3}(1-x^2)^{\frac{3}{2}}$

Now for the first one we have

$10\int\sqrt{1-x^2}dx$

now making the sub $x=\sin(\theta)\Rightarrow{dx=\cos(\theta)d\theta}$

we get

$10\int\sqrt{1-\sin^2(\theta)}\cos(\theta)d\theta=10\int\cos^2(\t heta)d\theta$

Now using the fact that

$\cos^2(\theta)=\frac{1+\cos(2\theta)}{2}$

we get

$5\int\bigg[1+\cos(2\theta)\bigg]dx$

giving us

$5\bigg[\theta+\frac{1}{2}\sin(2\theta)\bigg]$

now seeing

$\sin(2\theta)=2\sin(\theta)\cos(\theta)$

we need now to make the resub $x=\sin(\theta)\Rightarrow{\theta=\arcsin(x)}$
and using the fact that

$\cos^2(\theta)+\sin^2(\theta)=1\Rightarrow{\cos(\t heta)=\sqrt{1-\sin^2(\theta)}}$

we get

$5\bigg[\arcsin(x)+\sin(\arcsin(x))\cos(\arcsin(x))\bigg]=5\bigg[\arcsin(x)+$ $x\sqrt{1-(\sin(\arcsin(x))^2}\bigg]$ $=5\bigg[\arcsin(x)+x\sqrt{1-x^2}\bigg]$

So combinging them we get

$\int(5-x)2\sqrt{1-x^2}dx=\frac{2}{3}(1-x^2)^{\frac{3}{2}}+5\bigg[\arcsin(x)+x\sqrt{1-x^2}\bigg]+C$

4. Just to point this out

$10\int_{-1}^{1} \sqrt{1-x^2}dx$

Note that $y=\sqrt{1-x^2}$ is a semi circle centered at the orgin with a radius of 1. The limits of integration go from -1 to 1 so get the area of half of a cirlce so

$10\int_{-1}^{1} \sqrt{1-x^2}dx =10 \left( \frac{1}{2}\pi (1)^2\right)=5\pi$

5. Originally Posted by TheEmptySet
Just to point this out

$10\int_{-1}^{1} \sqrt{1-x^2}dx$

Note that $y=\sqrt{1-x^2}$ is a semi circle centered at the orgin with a radius of 1. The limits of integration go from -1 to 1 so we the area of half of a cirlce so

$10\int_{-1}^{1} \sqrt{1-x^2}dx =10 \left( \frac{1}{2}\pi (1)^2\right)=5\pi$
Haha, thanks a lot Empty Set, I always look for that, and the only time I dont see it

at least he can integrate it himself now xD

6. Originally Posted by Mathstud28
Haha, thanks a lot Empty Set, I always look for that, and the only time I dont see it

at least he can integrate it himself now xD
Hey they both work, and I miss stuff all the time.

That is why I love this place. It keeps me on my toes!!