Results 1 to 6 of 6

Math Help - Integration applied to Physics problem

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    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
    Last edited by galactus; November 24th 2008 at 05:38 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by galactus View Post
    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

    we see the answer is

    \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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by TheEmptySet View Post
    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Mathstud28 View Post
    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!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Applied problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 1st 2008, 05:44 PM
  2. Applied integration? Dosage problem:
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 4th 2008, 12:18 PM
  3. Tough Applied Integration Question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 7th 2008, 03:19 PM
  4. physics, acceleration, physics problem
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: September 29th 2007, 03:50 AM
  5. Applied Maths and Physics problem
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: June 26th 2007, 03:54 PM

Search Tags


/mathhelpforum @mathhelpforum