Calculus 2: Finding Work with Integrals

[iRaid]

Homework Statement

Not sure if this goes here or physics, but this is for my calculus 2 class so I decided here would be best.

#22

Homework Equations

W=∫Fdx

The Attempt at a Solution

I think the limits of integration are from 0 to 1 since the water is right under the spout, but not sure.

Would the integral be:
$$\int_0^1 \pi(3^{2})(62.5x)dx$$

 

[LCKurtz]

Homework Statement

Not sure if this goes here or physics, but this is for my calculus 2 class so I decided here would be best.

#22

Homework Equations

W=∫Fdx

The Attempt at a Solution

I think the limits of integration are from 0 to 1 since the water is right under the spout, but not sure.

Would the integral be:
$$\int_0^1 \pi(3^{2})(62.5x)dx$$

You haven't told us which problem you are working, although with the $\pi$ in there, I guess it's the second one. You also haven't told us what $x$ represents. Or where you got the 0 and 1 limits, or how you are analyzing the problem. Show us some steps and how you are setting it up and maybe we can help you.

[Edit] Now I see you indicated problem 22.

 

[HallsofIvy]

As LCKurtz implied, your formula is completely wrong and you don't say how you got it. The work required to lift "j" Newtons of water a distance h meters is jh Joules. The problem here, and the reason you need to use calculus, is that the amount of water varies from "height" to "height".

For simplicity "x" be the height above the center of the spherical tank, so that x varies from -3 to 3 to cover the entire tank. Then $x^2+ y^2+ z^2= 9$ is the equation of the sphere and at a given height, x, the "layer of water" is the circle $y^2+ z^2= 9- x^2$ which has radius $\sqrt{9- x^2}$ and area $\pi(9- x^2)$. Imagining this "layer" to have thickness dx, its volume is $\pi(9- x^2)dx$ and its weight is $\pi(9- x^2)dx$ times the density of water, $\delta$ (I will explain later why I am NOT using "62.4 pounds per cubic foot) $\delta\pi(9- x^2)dx$.

The distance from x to the top of the tank is $3- x$ so the distance that the "layer" of water has to be lifted is $$3- x+ 1= 4- x$$. That means that the work done in lifting that "layer" of water is $\delta\pi (4- x)(9- x^2)dx$.

Integrate that from the bottom of the tank to the top, from x= -3 to x= 3.

(I am puzzled that you are told to "use the fact that water weighs 62 lbs/ft³" when all the lengths are given in meters.)

 

[pasmith]

(I am puzzled that you are told to "use the fact that water weighs 62 lbs/ft³" when all the lengths are given in meters.)

To be fair, he's only told to use that for exercises 23 and 24, which are not shown but presumably have their linear dimensions given in feet. Presumably the fact that in SI units the density of water is $1,000\,\mathrm{kg}\,\mathrm{m}^{-3}$ was thought too obvious to be worth mentioning.

 

[iRaid]

As LCKurtz implied, your formula is completely wrong and you don't say how you got it. The work required to lift "j" Newtons of water a distance h meters is jh Joules. The problem here, and the reason you need to use calculus, is that the amount of water varies from "height" to "height".

For simplicity "x" be the height above the center of the spherical tank, so that x varies from -3 to 3 to cover the entire tank. Then $x^2+ y^2+ z^2= 9$ is the equation of the sphere and at a given height, x, the "layer of water" is the circle $y^2+ z^2= 9- x^2$ which has radius $\sqrt{9- x^2}$ and area $\pi(9- x^2)$. Imagining this "layer" to have thickness dx, its volume is $\pi(9- x^2)dx$ and its weight is $\pi(9- x^2)dx$ times the density of water, $\delta$ (I will explain later why I am NOT using "62.4 pounds per cubic foot) $\delta\pi(9- x^2)dx$.

The distance from x to the top of the tank is $3- x$ so the distance that the "layer" of water has to be lifted is $$3- x+ 1= 4- x$$. That means that the work done in lifting that "layer" of water is $\delta\pi (4- x)(9- x^2)dx$.

Integrate that from the bottom of the tank to the top, from x= -3 to x= 3.

(I am puzzled that you are told to "use the fact that water weighs 62 lbs/ft³" when all the lengths are given in meters.)

I see what you're saying, but are you sure that this is a 3d shape, not a circle? We haven't really dealt with spheres (other than rotating about an axis).

 

[Dick]

I see what you're saying, but are you sure that this is a 3d shape, not a circle? We haven't really dealt with spheres (other than rotating about an axis).

It's not a circle. For one thing they've said 'tank'. For another thing they've put suggestive shading in. For another they are giving densities in terms volume, like cubic meters or cubic feet instead of per square meter or square feet.

 

[iRaid]

It's not a circle. For one thing they've said 'tank'. For another thing they've put suggestive shading in. For another they are giving densities in terms volume, like cubic meters or cubic feet instead of per square meter or square feet.

Ok I see what you're saying.

Thanks for the help to everyone.