Revolving a tank around the y axis: Work needed to pump water

[think4432]

1. A tank is formed by revolving y = 3x^2, x = [0,2] around the y-axis is filled to the 4 feet point with water (w = 62.4 lb/ft^3). Find the work necessary to pump the water out of the tank over the top.

I got the integral being from a = 0 and b = 4 and integrating [(12y-y^2)/3]dy with w and pi as constants outside of the integral.

And after the integration I got:

224/9 pi (62.4)

And when I multiplied the 62.4 out it came out to be 1553.066 pi

Would this be correct? Just seems bit of a weird number?

[I asked this question earlier...and received help on it on an earlier thread, but I just wanted to see if the numbers I got are correct]

 

[HallsofIvy]

At each y, the figure rotated around the y-axis is a disk or radius x= (y/3)^{1/2} and so has area \pi x^2= (\pi/3)y and a very thin disk of thickness dy would have volume (\pi/3)y dy.

The weight of that disk is (\pi w/3)y dy and must be lifted the remaining 12- y feet so the work done in lifting just that disk is (force times distance) (\pi w/3) y(12- y)dy.

Integrate that from y= 0 to y= 4- that appears to be exactly what you have done!

 

[think4432]

At each y, the figure rotated around the y-axis is a disk or radius x= (y/3)^{1/2} and so has area \pi x^2= (\pi/3)y and a very thin disk of thickness dy would have volume (\pi/3)y dy.

The weight of that disk is (\pi w/3)y dy and must be lifted the remaining 12- y feet so the work done in lifting just that disk is (force times distance) (\pi w/3) y(12- y)dy.

Integrate that from y= 0 to y= 4- that appears to be exactly what you have done!

Wow. Ok.

Thanks! The number just seems really weird! Haha.

Thats all!

Thanks though! :]