Volume of Revolution: R around Y-Axis

[cmab]

R is bounded by the curves y=(x-1)^2 and y=2(x-1). Axis of revolution: y-axis.

How am i supposed to do this. I know how to do the washer method, I know how to apply it when it is revolved arround the x-axis, but I don't know how to do it when it is the y axis. Can anybody explain to me the steps to do ?

 

[OlderDan]

R is bounded by the curves y=(x-1)^2 and y=2(x-1). Axis of revolution: y-axis.

How am i supposed to do this. I know how to do the washer method, I know how to apply it when it is revolved arround the x-axis, but I don't know how to do it when it is the y axis. Can anybody explain to me the steps to do ?

Your integral will be with respect to y, and you need to find the inner and outer radius as a function of y. That involves solving your equations for x.

 

[cmab]

Your integral will be with respect to y, and you need to find the inner and outer radius as a function of y. That involves solving your equations for x.

I did the reciprocals. so its x= sqrt(x)+1 and x=y/2 +1

I did everything and the result is 16pie/12, and in the answer sheet it says 64pie/12

 

[OlderDan]

I did the reciprocals. so its x= sqrt(x)+1 and x=y/2 +1

I did everything and the result is 16pie/12, and in the answer sheet it says 64pie/12

You mean x= sqrt(y)+1 and x=y/2 +1. Is that what you did? The answer given is correct (reduces to 16pi/3). When you take the difference between the lerger area and the smaller area of each washer, two terms cancel and two terms survive that need to be integrated.

 

[cmab]

Thx man, I gotted the answer but I was too tired to realize it :rofl:

 

[cmab]

Answer sheet = 64pie/15
But I think the paper is wrong, cause it keeps giving me 64pi/12 (Unreduced)