MHB Find the area by using disk method

[jaychay]

The problem is to solve for the area R.

Can you please help me ?
I have tried to do it many times.

Thank you in advice.

 

[jaychay]

Volume V1 is on x - axis
Volume V2 is on y=1

 

[skeeter]

$\displaystyle V_1 - V_2 = \pi \int_1^4 [f(x)]^2 \, dx - \pi \int_1^4 [f(x)-1]^2 \, dx = 4\pi$

$\displaystyle \int_1^4 [f(x)]^2 \, dx - \int_1^4 [f(x)]^2 - 2f(x) + 1 \, dx = 4$

$\displaystyle \cancel{\int_1^4 [f(x)]^2 \, dx} - \cancel{\int_1^4 [f(x)]^2 \, dx} + \int_1^4 2f(x) - 1 \, dx = 4$

note $\displaystyle \int_1^4 f(x) \, dx = R + 3$
can you finish?

 

[jaychay]

$\displaystyle V_1 - V_2 = \pi \int_1^4 [f(x)]^2 \, dx - \pi \int_1^4 [f(x)-1]^2 \, dx = 4\pi$

$\displaystyle \int_1^4 [f(x)]^2 \, dx - \int_1^4 [f(x)]^2 - 2f(x) + 1 \, dx = 4$

$\displaystyle \cancel{\int_1^4 [f(x)]^2 \, dx} - \cancel{\int_1^4 [f(x)]^2 \, dx} + \int_1^4 2f(x) - 1 \, dx = 4$

note $\displaystyle \int_1^4 f(x) \, dx = R + 3$
can you finish?

Can you explain to me please where did (f(x)-1)^2 come from ? and why you have to put -1 behind f(x) ?

 

[skeeter]

https://www.physicsforums.com/attachments/10787

Can you explain to me please where did (f(x)-1)^2 come from ? and why you have to put -1 behind f(x) ?

rotating R about the line y = 1 using disks ... what is the radius of a disk in this case?