Volume of solid generated by revolving region

[MathsLearner]

I am trying to solve the problem

My attempt is the volume of the solid about the y axis is
[math] \pi {R(y)}^2 - \pi {r(y)}^2 [/math]the triangle equation and the diagram is

The Line1 equation is [math] y = x - 1[/math] but i am not clear of line 2 equation which is [math] x = 1 [/math] i am not able to write in terms of y. Without equation1 in the volume equation
[math] \int_0^1 \pi (y+1)^2 dy [/math]. The answer is [math] \frac {7\pi} 3 [/math]. But the actual answer is [math] \frac {4\pi} 3 [/math]. So, the answer difference is because of the missing equation? Please help

 

[Steven G]

Suppose you revolved line 1 about the y-axis. What would that volume give you, can you describe in words what volume you want to subtract off, that is what will it look like? That last sentence should help you find the volume without even integrating. Now back to using integrals, why do you need to write x=1 in terms of y?? If you draw any horizontal strip from x=0 to x=1 what is the length of this strip? Does it depend on y? Besides you are not trying to find y, but rather you want to find r. IF r is a function of x, then you need to write r as a function of y, but if r is a constant then r(y) equals that constant.

By the way, based on your r you found the volume if you revolved about the line x=1. Do you see that.

 

[HallsofIvy]

Your "r" is 1, the distance from the y-axis (x= 0) to x= 1. Yes the line through (1, 0) and (2, 1) is y= x-1 or x= y+ 1 so your "R" is y+ 1. You want to integrate \(\displaystyle (\pi R^2- \pi r^2)dy= (\pi (y+1)^2- \pi (1))dy= \pi (y^2+ 2y)dy\) with y from 0 to 1.