Volume about the x axis problem (1 Viewer)

[js992]

The area in the first quadrant under y=1 and above y=2-2/x is rotated about the x axis. Find the volume formed.

Chapter 12E Q14 Cambridge year 11 3U

I keep getting the answer wrong and am not sure if i'm doing it right or not.


2
1



The answer is 8 ln 2 - 4 -__-
​

 

[Drongoski]

Easier if you formulate solution via cylindrical shells.


You misquoted the book's answer.

 

[life92]

The area in the first quadrant under y=1 and above y=2-2/x is rotated about the x axis. Find the volume formed.

Chapter 12E Q14 Cambridge year 11 3U

I keep getting the answer wrong and am not sure if i'm doing it right or not.


2
1



The answer is 8 ln 2 - 4 -__-
​

This is actually obviously very close to the answer.
Btw you are missing a pi XD

But anyway, as you can see, you integrated from 1=>2, however, if you look at a diagram, there is still a space underneath y=1 in the first quadrant.
This is the missing volume that you are leaving out behind, which you can simply find by finding the area of the cylinder.

It has radius 1 (from y=0 to y=1) and height 1 (from x=0 to x=1), so therefore the volume of this cylinder is pi

Now the answer you got is pi (8ln2 - 5), and if you add pi, then you get pi(8ln2-4), which is the correct answer.

Hope that helps !

 

[frenzal_dude]

So are the limits from x=2 to x=0? I tried subbing those in, x=2 is fine, but you can't sub in x=0 because ln0 and 1/0 = error.

I tried just pretending they = 0, and I got the right answer. But is this the right way to do it?

 

[frenzal_dude]

The area in the first quadrant under y=1 and above y=2-2/x is rotated about the x axis. Find the volume formed.

Chapter 12E Q14 Cambridge year 11 3U

I keep getting the answer wrong and am not sure if i'm doing it right or not.


2
1



The answer is 8 ln 2 - 4 -__-
​

In your integral, how did you get 3-(4/x^2) + 4/x?

Doesn't 1^2 - (2-2/x)^2 = -3 + 8/x - 4/(x^2)?