# Thread: [SOLVED]Make x the subject in y = 2x - x^2.

1. ## [SOLVED]Make x the subject in y = 2x - x^2.

Hi it's late but I can't seem to single out x in the following equation:

$y = 2x - x^2$

Can it be done?

I want to evaluate the volume by using washers even though I can just use shells

2. You can't solve that equation for x can you be more specific about what you are trying to do?

3. I knew it!

I was trying to find the volume enclosed by this eq. by washers instead of shells.

From my book:

The volume V can also be found using washers; however, the calculations
would be more involved since the given equation would have to be solved
for x in terms of y.
The book, however, tells me that it is possible but does not elaborate

It's really more a question of surmounting such a simple looking equation than actually finding the volume which I can do fine via shells.

I knew it!

I was trying to find the volume enclosed by this eq. by washers instead of shells.

From my book:

The book, however, tells me that it is possible but does not elaborate

It's really more a question of surmounting such a simple looking equation than actually finding the volume which I can do fine via shells.
$y = 2x - x^2$

$x^2 - 2x = -y$

$x^2 - 2x + 1 = 1 - y$

$(x-1)^2 = 1 - y$

$x-1 = \pm \sqrt{1-y}$

$x = 1 \pm \sqrt{1-y}$

washers ...

$R(y) = 1 + \sqrt{1-y}$

$r(y) = 1 - \sqrt{1-y}$

$[R(y)]^2 = 2 + 2\sqrt{1-y} - y$

$[r(y)]^2 = 2 - 2\sqrt{1-y} - y$

$[R(y)]^2 - [r(y)]^2 = 4\sqrt{1-y}$

$V = \pi \int_0^1 4\sqrt{1-y} \, dy = \frac{8\pi}{3}$

shells ...

$V = 2\pi \int_0^2 x(2x-x^2) \, dx = \frac{8\pi}{3}$

5. Woah man you are my hero!

I should be thinking like that but it's 2am

Thank you