Hi it's late but I can't seem to single out x in the following equation:
$\displaystyle y = 2x - x^2$
Can it be done?
I want to evaluate the volume by using washers even though I can just use shells
Hi it's late but I can't seem to single out x in the following equation:
$\displaystyle y = 2x - x^2$
Can it be done?
I want to evaluate the volume by using washers even though I can just use 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 elaborateThe 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.
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.
$\displaystyle y = 2x - x^2$
$\displaystyle x^2 - 2x = -y$
$\displaystyle x^2 - 2x + 1 = 1 - y$
$\displaystyle (x-1)^2 = 1 - y$
$\displaystyle x-1 = \pm \sqrt{1-y}$
$\displaystyle x = 1 \pm \sqrt{1-y}$
washers ...
$\displaystyle R(y) = 1 + \sqrt{1-y}$
$\displaystyle r(y) = 1 - \sqrt{1-y}$
$\displaystyle [R(y)]^2 = 2 + 2\sqrt{1-y} - y$
$\displaystyle [r(y)]^2 = 2 - 2\sqrt{1-y} - y$
$\displaystyle [R(y)]^2 - [r(y)]^2 = 4\sqrt{1-y}$
$\displaystyle V = \pi \int_0^1 4\sqrt{1-y} \, dy = \frac{8\pi}{3}$
shells ...
$\displaystyle V = 2\pi \int_0^2 x(2x-x^2) \, dx = \frac{8\pi}{3}$