How do you find the volume of the solid if the region in the first quadrant bounded by the curves #x=y-y^2# and the y axis is revolved about the y axis?

Question

Paisley Johnson · Accepted Answer

I've edited your question to what I think you meant to ask. If I'm mistaken, please accept my apologies. The curve #x=y-y^2# is a parabola that opens to the left.  It has #y# intercepts #0# and #1#. Using discs, we see that the radius is #x# or #y-y^2#, the thickness is #dy#, so the volume or a representative disc is  #pir^2 d9y = pi (y-y^2)^2 dy# Integrate from #y=0# to #y=1# #int_0^1  pi (y-y^2)^2 dy = pi int_0^1   (y^2-3y^3+y^4) dy # Which I believe is #(17 pi)/60#

Evelyn Agard · Answer

undefined To find the volume of the solid generated by revolving the region bounded by the curve $x = y - y^2$ and the y-axis about the y-axis, we will use the method of cylindrical shells.

The formula for the volume of a solid generated by revolving a region about the y-axis using cylindrical shells is given by:

$$ V = 2\pi \int_a^b x \cdot h(y) \, dy $$

where $ h(y) $ represents the height of the shell, and $ a $ and $ b $ represent the limits of integration.

First, we need to find the limits of integration $ a $ and $ b $ by setting $ x = y - y^2 $ equal to zero:

$$ y - y^2 = 0 $$

$$ y(1 - y) = 0 $$

So, $ y = 0 $ and $ y = 1 $ are the bounds of integration.

Next, we express $ x $ in terms of $ y $:

$$ x = y - y^2 $$

Now, we need to find the height of the shell, which is the distance between the curve and the y-axis. This is simply $ x $.

So, our integral becomes:

$$ V = 2\pi \int_0^1 (y - y^2) \cdot (y) \, dy $$

$$ V = 2\pi \int_0^1 (y^2 - y^3) \, dy $$

$$ V = 2\pi \left[ \frac{y^3}{3} - \frac{y^4}{4} \right]_0^1 $$

$$ V = 2\pi \left( \frac{1}{3} - \frac{1}{4} \right) $$

$$ V = 2\pi \left( \frac{4 - 3}{12} \right) $$

$$ V = 2\pi \left( \frac{1}{12} \right) $$

$$ V = \frac{\pi}{6} $$

So, the volume of the solid generated by revolving the region about the y-axis is $ \frac{\pi}{6} $ cubic units.