How do you find the volume of the solid obtained by rotating the region bounded by the curves #x=y-y^2# and the y axis rotated around the y-axis?

Question

Charles Anctil · Accepted Answer

#pi/30#

The curve represents a horizontal parabola as seen in the picture

The region rotated about y axis is the shaded region.

The volume of the solid so generated would be(consider an elementary strip of length and thickness #delta#y. If it is rotated about x axis its volume would be #pix^2 dy#. The volume of the solid generated by rotating the whole shaded region would be

#int_0^1 pi x^2 dy#

=#int_0^1 pi (y-y^2)^2 dy#

=#int_0^1 pi(y^2 -2y^3 +y^4)dy#

=#pi(y^3 /3 -2y^4 /4 +y^5 /5)_0^1#

=# (pi)/30#

Luke Alvarez · Answer

undefined To find the volume of the solid obtained by rotating the region bounded by the curves x = y - y^2 and the y-axis around the y-axis, you can use the method of cylindrical shells.

The formula for the volume using cylindrical shells is:

V = ∫[a, b] 2πx * h(y) dy

Where:
- a and b are the limits of integration (the y-values where the curves intersect),
- x = y - y^2 is the expression for the radius of each shell, and
- h(y) is the height of each shell, which is the distance from the curve to the axis of rotation, in this case, the distance from x = y - y^2 to the y-axis.

First, find the limits of integration by setting the equations equal to each other: y - y^2 = 0. This gives y = 0 and y = 1 as the limits of integration.

Next, express x in terms of y: x = y - y^2.

Then, find the height h(y), which is simply x since we are rotating around the y-axis.

Now, you have all the components to set up the integral:

V = ∫[0, 1] 2π(y - y^2) * (y) dy

Integrate this expression with respect to y over the interval [0, 1] to find the volume of the solid of revolution.

Aurora Alvarado · Answer

undefined To find the volume of the solid obtained by rotating the region bounded by the curves $ x = y - y^2 $ and the y-axis around the y-axis, you can use the method of cylindrical shells.

The formula to find the volume using cylindrical shells is:

$$ V = 2\pi \int_{a}^{b} xf(y) \, dy $$

Where $ f(y) $ represents the distance from the axis of rotation to the outer curve, and $ a $ and $ b $ represent the limits of integration.

First, you need to find the limits of integration by solving the equation $ x = y - y^2 $ for $ y $, which yields $ y = \frac{1}{2} \pm \frac{\sqrt{5}}{2} $.

Next, you need to express $ x $ in terms of $ y $ as $ x = y - y^2 $. Then, plug $ x $ and $ f(y) $ into the formula and integrate from the lower limit to the upper limit.

$$ V = 2\pi \int_{\frac{1}{2} - \frac{\sqrt{5}}{2}}^{\frac{1}{2} + \frac{\sqrt{5}}{2}} (y - y^2)(y) \, dy $$

After integrating, you'll get the volume of the solid obtained by rotating the region.