# 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?

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

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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.

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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