# Find the volume using cylindrical shells? (Enclosed by x-axis and parabola #y=3x-x^2#, revolved about #x=-1#)

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So far, I have the bounds at 0 and 3.

...I think I've forgotten how to apply shells, exactly.

Assuming thickness is #dx# , and height is the area given (#3x-x^2# )... then what is the radius? I don't know what that would be (is it #x# ?)

What I have, plugged into equation:

#V=2\pi\int_0^3[r*(3-x^2)dx]#

So far, I have the bounds at 0 and 3.

...I think I've forgotten how to apply shells, exactly.

Assuming thickness is

What I have, plugged into equation:

In cylindrical shell method the slice should be parallel to the axis of revolution.

In your question the area bounded revolving around

where

show below the region (shaded) revolving around

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To find the volume using cylindrical shells for the region enclosed by the x-axis and the parabola (y = 3x - x^2) revolved about (x = -1), integrate (2\pi \cdot x \cdot f(x) , dx) from the lower bound of the region to the upper bound, where (f(x) = 3x - x^2).

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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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