Home > Calculus > Determining the Volume of a Solid of Revolution

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

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]#

Answer 1

Paisley Johnson

#color(blue)[V=2piint_0^3(x+1)(3x-x^2)*dx=(45pi)/2#

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

In your question the area bounded revolving around #x=-1# so our integral will given by :

#color(red)[V=2piint_0^3(x+1)(3x-x^2)*dx#

where #(x+1)# is the radius.

#V=2piint_0^3(x+1)(3x-x^2)*dx#

#V=2piint_0^3(3x^2-x^3+3x-x^2)*dx=2piint_0^3(2x^2-x^3+3x)*dx#

#2pi*[2/3*x^3-1/4*x^4+3/2*x^2]_0^3=(45pi)/2#

show below the region (shaded) revolving around #x=-1#:

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Answer 2

Scarlett Allison

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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Answer from HIX Tutor

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.