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

Question

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]#

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Answer 1 · 2018-06-06T09:14:42

#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#:

james