4U Volumes - Cylindrical shells (1 Viewer)

[soisorce26]

wow, it's my first post. anyways can someone help me with the following questions:

1. The region bounded by the curves y=1/(x+1) and y=1/(x+2), the ordinates x=0 and x=2 is rotated about the y-axis. Calculate the volumes generated.

2. The area b/w the curves y=1/(x²+1) and y=1/(x²+2), the ordinates x=-2 and x=2 is rotated about the y-axis. Find the volume formed.

 

[SeDaTeD]

Q1.

We shall do this by considering cylindrical shells of a small thickness, finding their volumes then summing to find the whole volume.

For a typical point x between x=0 and x=2, consider the cylindrical shell generated by rotating the region between the curves from x to x + δx, for small δx.

The height of the cylinder (at x) is h = [1/(x+1) - 1/(x+2)] = 1/(x+1)(x+2). We shall use this value to approximate the height in the interval x to x + δx. (This approximation becomes more accurate as δx becomes smaller).

Since we are rotating about the y-axis, the cylindrical shell will have inner radius x and outer radius x + δx. (Just the distance from the axis of rotation to the point). Thus, the volume of the shell (δV) is

δV = pi*(x+δx)^2*h - pi*x^2*h
= pi*h*[(x+δx)^2 - x^2]
= pi*h*(x+δx-x)(x+δx+x)
= pi*h*(2x+δx)δx
= 2pi*h*xδx, as (δx)^2 is negligible for δx small.
= [2pi*x/(x+1)(x+2)] δx

Now, if we sum all the volumes (δV) from x=0 to x=2 and take the limit as δx->0, this will give us the volume of the solid.

V = lim δx->0 Σ {x=0 to x=2} δV
= ∫ {x=0 to x=2} dV, (by definition of the (Riemann) Integral)
= ∫ {x=0 to x=2} [2pi*x/(x+1)(x+2)] dx
= 2pi∫ {x=0 to x=2} (2/(x+2) - 1/(x+1)) dx, (by partial fractions - or inspection)
= 2pi [2ln(x+2) - ln(x+1)] {x=0 to x=2}
= 2pi*ln(4/3), after some simplification.

(Hoping I didn't make any errors)

 

[soisorce26]

oh i see, thanks alot ^^