Wednesday, December 21, 2011

Calculus: Early Transcendentals, Chapter 6, 6.3, Section 6.3, Problem 6

The shell has the radius x , the cricumference is 2pi*x and the height is 4x - x^2 - x , hence, the volume can be evaluated, using the method of cylindrical shells, such that:
V = 2pi*int_(x_1)^(x_2) x*(3x - x^2) dy
You need to evaluate the endpoints x_1 and x_2 , such that:
4x - x^2= x => 3x -x^2 = 0 => x(3 - x) = 0 => x = 0 and 3-x = 0 => x = 3
V = 2pi*int_0^3 x*(3x - x^2) dy
V = 2pi*(int_0^3 3x^2 dx - int_0^3 x^3dx)
Using the formula int x^n dx = (x^(n+1))/(n+1) yields:
V = 2pi*(3x^3/3 - x^4/4)|_0^3
V = 2pi*(x^3 - x^4/4)|_0^3
V = 2pi*(3^3 - 3^4/4)
V = 2pi*(3^3)/4
V = (27pi)/2
Hence, evaluating the volume, using the method of cylindrical shells, yields V = (27pi)/2.

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