Monday, September 29, 2014

Calculus of a Single Variable, Chapter 7, 7.3, Section 7.3, Problem 12

To be able to use the shell method, a a rectangular strip from the bounded plane region should be parallel to the axis of revolution.
By revolving multiple rectangular strip, it forms infinite numbers of hollow pipes or representative cylinders.

In this method, we follow the formula: V = int_a^b (length * height * thickness)
or V = int_a^b 2pi * radius*height*thickness
where:
radius (r) = distance of the rectangular strip to the axis of revolution
height (h) = length of the rectangular strip
thickness = width of the rectangular strip as dx or dy .
For the bounded region, as shown on the attached image, the rectangular strip is parallel to y-axis (axis of rotation). We can a let:
r=x
h=f(x) or h=y_(above)-y_(below)
h = -x^2+1 -0
h = -x^2+1
thickness = dx with boundary values from a=0 to b=1 .
Plug-in the values on V = int_a^b 2pi * radius*height*thickness , we get:
V = int_0^1 2pi*x*(-x^2+1)*dx
V =int_0^1 2pi(-x^3+x)dx
Apply basic integration property: intc*f(x) dx = c int f(x) dx and
int (u+v) dx=int (u) dx+ int (v) dx .
V = 2pi int_0^1 (-x^3+x)dx
V = 2pi[ int_0^1 (-x^3)dx +int_0^1 (x)dx]
Apply power rule for integration: int x^n dy= x^(n+1)/(n+1)
V = 2pi[ -x^(3+1)/(3+1) + x^(1+1)/(1+1)]|_0^1
V = 2pi[ -x^4/4 + x^2/2]|_0^1
Apply definite integration formula: int_a^b f(y) dy= F(b)-F(a) .
V = 2pi[ -(1)^4/4 + (1)^2/2] -2pi[ -(0)^4/4 + (0)^2/2]
V =2pi[-1/4+1/2]-2pi[0+0]
V =2pi[1/4]-2pi[0]
V =pi/2-0
V =pi/2 or 1.57 (approximated value)

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