Saturday, April 27, 2013

Suppose you have an region bounded by the curves y = x^3 an y = x^2.  Now you rotate the region about the x axis to for a three dimension figure. What is the volume of this newly formed figure?

To calculate the volume,  you need to know the equation of the curve that forms the outer edge of the shape minus the equation of the curve that forms the inner edge of the shape. The bounds of integration are the points of intersection of the curves.

Set the 2 equations equal to each other and solve for x.

x^3 = x^2

x^3 - x^2 = 0

x^2(x - 1) = 0

x = 0 and x = 1. This can also be seen by a graph of the two equations.   

Now that we have the points of intersection, we take pi times the integral of the outer radius squared minus the inner radius squared

integrate from 0 to 1 of Pi[(x^2)^2 - (x^3)^2]

integral from 0 to 1 of Pi[x^4 - x^6]

Pi[x^5/5 - x^7/7] evaluated from 0 to 1

Pi(1/5 - 1/7) = Pi(7/35 - 5/35) = Pi(2/35)

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