How do you find the volume of the solid obtained by rotating the region bounded by the curves y=x, x=0, and y=(x^2)-6 rotated around the y=3?

1 Answer
Aug 2, 2015

piint_0^3(3-(x^2-6))^2-(3-x)^2dx=(603pi)/5

Explanation:

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The grey region is what we will be rotating around the horizontal line y=3.

The outer radius is 3-(x^2-6)

The inner radius is 3-x

Using the method of washers

piint_0^3(3-(x^2-6))^2-(3-x^2)^2dx

piint_0^3(9-x^2)^2-(3-x)^2dx

piint_0^3 81-18x^2+x^4-(9-6x+x^2)dx

piint_0^3 81-18x^2+x^4-9+6x-x^2dx

piint_0^3 72-19x^2+x^4+6xdx

Integrating

pi[72x-19/3x^3+x^5/5+3x^2]

pi[72(3)-19/3(3)^3+3^5/5+3(3)^2]

pi[216-171+243/5+27]

pi[72+243/5]

pi[360/5+243/5]=(603pi)/5