Question
1) A region is bounded by the line y = x and the parabola y = x2 - 6x + 10. What is the volume of the solid generated by revolving the region about the x-axis?
Answers
find their intersection
x^2 - 6x + 10 = x
x = 2 or x = 5
volume = π [integral](x^2 - (x^2 - 6x + 10)^2 dx from 2 to 5
I would expand (x^2 - 6x + 10)^2
after that it just becomes a routine integration of simple terms and careful arithmetic.
x^2 - 6x + 10 = x
x = 2 or x = 5
volume = π [integral](x^2 - (x^2 - 6x + 10)^2 dx from 2 to 5
I would expand (x^2 - 6x + 10)^2
after that it just becomes a routine integration of simple terms and careful arithmetic.