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#?

Question

Emma Aldridge · Accepted Answer

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

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#

Luke Alvarez · Answer

undefined To find the volume of the solid obtained by rotating the region bounded by the curves $ y = x $, $ x = 0 $, and $ y = x^2 - 6 $ around the line $ y = 3 $, you can use the method of cylindrical shells.

The formula for the volume of a solid generated by revolving the region bounded by $ f(x) $, $ g(x) $, and the lines $ x = a $ and $ x = b $ around the line $ y = c $ is:

$$ V = 2\pi \int_a^b (x - c) \cdot |f(x) - g(x)| \, dx $$

Here, $ a $ and $ b $ are the x-values of the intersection points of $ y = x $ and $ y = x^2 - 6 $, which are the solutions of the equation $ x = x^2 - 6 $.

1. First, find the intersection points by solving $ x = x^2 - 6 $.
   $ x = x^2 - 6 $
   $ x^2 - x - 6 = 0 $
   $ (x - 3)(x + 2) = 0 $
   $ x = 3 $ or $ x = -2 $

2. Now, set up the integral:
   $ V = 2\pi \int_{-2}^3 (x - 3) \cdot |(x) - (x^2 - 6)| \, dx $

3. Compute the absolute value:
   When $ x $ is in the range $[-2, 0]$, $ |(x) - (x^2 - 6)| = (x^2 - x + 6) - x = x^2 - 2x + 6 $
   When $ x $ is in the range $[0, 3]$, $ |(x) - (x^2 - 6)| = (x) - (x^2 - 6) = 6 - x^2 + x $

4. Evaluate the integral:
$$ V = 2\pi \left( \int_{-2}^0 (x - 3)(x^2 - 2x + 6) \, dx + \int_0^3 (x - 3)(6 - x^2 + x) \, dx \right) $$

After evaluating this integral, you will get the volume of the solid.