- #1
- 2,704
- 19
Find the volume created by revolving:
[tex] y=x^2 +x - 2 [/tex]
and y=0 about the x axis.
[tex] y=x^2 +x - 2 [/tex] intersects the x-axis at -1 and 1 so those are the bounds of integration.
the radius of the figure = -([tex] y=x^2 +x - 2 [/tex])
so:
V = [tex] - \pi \int_{-1}^1 (x^2 + x - 2)^2 dx [/tex]
After integrating that I get 18[tex]\pi[/tex]/5. The actual answer is 81[tex]\pi[/tex]/10. What I want to know is if i set the integral up right. If I did then this is only a stupid integration mistake. If not then I hope someone finds my mistake. Thanks for the help.
[tex] y=x^2 +x - 2 [/tex]
and y=0 about the x axis.
[tex] y=x^2 +x - 2 [/tex] intersects the x-axis at -1 and 1 so those are the bounds of integration.
the radius of the figure = -([tex] y=x^2 +x - 2 [/tex])
so:
V = [tex] - \pi \int_{-1}^1 (x^2 + x - 2)^2 dx [/tex]
After integrating that I get 18[tex]\pi[/tex]/5. The actual answer is 81[tex]\pi[/tex]/10. What I want to know is if i set the integral up right. If I did then this is only a stupid integration mistake. If not then I hope someone finds my mistake. Thanks for the help.
