- #1

Flaneuse

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I am trying to find the volume of the solid created by revolving the area bounded by these two equations

y = 6 - 2x - x[tex]^{2}[/tex]

y = x + 6

around the x-axis, and around y = 3.

First, I found the intersection points of the two equations, which are at (-3,3) and (0,6)

For the revolution around the x-axis, I tried doing

[tex]\pi \int ^{0}_{-3}((6 - 2x - x[^2}[/tex])[tex]^{2}[/tex] - (x+6)[tex]^{2}[/tex] )dx,

simplified to [tex]\pi [/tex] \int ^{6}_{0}(-32x -13x[tex]^{2}[/tex] +4x[tex]^{3}[/tex] + x[tex]^{4}[/tex] )dx ,

and got a final answer of -27[tex]\pi[/tex]/5

The correct answer is 243[tex]\pi[/tex] /5

For the revolution around y = 3, I got -3-x^{2} from subtracting the lower function, x + 6, from the top function, 6 - 2x - x^{2}, and then tried

[tex]\pi \int^{6}_{0}(6-(-3-x^{2}))^{2} dx, simplified to \pi^{6}_{0}\int (6+3x+x^{2} ) dx, and ended up with 29376[tex]\pi[/tex]/5.

The answer for this one is supposed to be 108[tex]\pi[/tex]/5

Again, I think my real problems are in the setup, probably with the "top minus bottom" part.

Also, I am new to Physics Forum, and can't seem to get rid of how, in the second integral, my words ended up included in the integral. I would appreciate it if anyone could tell me how to fix/avoid that.

Thanks!