- #1
Flaneuse
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Can anyone help me with this calculus problem? I have tried it a couple of different ways, but keep getting various incorrect answers; I think my problem is probably in the setup.
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!
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!