El_Arrow
  • El_Arrow
Need help please!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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El_Arrow
  • El_Arrow
find the volume of the solid formed by revolving the region bounded by the graph of f(x) = x^2 and g(x) = 12-x to the right of x=1 about the y-axis
El_Arrow
  • El_Arrow
first thing i did was set f(x) and g(x) equal to one another and i got -4 and 3
El_Arrow
  • El_Arrow
|dw:1440982705148:dw|

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freckles
  • freckles
\[x^2=12-x \\ x^2+x-12=0 \\ (x+4)(x-3)=0 \\ x=-4 \text{ or } x=3 \] (-4)^2=16 (3)^2=9 |dw:1440982735696:dw|
freckles
  • freckles
|dw:1440982829162:dw| we are looking at this region (the shaded region) right?
El_Arrow
  • El_Arrow
yes
El_Arrow
  • El_Arrow
so it would be the integral from 1 to 3 @freckles ?
freckles
  • freckles
yea
freckles
  • freckles
so it looks like you were trying to use shells method
El_Arrow
  • El_Arrow
yes thats correct
freckles
  • freckles
|dw:1440983126638:dw|
El_Arrow
  • El_Arrow
so something like this |dw:1440983268893:dw|
freckles
  • freckles
yes. |dw:1440983363623:dw|
El_Arrow
  • El_Arrow
but why do you have the square root over the x why not x^2
freckles
  • freckles
lol
freckles
  • freckles
I forgot what function we were using
freckles
  • freckles
sqrt(x) is suppose to be x^2
El_Arrow
  • El_Arrow
oh okay lol
freckles
  • freckles
i'm sorry i do idiot things sometimes
El_Arrow
  • El_Arrow
its okay
freckles
  • freckles
|dw:1440983615720:dw|
El_Arrow
  • El_Arrow
the answer is 116pi/3 but i got 16pi?
freckles
  • freckles
hmm so the radius should have been just x
El_Arrow
  • El_Arrow
i think so let me see if it works
freckles
  • freckles
|dw:1440983787477:dw|
freckles
  • freckles
|dw:1440983907300:dw| so you were right to begin with you just needed 1 to 3 instead of 0 to 1
El_Arrow
  • El_Arrow
yeah i got right now thanks alot
freckles
  • freckles
it does make sense I think if I had looked at that whole thing from 0 to 3 and subtract out the part that is 0 to 1 we would have seen it better \[\int\limits_0^3 2\pi(x)(12-x-x^2) dx-\int\limits_0^1 2\pi(x)(12-x-x^2)dx \\ =\int\limits_1^3 2 \pi (x) (12-x-x^2) dx\]

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