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if i see it right; you went ahead and subtracted 2 to get it to the x axis right?
y = 4 -2x -x^2 y = x+4 and spin that around the x axis (y=0)
is your interval good? 6-2x-x^2 -x-6 -x^2 -3x = 0 -x(x+3) = 0; x= [-3,0] is good
Since it is being revolved around y=2, I was subtracting 2 to eliminate that much verticasl distance from the radius.
then we can integrate: subtract one function from the other... 4-2x-x^2 - x-4 pi [S] [-x^3 -3x]^2 dx ; [-3,0]
-x^2....typoed it :)
pi [S] (x^6 +9x^2 +6x^4) dx right?
-2-3x-x^2 is your function; why?
That was my radius
it doesnt match my radius tho :) so whos right?
yours is -2 larger
you see it?
y=6-2x-x^2 (-2) = 4-2x -x^2 y=x+6 (-2) = x+4
4 -2x -x^2 -[x+4] -3x -x^2 right?
so I need to subtact 2 from both functions?
Yes, becasue you need to move both function pi [S] (x^6 +9x^2 +6x^4) dx pi (x^7/7 +3x^3 + 6x^5/5) ; [-3,0]
How are you getting thisx^6 +9x^2 +6x^4) dx
you gotta squre your radius
pi [S] [f(x)]^2 dx
[f(x)]^2 = [-2x-x^2]^2
If the radius is (-3x-x^2) and we square it, don't we get 9x^2-6x^3=x^4?
-3x(-3x) = 9x^2 -x^2(-x^2) = x^4 -3x(-x^2)(2) = +6x^3
we could go with -1(3x+x^2) as well i htink
but then we get - everything
That 's OK b/c the def integral goes form -3 to 0
wanna go with - everything? i dont think it makes a diff :)
pi (-3x^3 -x^5/5 -3x^4/2) then right?
I get 8.1
pi * (((-3) * ((-3)^3)) - (((-3)^5) / 5) - ((3 * ((-3)^4)) / 2)) = 25.4469005 is what google spits out ....
which is 8.1 pi :)
So do you think that's right?
i think so; you can double check it if you want.... but im pretty sure we worked the kinks out of it :)
Thanks so much!!!
try to integrate one function at a time and subtract one from the other in the end to verify; kinda like making a donut and then cutting out the hole
youre welcome :)
Nice way to check! Thanks again!