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El_Arrow
 one year ago
Need help please!
El_Arrow
 one year ago
Need help please!

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El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.1find the volume of the solid formed by revolving the region bounded by the graph of f(x) = x^2 and g(x) = 12x to the right of x=1 about the yaxis

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.1first thing i did was set f(x) and g(x) equal to one another and i got 4 and 3

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.1dw:1440982705148:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[x^2=12x \\ x^2+x12=0 \\ (x+4)(x3)=0 \\ x=4 \text{ or } x=3 \] (4)^2=16 (3)^2=9 dw:1440982735696:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.3dw:1440982829162:dw we are looking at this region (the shaded region) right?

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.1so it would be the integral from 1 to 3 @freckles ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so it looks like you were trying to use shells method

freckles
 one year ago
Best ResponseYou've already chosen the best response.3dw:1440983126638:dw

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.1so something like this dw:1440983268893:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.3yes. dw:1440983363623:dw

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.1but why do you have the square root over the x why not x^2

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I forgot what function we were using

freckles
 one year ago
Best ResponseYou've already chosen the best response.3sqrt(x) is suppose to be x^2

freckles
 one year ago
Best ResponseYou've already chosen the best response.3i'm sorry i do idiot things sometimes

freckles
 one year ago
Best ResponseYou've already chosen the best response.3dw:1440983615720:dw

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.1the answer is 116pi/3 but i got 16pi?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3hmm so the radius should have been just x

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.1i think so let me see if it works

freckles
 one year ago
Best ResponseYou've already chosen the best response.3dw:1440983787477:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.3dw:1440983907300:dw so you were right to begin with you just needed 1 to 3 instead of 0 to 1

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.1yeah i got right now thanks alot

freckles
 one year ago
Best ResponseYou've already chosen the best response.3it 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 \[\int\limits_0^3 2\pi(x)(12xx^2) dx\int\limits_0^1 2\pi(x)(12xx^2)dx \\ =\int\limits_1^3 2 \pi (x) (12xx^2) dx\]
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