Let Vx and Vy be the volumes of the solids that result when the region enclosed by y=1/x y=0 x=1/2 and x=1/2 (b > 1/2) is revolved about the x-axis and the y-axis, respectively. Is there a value of b for which Vx = Vy?
@sourwing i accidentally closed my question. would you be able to help ? :/
well, you see, the question doesn't ask for what it is. It's yes/no question :DDD
@sourwing hehe i wish it was as simple as that! the answer manual actually gives an actual number so I think i have to support the yes
you have the solution manual and still don't know the answer? O.O
Well Idk how they got the answer it only give me the answer
@sourwing
it's revolved around the x-axis, what is you setup?
well this is where I have my doubts. Im not sure where to start!
V_x = pi ∫ radius^2 [disk method] V_x = pi ∫(1/x)^2 dx, from 1/2 to b around the y-axis: V_y = 2pi ∫(radius) (height) [shell method] V_y = 2pi ∫ x (1/x) dx from 1/2 to b
set V_x = V_y and solve for b
oh thank you! let me try it out
for Vx I got..\[2\pi \left( \frac{ 4b ^{2}-1 }{ b ^{2} } \right)\]
and my Vy=2pi
wait for the Vx just pi not 2pi!
can you exaplain ur question to me
why are there 2, x=1/2s
oops I'm sorry i may have wrote that twice it is supposed to say…
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