anonymous
  • anonymous
Let R be the region enclosed by the graph of y=ln x, the line x=3, and the x-axis. (a) Fine the area of region R. (b) Find the volume of the solid generated by revolving region R about the x-axis. (c) Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid generated by revolving region R about the line x=3.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
ok so you firstly have to integrate lnx to find the answer to the first question between 0 and 3
anonymous
  • anonymous
do you know how to do this??
anonymous
  • anonymous
I thought it would be from 1 to 3 because of the x-axis bound. The graph crosses the x-axis at (1,0).

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anonymous
  • anonymous
So I took the integral of lnx and I got an answer of 1.296 after plugging in my bounds. I think this was the answer to part (a). Where I get confused is at part (b). I believe I have to integrate (lnx)² dx and multiply by pi. I got stuck at that step. How do you integrate that?
anonymous
  • anonymous
what was your general integral?
anonymous
  • anonymous
In part (a) I used S lnx dx from [1,3] and in part (b) I'm using (pi) S (lnx)² dx from [1,3]. (S meaning integral)
anonymous
  • anonymous
y 5??
anonymous
  • anonymous
S for integral, there are no 5's
anonymous
  • anonymous
ah sorry
anonymous
  • anonymous
yh i don't know how you got 1.296 tho
anonymous
  • anonymous
for part b, try integration by parts to compute \[\int\limits_{1}^{3}\]pi*(lnx)^2dx
anonymous
  • anonymous
Well, I integrated S lnx dx, and I got xlnx-x. Then I plugged in my bounds [1,3]. So I got (3ln3-3)-(ln1-1).
anonymous
  • anonymous
part a is 1.296 your right about it
anonymous
  • anonymous
Yeah, one of my friends told me to use integration by parts, but I don't remember how to do it... :P
anonymous
  • anonymous
yh sorry .. i'm out of it .. trying 2 do too many things at once . i'm gonna sign out come bck on later .. sorrry for the confusion
anonymous
  • anonymous
Okay! Thanks helping :)
anonymous
  • anonymous
well u= lnx, du= dx/x, dv= lnx, so v= xlnx -x and int(udv) = uv- int(vdu)
anonymous
  • anonymous
that gives us lnx( xlnx -x) - (xlnx-2x)
anonymous
  • anonymous
okay, I think I'm following
anonymous
  • anonymous
So in the way that you set up the u,du and v,dv, is it like setting it up (lnx)(lnx) with one being u and the other being dv?
anonymous
  • anonymous
If we're using the formula uv - int v du, then wouldn't it be lnx (xlnx-x) - int xlnx -x (1/x) dx?
anonymous
  • anonymous
ai, are you still there?
anonymous
  • anonymous
sorry was posting at another place ...yes! then divide (xlnx-x)/x and get "lnx-1" integrate it
anonymous
  • anonymous
Okay, I got lnx(xlnx-x)-xlnx+2x +c [1,3]
anonymous
  • anonymous
Alright! I got the answer :) it's 1.029 pi, or 3.233 if you multiply pi in. Do you know how to do part (c)?
anonymous
  • anonymous
I know that in order to revolve the function about x=3, we must subtract 3 from our equation. I don't really understand how to set it up though.
anonymous
  • anonymous
my replies are not going through :\ int(4-e^y) over the region ..
anonymous
  • anonymous
Could you explain that for me? I don't really understand how you got that.
anonymous
  • anonymous
correction : int(3-e^y) ! well we have to rotate it around y-axis now i.e.at x=3..so write the function in terms of y, by that we get x= e^y.... the radius of rotation, if you visualize it would be 4-f(y).......... PS the answer in reality doesn't makes sense as the given function on rotating on x=3 would overlap itself :\
anonymous
  • anonymous
why is it 3-e^y? I understand how you switched from in terms of x to in terms of y x=e^y. but I'm confused again after that :S
anonymous
  • anonymous
can someone plx explain part c?

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