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.

- anonymous

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- anonymous

ok so you firstly have to integrate lnx to find the answer to the first question between 0 and 3

- anonymous

do you know how to do this??

- 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

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

what was your general integral?

- 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

y 5??

- anonymous

S for integral, there are no 5's

- anonymous

ah sorry

- anonymous

yh i don't know how you got 1.296 tho

- anonymous

for part b, try integration by parts to compute \[\int\limits_{1}^{3}\]pi*(lnx)^2dx

- 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

part a is 1.296
your right about it

- anonymous

Yeah, one of my friends told me to use integration by parts, but I don't remember how to do it... :P

- 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

Okay! Thanks helping :)

- anonymous

well u= lnx, du= dx/x, dv= lnx, so v= xlnx -x and int(udv) = uv- int(vdu)

- anonymous

that gives us lnx( xlnx -x) - (xlnx-2x)

- anonymous

okay, I think I'm following

- 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

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

ai, are you still there?

- anonymous

sorry was posting at another place ...yes! then divide (xlnx-x)/x and get "lnx-1" integrate it

- anonymous

Okay, I got lnx(xlnx-x)-xlnx+2x +c [1,3]

- 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

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

my replies are not going through :\ int(4-e^y) over the region ..

- anonymous

Could you explain that for me? I don't really understand how you got that.

- 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

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

can someone plx explain part c?

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