The Student Room Group

Reply 1

Futurdoc
Need help with this annoying question :|

Q: A drinking glass is modelled by rotating the graph of y = e^x about the y-axis, for 1<y<5. Find the volume of the glass.

- I'm going to make x the subject if we're rotating around the y axis. Thus x = (ln y) ^ 2

Now I don't know how to integrate that.

I tried substituion..

But du/dx = 1/y :s-smilie:

Help?


Bring the 2 to the front of the ln and then to the front of the integral so you are left with 2π15lnydy2\pi \displaystyle\int_1^5 lnydy
then do the integration by parts, with u=lny and dv/dy=1

Reply 2

You can integrate natural logs using integration by parts, using a constant as other functions.

That means you can differentiate lny and integrate the constant in the calculation. I hope that helps a little?

Reply 3

Futurdoc

But du/dy = 1/y :s-smilie:




You're on the right lines. So what's y in terms of u and carry on. I suspect it's just a little messier than you were expecting.

Edit: Corrected in that I assume you used u = ln y as the substitution.

Reply 4

But I thought we could only bring the two down to the front if it was ln (y^2). :s-smilie:

Reply 5

I changed the limits into terms of x as opposed to y. And followed it through. I got 12(pi) units ^3.

Has anyone done it ?

Reply 6

Futurdoc
I changed the limits into terms of x as opposed to y. And followed it through. I got 12(pi) units ^3.

Has anyone done it ?


I made have made a slip, but I made it:

5(ln5)210ln5+85 (\ln 5)^2 -10\ln 5 +8 times pi!


from

π0ln5u2eu  du\displaystyle\pi\int_0^{\ln 5} u^2 e^u\;du

Reply 7

ghostwalker
I made have made a slip, but I made it:

5(ln5)210ln5+85 (\ln 5)^2 -10\ln 5 +8 times pi!


from

π0ln5u2eu  du\displaystyle\pi\int_0^{\ln 5} u^2 e^u\;du



(e^x) ^ 2 = e ^2x don't it ?

Reply 8

Futurdoc
(e^x) ^ 2 = e ^2x don't it ?


Granted, but I don't see the relevance.

Reply 9

thank guys , i think im okay now

Reply 10

Are you a doctor now?

Reply 11

Original post by Moonblue05
Are you a doctor now?

Please don't bump 11 year old threads! It looks like the OP hasn't been online for about 5 years :smile: