# [SOLVED] finding surface area using parametric equations

• Mar 20th 2010, 06:50 PM
mlckb1
[SOLVED] finding surface area using parametric equations
Find the surface area generated by rotating x=e^t -t & y=4e^t for 0 to 1 (this is in terms of t) about the y axis.

So i've managed to get the integral as far as

(e^(t)-t)((e^(2t)+1)^1/2)

i'm just not sure what to do after that i thought about multiplying out the square root but even if i were to do that i can't think of a proper u-sub that could work. So what do you think am i on the right track or is there an easier way that i'm just not seeing.
• Mar 20th 2010, 11:39 PM
Pulock2009
suggestion
Quote:

Originally Posted by mlckb1
Find the surface area generated by rotating x=e^t -t & y=4e^t for 0 to 1 (this is in terms of t) about the y axis.

So i've managed to get the integral as far as

(e^(t)-t)((e^(2t)+1)^1/2)

i'm just not sure what to do after that i thought about multiplying out the square root but even if i were to do that i can't think of a proper u-sub that could work. So what do you think am i on the right track or is there an easier way that i'm just not seeing.

i wonder whether you could use a by -parts and perhaps obtain a reduction formulae. but even then the expression gets too complicated. please recheck whether the integral you obtained is correct.
• Mar 21st 2010, 09:04 AM
mlckb1
Quote:

Originally Posted by Pulock2009
i wonder whether you could use a by -parts and perhaps obtain a reduction formulae. but even then the expression gets too complicated. please recheck whether the integral you obtained is correct.

I'll give a bit more of what i did in the beginning of the problem so you can see if what i'm doing is correct.

for this problem i'm supposed to be using the integral of x(t) times ds where
ds=((dx/dt)^2 + (dy/dt)^2)^(1/2)

After i did this i got (e^(2t)+1)^(1/2).....
...And now i just realized where i went wrong, i forgot to square the two resulting in something zeroing out where it shouldn't have. oh well thanks for your help anyway. (Smile)