# Thread: Surface Area Of A Parametric Curve?

1. ## Surface Area Of A Parametric Curve?

Does anybody know how to get the surface area under this parametric curve about the x axis? I've been trying for hours..

x= 8(t^2)
y= 64t - (t^3)/3
0 <= t <= 8sqrt(3)

My work:
S = 2pi(integral(((64t - (t^3)/3)sqrt((16t^2 + (64-t^2)^2)dt
from 0 to 8sqrt(3)

S = 2pi(integral(((64t - (t^3)/3)sqrt((128t^2+4096t^4)dt
from 0 to 8sqrt(3)

2. Originally Posted by AlphaRock x= 8(t^2)
y= 64t - (t^3)/3
0 <= t <= 8sqrt(3)

My work:
S = 2pi(integral(((64t - (t^3)/3)sqrt((16t^2 + (64-t^2)^2)dt
from 0 to 8sqrt(3)
certainly, $\displaystyle x'(t)=16t$ thus $\displaystyle \big(x'(t)\big)^2=256t^2.$

3. ## didnot understand Originally Posted by Krizalid certainly, $\displaystyle x'(t)=16t$ thus $\displaystyle \big(x'(t)\big)^2=256t^2.$
i didnt get you. could you please elaborate????
why cant we simply eleminate the parameter then integrate.

4. the OP got a typo, that's all.

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