# Thread: Surface Area of a Revolved Parametric Equations

1. ## Surface Area of a Revolved Parametric Equations

I have worked this problem over but still can not get a reasonable answer. I would like some help, please.

I am given the parametric equations: x=e^t and y=(1/3)e^3t. I am to revolve the resulting shape around the x-axis and calculate the surface area. I know that I must use
2*pi*the radius*sqrt((dx/dt)^2 + (dy/dt)^2)
in order to find the surface area. However, I'm not sure what my radius is and the integral is difficult because of the tricks it needs.

I could really use some assistance ASAP.

2. Originally Posted by Ekeonee888
I have worked this problem over but still can not get a reasonable answer. I would like some help, please.

I am given the parametric equations: x=e^t and y=(1/3)e^3t. I am to revolve the resulting shape around the x-axis and calculate the surface area. I know that I must use
2*pi*the radius*sqrt((dx/dt)^2 + (dy/dt)^2)
in order to find the surface area. However, I'm not sure what my radius is and the integral is difficult because of the tricks it needs.

I could really use some assistance ASAP.
Do you have to do it parametrically? The y(x) formula is very simple.

-Dan

3. $x=e^t$

$y=\frac{1}{3}e^{3t}$

$y=\frac{1}{3}x^3$

This will make it a lot easier.

4. But one must put it in terms of t with the bounds 0<=t>=(ln3)/4

5. If you are given bounds for t then you know the bounds of x, as x is defined as a function of t.

6. As it's easier to work with respect to x, put the bounds for t in $x=e^t$ to find what they correspond for x and then integrate using these bounds.

7. Originally Posted by Ekeonee888
I have worked this problem over but still can not get a reasonable answer. I would like some help, please.

I am given the parametric equations: x=e^t and y=(1/3)e^3t. I am to revolve the resulting shape around the x-axis and calculate the surface area. I know that I must use
2*pi*the radius*sqrt((dx/dt)^2 + (dy/dt)^2)
in order to find the surface area. However, I'm not sure what my radius is and the integral is difficult because of the tricks it needs.

I could really use some assistance ASAP.
$\frac{dx}{dt} = e^t$

$\frac{dy}{dt} = e^{3t}$

So
$A = 2 \pi \int_0^{ln(3)/4} e^t \sqrt{e^{2t} + e^{6t}}~dt$

$= 2 \pi \int_0^{ln(3)/4} e^{2t} \sqrt{1 + e^{4t}}~dt$

Let $u = e^{2t} \implies du = 2 e^{2t}~dt$, so
$A = \pi \int_1^{3/2} \sqrt{1 + u^2}~du$

Can you finish this now?

-Dan