# Thread: Finding the Surface Area of Revolution and Arc Length

1. ## Finding the Surface Area of Revolution and Arc Length

a.) Calculate the arc length of y=x^2 from [0,1]

- For this one I used the formula the for arc length and got stuck once I used trig substitution.

b.) Compute the surface area of revolution about the x-axis for y=x^2 from [0,4]

- For this problem I used the formula surface area of revolution, but when I plugged my results in my calculator I got -6,036, which I'm assuming is incorrect.

2. Originally Posted by miserablestudent
a.) Calculate the arc length of y=x^2 from [0,1]

- For this one I used the formula the for arc length and got stuck once I used trig substitution.

b.) Compute the surface area of revolution about the x-axis for y=x^2 from [0,4]

- For this problem I used the formula surface area of revolution, but when I plugged my results in my calculator I got -6,036, which I'm assuming is incorrect.
$\displaystyle \displaystyle \int_0^1 \sqrt{1+4x^2} \, dx$

$\displaystyle \displaystyle x = \frac{\tan{t}}{2}$

$\displaystyle \displaystyle dx = \frac{\sec^2{t}}{2} \, dt$

$\displaystyle \displaystyle \frac{1}{2} \int_0^{\arctan(2)} \sec^3{t} \, dt$

use integration by parts ... $\displaystyle u = \sec{t}$ , $\displaystyle dv = \sec^2{t} \, dt$

note that you'll get $\displaystyle \displaystyle \int \sec^3{t} \, dt = \frac{1}{2}\left[\sec{t}\tan{t} + \int \sec{t} \, dt\right]$

surface area problem ...

$\displaystyle \displaystyle A = 2\pi \int_0^4 x^2 \sqrt{1 + 4x^2} \, dx \approx 817$

3. I was able to get the same answer for the second one, but I wasn't able to find the correct anti-derivative.