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Thread: Finding the Surface Area of Revolution and Arc Length

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by miserablestudent View Post
    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 ...

    $\displaystyle \displaystyle A = 2\pi \int_0^4 x^2 \sqrt{1 + 4x^2} \, dx \approx 817$
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  3. #3
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    I was able to get the same answer for the second one, but I wasn't able to find the correct anti-derivative.
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