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Math Help - Surface Area of curves

  1. #1
    Junior Member
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    Surface Area of curves

    Find the area of the surface generated by revolving the curve y = \sqrt{x + 1} , 1 \leq x \leq 5 aroung the x-axis.


    S = \int^{b}_{a} 2\pi y \sqrt{1 + (dy/dx)^{2}} dx
    S = \int^{5}_{1} 2\pi \sqrt{x + 1} \sqrt{1 + (1/2\sqrt{x + 1})^{2}} dx
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    I used the substitution rule
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    S = 49\pi /3

    Am I correct?

    Thanks.
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello temaire
    Quote Originally Posted by temaire View Post
    Find the area of the surface generated by revolving the curve y = \sqrt{x + 1} , 1 \leq x \leq 5 aroung the x-axis.


    S = \int^{b}_{a} 2\pi y \sqrt{1 + (dy/dx)^{2}} dx
    S = \int^{5}_{1} 2\pi \sqrt{x + 1} \sqrt{1 + (1/2\sqrt{x + 1})^{2}} dx
    .
    .
    .
    I used the substitution rule
    .
    .
    .
    S = 49\pi /3

    Am I correct?

    Thanks.
    That's exactly what I got. Good job!

    Incidentally, you can use a little trick to avoid some of the fractional powers, and make the working a bit easier:
    y=\sqrt{x+1}

    \Rightarrow y^2 = x+1

    \Rightarrow 2y\frac{dy}{dx}=1

    \Rightarrow \left(y\frac{dy}{dx}\right)^2 = \tfrac14
    Now write
    S = \int^{b}_{a} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}} \;dx
    as:
    S = 2\pi\int^{b}_{a}  \sqrt{y^2 + \left(y\frac{dy}{dx}\right)^{2}} \;dx
    =2\pi\int_1^5\sqrt{x+\tfrac54}\;dx

    = ... etc
    Grandad
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    I definitely recommend to use Wolfram to check if you found the correct value of the integral.

    When we know how to compute an integral, in particular a definite one, we need some of speed so there's when an online math tool comes handy for these stuff.
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