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Math Help - surface area of revolution

  1. #1
    Senior Member slevvio's Avatar
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    surface area of revolution

    Find the surface area of revolution generated by the following curve (given in parametric form):

     x = cos\theta(1+cos\theta), y=sin\theta(1+cos\theta), \left(0\le\theta\le\pi\right).

    I managed to get the following:

    Surface Area = 2\pi\int_0^{\pi} 2sin\theta cos\frac{\theta}{2}(1+cos\theta)d\theta

    How on Earth do I integrate this? Thanks for any help
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by slevvio View Post
    Find the surface area of revolution generated by the following curve (given in parametric form):

     x = cos\theta(1+cos\theta), y=sin\theta(1+cos\theta), \left(0\le\theta\le\pi\right).

    I managed to get the following:

    Surface Area = 2\pi\int_0^{\pi} 2sin\theta cos\frac{\theta}{2}(1+cos\theta)d\theta

    How on Earth do I integrate this? Thanks for any help
    I would first start of by rewriting the integrand. Recall that

    cos{\frac{\theta}{2}}=+-\sqrt{\frac{1+cos{\theta}}{2}}.

    I will take the positive square root since surface area must be positive. Therefore, the integrand can be simplified to:

    2sin{\theta}\sqrt{\frac{1+cos{\theta}}{2}}(1+cos{\  theta})=\sqrt{2}sin{\theta}(1+cos{\theta})^{3/2}

    Thus the integral becomes:

     2\pi\int_{0}^{\pi}\sqrt{2}sin{\theta}(1+cos{\theta  })^{3/2}\,d\theta

    Apply a substitution...primarily u=1+cos\theta. Thus du=-sin\theta d\theta. Solving for d\theta, we get d\theta=-\frac{du}{sin\theta}.

    Before we substitute, we can change the limits of integration as well. The upper limit will become u_2=1+cos\pi=0 and the lower limit will become u_1=1+cos(0)=2.

    Now, the integral can be rewritten as:

    -2\pi\int_{2}^{0}\sqrt{2}u^{3/2}\,du=2\pi\int_{0}^{2}\sqrt{2}u^{3/2}\,du

    evaluating the integral we get:

    \frac{4\pi}{5}\sqrt{2}(2^{5/2}-0^{3/2})=\frac{4\pi}{5}(2^3)=\frac{32\pi}{5}.

    Hope that helped you out!
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