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Thread: surface area and integration

  1. #1
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    surface area and integration

    Calculate the area of the portion E+ of the ellipsoid that lies above the xy plane by making use of the parameterisation $\displaystyle r_{\to} (\theta, \phi)= (2cos \phi sin \theta , 2 sin \phi cos \theta , cos \theta )$.

    Hint: $\displaystyle \int \sqrt {a^2 + b^2 u^2} du$ = $\displaystyle \frac {1}{2} \sqrt {a^2 + b^2 u^2} + \frac {a^2} {2b} log(bu+\sqrt {a^2 + b^2 u^2} ) $

    I know that I need to evaluate $\displaystyle A=\int \int_{E+} || r_{\theta} \times r_{\phi} || $ for $\displaystyle 0 \leq \theta \leq \pi/2$ and $\displaystyle 0 \leq \phi \leq 2\pi $

    Now, I have found $\displaystyle || r_{\theta} \times r_{\phi} || $ = $\displaystyle 2sin \theta \sqrt{(sin \theta)^2 + 4(cos \theta)^2} $ but am stuck on how to use the hint?
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  2. #2
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    Quote Originally Posted by nmatthies1 View Post
    Calculate the area of the portion E+ of the ellipsoid that lies above the xy plane by making use of the parameterisation $\displaystyle r_{\to} (\theta, \phi)= (2cos \phi sin \theta , 2 sin \phi cos \theta , cos \theta )$.

    Hint: $\displaystyle \int \sqrt {a^2 + b^2 u^2} du$ = $\displaystyle \frac {1}{2} \sqrt {a^2 + b^2 u^2} + \frac {a^2} {2b} log(bu+\sqrt {a^2 + b^2 u^2} ) $

    I know that I need to evaluate $\displaystyle A=\int \int_{E+} || r_{\theta} \times r_{\phi} || $ for $\displaystyle 0 \leq \theta \leq \pi/2$ and $\displaystyle 0 \leq \phi \leq 2\pi $

    Now, I have found $\displaystyle || r_{\theta} \times r_{\phi} || $ = $\displaystyle 2sin \theta \sqrt{(sin \theta)^2 + 4(cos \theta)^2} $ but am stuck on how to use the hint?
    Seems you want to calculate

    $\displaystyle \int || r_{\theta} \times r_{\phi} || \; d\theta =\int 2\sin \theta \sqrt{\sin^2 \theta + 4\cos^2 \theta} \; d\theta=-2\int \sqrt{1 + 3\cos^2 \theta} \; d \cos \theta$$\displaystyle = -2\int \sqrt{1 + 3u^2} \; du$ (Let $\displaystyle u=\cos \theta$)

    Then use the hint to proceed ....
    Last edited by mr fantastic; Sep 18th 2009 at 07:53 AM. Reason: Restored original post
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  3. #3
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    Quote Originally Posted by nmatthies1 View Post
    Calculate the area of the portion E+ of the ellipsoid that lies above the xy plane by making use of the parameterisation $\displaystyle r_{\to} (\theta, \phi)= (2cos \phi sin \theta , 2 sin \phi cos \theta , cos \theta )$.
    This does NOT give an ellipsoid. $\displaystyle \vec{r}(\theta, \phi)= (2cos\phi sin\theta, 2 sin\phi sin\theta, cos\theta)$ gives the ellipsoid $\displaystyle \frac{x^2}{4}+ \frac{y^2}{4}+ z^2= 1$.

    Hint: $\displaystyle \int \sqrt {a^2 + b^2 u^2} du$ = $\displaystyle \frac {1}{2} \sqrt {a^2 + b^2 u^2} + \frac {a^2} {2b} log(bu+\sqrt {a^2 + b^2 u^2} ) $ .

    I know that I need to evaluate $\displaystyle A=\int \int_{E+} || r_{\theta} \times r_{\phi} || $ for $\displaystyle 0 \leq \theta \leq \pi/2$ and $\displaystyle 0 \leq \phi \leq 2\pi $

    Now, I have found $\displaystyle || r_{\theta} \times r_{\phi} || $ = $\displaystyle 2sin \theta \sqrt{(sin \theta)^2 + 4(cos \theta)^2} $ but am stuck on how to use the hint?
    In the form I have given, $\displaystyle \phi$ is the "longitude" and ranges from 0 to $\displaystyle 2\pi$, and $\displaystyle \theta$ is the "co-latitude" and ranges from 0 to $\displaystyle \pi$.
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