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Math Help - Polar

  1. #1
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    Polar

    Hi just want to check if my solution to this question is correct:

    Evaluate \int_{R}\int \frac{1}{(x^2+y^2+1)^\frac{7}{2}} dA where R is the region bounded by the y-axis, y = x and x^2+y^2 = 4.

    edit: sorry the title is supposed to be Polar coordinates + double integral




    My solution

    By using polar coordinates:
    x = r cos\theta

    y = r sin\theta

    Thus drawing the region R on the x-y plane, we deduce
    R = \{(r,\theta): 0\leq r\leq 2 , \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}\}

    Then
    \int_{R}\int \frac{1}{(x^2+y^2+1)^\frac{7}{2}} dA

    = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{2}<br />
\frac{r}{(r^2+1)^\frac{7}{2}} dr d\theta

    = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1}{5}(1 - \frac{1}{25\sqrt{5}}) d\theta

    = \frac{\pi}{20}(1 - \frac{1}{25\sqrt{5}})
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  2. #2
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    Quote Originally Posted by shinn View Post
    Hi just want to check if my solution to this question is correct:

    Evaluate \int_{R}\int \frac{1}{(x^2+y^2+1)^\frac{7}{2}} dA where R is the region bounded by the y-axis, y = x and x^2+y^2 = 4.

    edit: sorry the title is supposed to be Polar coordinates + double integral




    My solution

    By using polar coordinates:
    x = r cos\theta

    y = r sin\theta

    Thus drawing the region R on the x-y plane, we deduce
    R = \{(r,\theta): 0\leq r\leq 2 , \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}\}

    Then
    \int_{R}\int \frac{1}{(x^2+y^2+1)^\frac{7}{2}} dA

    = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{2}<br />
\frac{r}{(r^2+1)^\frac{7}{2}} dr d\theta

    = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1}{5}(1 - \frac{1}{25\sqrt{5}}) d\theta

    = \frac{\pi}{20}(1 - \frac{1}{25\sqrt{5}})
    It looks OK.
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