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Thread: Area under polar graph

  1. #1
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    Area under polar graph

    Find the area bounded by:

    $\displaystyle r=\frac{1}{\sqrt {1+\theta } }\\$

    $\displaystyle 0\leqslant \theta \leqslant \pi $

    I have so far set up the integral like this:

    $\displaystyle \int\nolimits_{0}^{\pi } {d\theta }\int\nolimits_{0}^{\frac{1}{\sqrt {1+\theta } }} {rdr}$

    Looks alright?
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  2. #2
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    Hello, thepongofping!

    Why are you using a double integral?



    Find the area bounded by: .$\displaystyle r\:=\:\dfrac{1}{\sqrt {1+\theta}},\;\;0 \le \theta \le\pi $

    $\displaystyle \text{Formula: }\;\displaystyle{A \;=\;\tfrac{1}{2}\!\int^{\beta}_{\alpha}\!\! r^2\,d\theta}$


    $\displaystyle \displaystyle{\text{We have: }\;A \;=\;\tfrac{1}{2}\!\int^{\pi}_0\! \left(\frac{1}{\sqrt{1+\theta}}\right)^2 d\theta \;=\; \tfrac{1}{2}\!\int^{\pi}_0\frac{d\theta}{1+\theta} } $

    . . $\displaystyle A \;=\;\frac{1}{2}\ln|1+\theta|\,\bigg]^{\pi}_0 \;=\;\frac{1}{2}\ln(1+\pi) - \frac{1}{2}\ln(1+0) $

    . . $\displaystyle A \;=\;\frac{1}{2}\ln(1+\pi)$
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