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Find the area of the region that is bounded by the given curve and lies in the specified sector. r = e^−θ/12, π/2 ≤ θ ≤ π
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here you have to go in polar cordinates, so you have to compute this integral: \[\iint {r\;dr\;d\theta }\]
How do I do that?
here are your steps: \[\iint {r\;dr\;d\theta } = \int_{\pi /2}^\pi {d\theta } \int_0^\rho {r\;dr} \] where \(\rho = e^{-\pi/12}\)
oops..where \(\rho = e^{ - \theta /12} \)
I got this result: \[\huge \begin{gathered} \iint {r\;dr\;d\theta } = \int_{\pi /2}^\pi {d\theta } \int_0^\rho {r\;dr} = \hfill \\ \hfill \\ = 3\frac{{{e^{\pi /12}} - 1}}{{{e^{\pi /6}}}} \hfill \\ \end{gathered} \]
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