# Find the area of the shaded region between $r=e^{\theta/2}$ and $r=θ$ .

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We have to find the area of the region $R$ between $r=\theta$ and $r=e^{\theta/2}$, and between $\theta =0$ and $\theta =\pi /2$. Since the Jacobian of the transformation from Cartesian to polar coordinates is $r$, the area $A$ is given by the double integral (see Explain $\iint \mathrm dx\,\mathrm dy = \iint r \,\mathrm \,d\alpha\,\mathrm dr$): $$\begin{eqnarray*} A &=&\iint_{R}dA=\int_{0}^{\pi /2}\left( \int_{\theta }^{e^{\theta /2}}r\;dr\right)\ d\theta \\ &=&\int_{0}^{\pi /2}\big( \frac{e^{\theta }}{2}-\frac{\theta ^{2}}{2} \big)\ d\theta \\ &=&\frac{1}{2}\int_{0}^{\pi /2}e^{\theta }d\theta -\frac{1}{2}\int_{0}^{\pi /2}\theta ^{2}\ d\theta \\ &=&\big( \frac{e^{\pi /2}}{2}-\frac{1}{2}\big) -\frac{\pi ^{3}}{48}. \end{eqnarray*}$$

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### barndog

I code sometimes. Formerly @apple.

Updated on May 29, 2020

The poser of the problem was nice to you because the lines terminating the area point to the right and upward from the origin, so the limits of integration for the area will be $\ \theta = 0 \$ and $\ \theta = \frac{\pi}{2} \$ . [Don't forget that you want to integrate $\ (\frac{1}{2} r(\theta)^2_{outer}) \ - \ (\frac{1}{2} r(\theta)^2_{inner}) \ . ]$