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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Updated on May 29, 2020

Comments

  • barndog
    barndog over 3 years

    enter image description here

    That's the picture of the shaded region I have to find the area of. I'm totally stuck on this problem mainly because these two curves don't intersect so I'm not sure how to find the bounds of integration. Once I find those my guess would be to the outside curve - the inside curve and just do the integration from there.

    • colormegone
      colormegone over 10 years
      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}) \ . ] $
    • André Nicolas
      André Nicolas over 10 years
      You presumably know the formula for area in polar coordinates. Here we have the difference of two areas that are not hard to find.
    • barndog
      barndog over 10 years
      Oh. Well that's stupid easy. facepalm thanks people.
    • colormegone
      colormegone over 10 years
      It's all right -- polar area integration problems aren't always done between intersection points...