calculusdefinite integralsintegrationmultivariable-calculuspolar coordinates
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.
Yes area of $BCD$ is $f(\theta)=\frac12r^2\tan\theta-\frac12r^2\theta=72(\tan\theta-\theta)$
For domain:
For the limit: $$\lim_{\theta\to\pi/2}f(\theta)=72\lim_{\theta\to\pi/2}(\tan\theta-\theta)=\infty$$ As $\lim_{\theta\to\pi/2}\tan\theta=\infty$ and $\lim_{\theta\to\pi/2}\theta=\pi/2$
Hints:
$x_B$ can be found by solving $3x-x^2=0$. Now you can work out the area of the right triangle colored red in the figure.
Now the area of the shaded region can be represented as
$$A=\int_{x_A}^{x_B} 3x-x^2\,dx-\text{area of the triangle}$$
Best Answer
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 $\iint \mathrm dx\,\mathrm dy = \iint r \,\mathrm \,d\alpha\,\mathrm dr$): $$ \begin{eqnarray*} A &=&\iint_{R}dA=\int_{0}^{\pi /2}\left( \int_{\theta }^{e^{\theta /2}}r\;dr\right)\ d\theta \\ &=&\int_{0}^{\pi /2}\big( \frac{e^{\theta }}{2}-\frac{\theta ^{2}}{2} \big)\ d\theta \\ &=&\frac{1}{2}\int_{0}^{\pi /2}e^{\theta }d\theta -\frac{1}{2}\int_{0}^{\pi /2}\theta ^{2}\ d\theta \\ &=&\big( \frac{e^{\pi /2}}{2}-\frac{1}{2}\big) -\frac{\pi ^{3}}{48}. \end{eqnarray*} $$