Find the area of the triangle bounded by the $x$-axis, the $y$-axis, and the line $r = 4 \sec\left(\theta - \frac{\pi}{4}\right)$ using an integral in polar coordinates. Then, check your answer using geometry.
Drawing out the picture shows that this line is in the first quadrant, so the boundaries are at $\theta = 0$ and $\theta = \frac{\pi}{2}$.
The integral we get here is $\displaystyle \frac{1}{2} \int_0^{\pi/2} (4 \sec\left(\theta - \frac{\pi}{4}\right)^2\ d\theta$
Using a substitution $u = \theta - \frac{\pi}{4}$, the integral becomes $8 \tan\left(\theta - \frac{\pi}{4}\right)$ from $0$ to $\pi/2$, which comes out to $16$.
For this triangle, the point on the $x$ axis is at $\theta = 0$, where $r = 4 \sec{-\frac{\pi}{4}} = 4\sqrt{2}$, and the same holds for the point along the $y$ axis. Thus, the area is $\frac{1}{2} (4 \sqrt{2})(4 \sqrt{2}) = 16$
$16$
Find the area inside one petal of the curve $r = \sin(4\theta)$.
First, we need to figure out where a petal starts and stops. Where does this curve come back to the origin?
This starts at the origin at $\theta = 0$, and the first time it comes back is when $4\theta = \pi$, or $\theta = \frac{\pi}{4}$.
Thus, the integral we need here is $\displaystyle \int_0^{\pi/4} \frac{1}{2} (\sin(4\theta))^2\ d\theta$.
This integral can be evaluated using our half angle formulas from before.
$\displaystyle \frac{\pi}{16} $
Find the area inside the curve $r = 2 + \sin(2\theta)$ and outside the curve $r = \sin(2\theta)$.
Since $\sin(2\theta)$ is always between $-1$ and $1$, we know that the curve $r = 2 + \sin(2\theta)$ is always outside $r = \sin(2\theta)$. Thus, we just need to find the area inside each curve and subtract them.
For $2 + \sin(2\theta)$, the radius is always positive, so there aren't any petals. Thus, to find the area, we just need to integrate from $0$ to $2\pi$.
For the inside curve, the radius does go to zero, so there are petals. By plotting the curve, we know that there are 4 petals, and the first one goes up to $\theta = \frac{\pi}{2}$.
The outside area is $\displaystyle \frac{1}{2} \int_0^{2\pi} (2 + \sin(2\theta))^2\ d\theta$, and the inside area is four times $\displaystyle \frac{1}{2} \int_0^{\pi/2} (\sin(2\theta))^2\ d\theta$.
$\displaystyle \frac{9\pi}{2} - \frac{\pi}{2} = 4\pi $
Find the length of the curve $r = e^\theta$ for $0 \leq \theta \leq 2\pi$
We have a standard formula for the length of a polar curve that we can use to find this length.
Since our function is $f(\theta) = e^\theta$, then our integral becomes $\displaystyle \int_0^{2\pi} \sqrt{f(\theta)^2 + f'(\theta)^2}\ d\theta = \displaystyle \int_0^{2\pi} \sqrt{e^{2\theta} + e^{2\theta}}\ d\theta$.
This integral simplifies to $\displaystyle \int_0^{2\pi} \sqrt{2} e^\theta\ d\theta$
$\displaystyle \sqrt{2}(e^{2\pi} - 1) = 755.89$
Find the length of the curve $r = 1 + \theta$ over the range $0 \leq \theta \leq \pi$.
Again, this goes back to the standard formula for length of polar curves.
Since our function is $f(\theta) = 1+\theta$, then our integral becomes $\displaystyle \int_0^{\pi} \sqrt{f(\theta)^2 + f'(\theta)^2}\ d\theta = \displaystyle \int_0^{\pi} \sqrt{(1+\theta)^2 + 1}\ d\theta$.
This integral can be left as is, and use $1 + \theta = \tan(\phi)$ as a trigonometric substitution. This will turn into a reduction formula, needing to integrate $\sec^3{\phi}$.
Integrating this out, we get that $\displaystyle \int \sec^3{\phi} d \phi = \frac{1}{2}\sec{\phi}\tan{\phi} + \frac{1}{2} \ln{|\sec{\phi} + \tan{\phi}|}$. This can then be converged back to $\theta$, and then plug in $0$ and $\pi$
$\displaystyle \left( \frac{1}{2} (1+\theta)(\sqrt{\theta^2 + 2\theta + 2} + \frac{1}{2} \ln( \sqrt{\theta^2 + 2\theta + 2} + 1 + \theta)\right) \mid_0^\pi = \frac{1}{2}\left( (\pi+1)\sqrt{\pi^2 + 2\pi + 2} + \ln(1 + \pi + \sqrt{\pi^2 + 2\pi + 2} - \sqrt{2} - \ln{\sqrt{2} + 1}\right) = 8.739$
Find the area of the region inside the outer loop but outside the inner loop of the function $r = 2\cos(\theta) - 1$.
Set up an integral (but do not evaluate) for the length of the polar curve $r = \sin^3(\theta)$ over the range $0 \leq \theta \leq \pi$.