polar coordinates
I just want to ask if the polar form of the unit circle is
$$\cos^2\theta+\sin^2\theta=1$$
So, if I were to try and find the area of the circle using an integral, I would get
$$\int_0^{2\pi}\left(\cos^2\theta+\sin^2\theta-1\right)d\theta$$
But this equals $0$; therefore, you can't find the area using polar coordinates. Or am I wrong?
The area of a circular sector of radius $r$ and angle $d\theta$ is $\pi r^2 \frac{d\theta}{2\pi} = \frac{1}{2} r^2 d\theta$.
The right side of the square ($0 < \theta < \pi/4$) is the line $x = 5$, which in polar coordinates is $r = \frac{5}{\cos \theta}$ (not $\sin$).
Putting this together, the integrand should be $\frac{1}{2}(\frac{5}{\cos \theta})^2 d\theta = \frac{25}{2 \cos^2 \theta} d\theta$.
This works out to $12.5$. Since it only covers half of the square, double it to get the $25$ you expect.
Finding the area by polar coordinates without any further transformations is not the easy way to do it. That said, your integral has several errors, the most important being that you need to split the defined region into three parts.
The graph shows the region and its limits, though you can find this purely analytically. Region 1's angle $\theta$ goes from $-\frac{\pi}2$ to $\operatorname{atan}\left(-\frac 23\right)$ and its $r$ goes from $0$ to $\frac{4\cos\theta}{\sin^2\theta}$. You almost had these limit right. Region 3's angle goes from $\operatorname{atan}(2)$ to $\frac{\pi}2$ and has the same limits on $r$. Region 2's angle goes from $\operatorname{atan}\left(-\frac 23\right)$ to $\operatorname{atan}(2)$, and $r$ starts at zero with its upper limit coming from
$$y-3 = x$$ $$r\sin\theta = 3-r\cos\theta$$ $$r = \frac{3}{\sin\theta+\cos\theta}$$
The final expression, the sum of three double integrals, should be clear. If you really need just one double integral, you could make $-\theta$ from $\frac{\pi}2$ to $\frac{\pi}2$ and $r$ from $0$ to the minimum of the expressions for the parabola and for the line. But this would be an improper integral, since the max value of $r$ would not be properly defined for the parabola at $\theta=0$.
The problem is much easier in Cartesian coordinates. I think it would also be easier in polar coordinates if you translated the region $3$ units to the left, so the line would set the limits on $\theta$ and the parabola would set the limits on $r$. If you are allowed to go that route, you should. I'll leave the details to you. (I have not worked out the details myself.)
Best Answer
@Heavenly96, see the very first formula in teal here:
The derivation for this is mainly due to an analogy with the area of a circle. What are the areas of a full, semi, quarter, and eighth circle respectively? They are $\pi r^2, {\pi\over2}r^2, {\pi\over4}r^2$, and ${\pi\over8}r^2$. Notice something interesting here? The area of each division ends up yielding half the angle required to find such area: $$A_{\Delta\theta}\approx{\Delta\theta\over2}r^2.$$ As $\Delta \theta \to 0$, summing up each of the pieces to gain the full area gives us the formula $$A_{r(\theta)} = \int_\alpha^\beta{d\theta\over2}r^2.$$
This is how the ${1\over 2}$ comes about.