Find the area using double integral and polar coordinates. I need to find the area using double integral and polar coordinates.
$$y=3-x$$
$$y^2=4x$$
This is what i figured already:
$${r\cos{\theta}+r\sin{\theta}} = 3$$
$$r=0, r=3, \theta=0, \theta=\pi/2$$
$${r^2\sin^2{\theta}-4r\cos{\theta}} = 0$$
$$r=0, r=\frac{4cos\theta}{sin^2\theta}$$
Any help would appreciated.
 A: 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.)
