# 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.

• What region are you integrating over? – John_dydx Jul 25 '15 at 16:29
• You wrongfully find bounds – Michael Galuza Jul 25 '15 at 16:29
• area between the curves. – Blay Heshbon Jul 25 '15 at 16:58

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.)