# Double integral polar coordinates for area bounded by curves

I need to find the area bounded by the curves :

$$x^2 + y^2 = 1, \ y^2= x\sqrt3, \ x \geqslant \frac{y^2}{\sqrt3}$$

My attempt:

$$\int^{\frac{\pi}{2}}_{\frac{3\pi}{2}} \int^{\text{some complicated number}}_{-\text{some complicated number}} \left( r\sin{\theta^2} -\sqrt3r\cos{\theta} \right)\ \mathrm{d}r \ \mathrm{d}\theta$$

I think that plus or minus $$\text{some complicated number}$$

is a point where $$x^2 + y^2 = 1$$ and $$y^2= x\sqrt3$$

intersect.

Am I on the right track with this question? Don't give me the answer please I want to figure it out myself, but I'd like to know if I am on the right track.

My understanding is that this is a circle, with a hill shape cut into the side of it and I'm integrating the area between the circle and the other shape.

The first integral refers to the quadrant of the area, the second integral is the specific area in the quadrant being integrated (i.e. the limits) and the function (i.e. the hill) gets turned into the polar coordinates function.

First of all you should draw a graph to see what region is bounded by those curves. So we will draw a graph of $$S= \{ (x, y) \in \mathbb{R}^2 \colon \ x^2 + y^2 \leqslant 1 \ \& \ x \geqslant \frac{y^2}{\sqrt{3}}\}.$$ Now you can find that those two plots intersect at $x = \frac{\sqrt{7}-\sqrt{3}}{2}$ and so you can write down the corresponding integrals $$\text{Area of } S = \int_{\frac{\sqrt{7}-\sqrt{3}}{2}}^{1} \int_{- \sqrt{1-x^2}}^{\sqrt{1-x^2}} \mathrm{d} y \ \mathrm{d} x + \int_{0}^{\frac{\sqrt{7}-\sqrt{3}}{2}} \int_{- \sqrt{x\sqrt{3}}}^{\sqrt{x\sqrt{3}}} \mathrm{d} y \ \mathrm{d} x.$$ Please see the picture below for details: This integral is easy to calculate. However, If you don't want to calculate the first one by using the standard integration, we can write it as

$$\int_{\frac{\sqrt{7}-\sqrt{3}}{2}}^{1} \int_{- \sqrt{1-x^2}}^{\sqrt{1-x^2}} \mathrm{d} y \ \mathrm{d} x = \iint_{D} \mathrm{d}(x,y),$$

where $D= \{ (x, y) \colon \ x^2+y^2 \leq 1 \ \& \ \frac{\sqrt{7}-\sqrt{3}}{2} \leq x\leq 1\}$ and use the polar coordinates substitution, that is, $$\begin{cases}x= r\cos\phi \\ y = r\sin\phi \end{cases}.$$ These will give us

$$\iint_{D} \mathrm{d}(x,y) = \int_{\alpha}^{\beta} \int_{0}^{1} r \ \mathrm{d} r \ \mathrm{d} \phi,$$ where $\alpha$ and $\beta$ are angles which you have to determine from the form of $D$ and a substitution, $r$ in the integral stands for the the Jacobian determinant of the coordinate conversion.

From the above you can see that in this case it is better to calculate this integral without using polar coordinates, however if you determine the values of $\alpha$ and $\beta$ correctly you obtain the same, correct result.

• thanks I don't get exactly why the limit of integration is from $${\frac{\sqrt{7}-\sqrt{3}}{2}}$$ it looks as though you are splitting the shape between the curves down the middle and adding up the two halves. How do you know it is perfectly a half ? – user3528592 Oct 26 '14 at 1:31
• How come you can't just go from 0 to $${\frac{\sqrt{7}-\sqrt{3}}{2}}$$ – user3528592 Oct 26 '14 at 1:36
• This is actually easier if you switch the order of integral from $dy\,dx$ to $dx\,dy$, so you only get one integral instead of a sum of 2 $$Area = \int_{y_1}^{y_2} \int_{\frac{y^2}{\sqrt{3}}}^{\sqrt{1-y^2}} dx\,dy$$ – Dylan Oct 31 '14 at 1:46