# Area of the region: $\;x ≥ 0; \;−x\sqrt3 ≤ y ≤ x\sqrt3;\,\;(x−1)^2 + y^2 ≤ 1$.

Can anyone please explain how to set up the needed integral?

I need to calculate the area of the region given by:

• $x ≥ 0,$

• $-x\sqrt3 ≤ y ≤ x\sqrt3,$

• $(x−1)^2 + y^2 ≤ 1$.

• Apparently, this is homework – Tunk-Fey Aug 14 '14 at 14:13
• you have mentioned 3 conditions, pls tell which condition is hard to understand? – Vikram Aug 14 '14 at 14:17
• Well i understand that i have to integrate a double integral with a certain function but I'm not sure which function and what boundaries. – Jody Aug 14 '14 at 14:24
• take a look at this: wolframalpha.com/input/… – Varun Iyer Aug 14 '14 at 14:25
• I don't think double integral is necessary.Find the intersection points of the circle and the pair of straight lines. Then upto the intersection point find area between the two straigthlines and after that find the area of the segment of the circle. – Soham Aug 14 '14 at 14:29

## 3 Answers

Here's a graph of the given bounds: The graph is symmetrical about the $x$-axis. So we can consider finding twice the desired area that lies at/above the $x$-axis.

And we can do this all using a single variable with respect to which we are integrating!

Note that the circle and the line $y = \sqrt 3x$ intersect at $x = 0$ and at \begin{align} \sqrt 3 x = \sqrt{1-(x-1)^2} &\implies 3x^2 = 1-(x^2 - 2x + 1) \\ \\ & \iff 4x^2 - 2x = 2x(2x -1)= 0 \\ \\ &\iff x = 0, \;\text{ or } x= \frac{1}{2}.\end{align} And note that we also need for $x\leq 2,$ since the circle with diameter lying along the x-axis intersects the x-axis on the right at $x=2$.

So we can integrate over two regions:

• $0\leq x \leq 1/2$ for the area between bounded above by $y =\sqrt 3x$ and below by $y = 0,$
• and $\; 1/2 \leq x \leq 2$, for the area bounded above by $y = \sqrt{1-(x-1)^2}$, and below by $y = 0$.

$$2\left(\int_0^{1/2} \sqrt 3x\,dx\quad + \quad\int_{1/2}^2 \sqrt{1-(x-1)^2}\,dx\right)$$

The left-most integral can easily be integrated and evaluated. The right-most integral can be solved using the trigonometric substitution $$(x-1) = \sin\theta\implies dx = \cos \theta\,d\theta,\;\text{ with } \,\theta = \arcsin(x-1)$$ and the corresponding bounds of integration changing to $\;-\pi/6 \leq \theta \leq 0.$

At first sketch the graph of the two lines and circle. By a simple double integrals in polar coordinates you can calculate the desired area. $(x-1)^2+y^2=1~$ reduces to $r=2\cos\theta~$. $Area=2\int_0^{\frac{\pi}{3}}\int_0^{2\cos\theta}rdrd\theta$.

• Thanks! that's what I was looking for! – Jody Aug 14 '14 at 14:58

First, draw the region.

Link: here

Now, we can use a double integral to solve this, or since its a circle, we can use simple geometry:

Notice that the region consists of $2$ equilateral triangles, meaning the area is simply $2/3$ the area of the circle plus the area of the two triangles, and their side length is equal to the radius of the circle, $1$.

So,

$$A = \frac{2}{3}*\pi*(1)^2 + \frac{1^2\sqrt{3}}{4}*2$$ $$A = \frac{2\pi}{3} + \frac{2\sqrt{3}}{4}$$

• Why was this needed to be stated – Varun Iyer Aug 14 '14 at 14:41
• Thnks, can you also show me how to do this by integration? – Jody Aug 14 '14 at 14:55