Calculate AB, Find the length, Find the area of the shaded region (A long one)

Calculate AB

Calculate the length of arc AB

Calculate the area of the shaded region 

However I do not know how, and any help is appreciated, please and thank you!

I couldn't resist putting you picture in here.  It is easy done with Desmos Graphing calculator.

Plus, "How to upload an Image"  is one of the sticky topics, in the bottom right of the page.

Are you happy with your answer?

This is the poster, the proper image of the problem is here

http://imgur.com/Ket2iKe

Assuming your points of intersection are correct - I haven't checked them- we can use the distance formula to find the length of chord AB....we have

( (2.68 - .52)^2 + (1.36 + 2.96)^2)^(.5) =

$${\left({\left({\mathtt{2.68}}{\mathtt{\,-\,}}{\mathtt{0.52}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\mathtt{1.36}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2.96}}\right)}^{{\mathtt{2}}}\right)}^{\left({\mathtt{0.5}}\right)} = {\mathtt{4.829\: \!906\: \!831\: \!399\: \!545\: \!7}}$$

We'll just call it 4.83 to simplify things

Next, we can use the Law of Cosines to find the (minor) angle formed by the two radii drawn to the points of intersection.  This will help us find the arc length.  So we have

AB^2 = 3^2 + 3^2 - 2(3)(3)cos (theta)

(4.83)^2 = 18 - 18cos (theta)

cos(theta) = -( (4.83)^2 -18)/18

cos (theta) =  $${\mathtt{\,-\,}}{\frac{\left({\left({\mathtt{4.83}}\right)}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{18}}\right)}{{\mathtt{18}}}} = -{\mathtt{0.296\: \!05}}$$

Now, taking the cosine inverse of this will give us the angle....

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left(-{\mathtt{0.296\: \!05}}\right)} = {\mathtt{107.220\: \!510\: \!644\: \!933^{\circ}}}$$

=  about 107.22 degrees....this seems reasonable.....

Now, we need to convert this to radians to find the arc length

 This is given by  .....107.22 x (pi)/180 =

$${\frac{{\mathtt{107.22}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{180}}}} = {\mathtt{1.871\: \!342\: \!023\: \!988\: \!320\: \!2}}$$

= about 1.87 radians

And the "formula" for the arc length (s) is given by

s = r x (theta in radians)   =     3 x 1.87 = 5.61

Now, to find the area of the shaded region, we can find the area of the (minor) sector formed by the two radial lines less the area of the triangle formed by the two radii and AB.

We have..... (1/2)(r^2) (theta in radians) - (1/2)(r^2)sin(theta in rads)

= (1/2)(r^2) (theta in radians - sin(theta in radians))=

(1/2)(9) (1.87 - sin(1.87)) = about 4.11 square units

I think that's it....if I haven't made any errors!!