# Area of the region ????

#### goku900

##### New member
I have the region r = 2 + cos(theta) . I know the area should be 18.64.

I set it = 0 and then solve for theta.

So theta = 0 and theta = 2pi

I set up my integral [0, 2pi] 1/2(r)^2 dThetaA

After simplification I got 1/4 integral cos2theta + 4costheta + 5 but my answer does not come out right after integrating ?

#### Sudharaka

##### Well-known member
MHB Math Helper
I have the region r = 2 + cos(theta) . I know the area should be 18.64.

I set it = 0 and then solve for theta.

So theta = 0 and theta = 2pi

I set up my integral [0, 2pi] 1/2(r)^2 dThetaA

After simplification I got 1/4 integral cos2theta + 4costheta + 5 but my answer does not come out right after integrating ?
Hi goku900, I don't think the given answer is correct. I get 14.14 as the answer. Your approach for solving the problem is correct.

$A=\frac{1}{2}\int_{0}^{2\pi}r^2\,d\theta=14.137$

To verify you can try drawing the graph using one of the may tools available online (I recommend Desmos). As you see you can even find a rectangle enclosing the figure with area $$4.5\times 4=18$$. So presumably the area should be less than 18. [GRAPH]xeao52688t[/GRAPH]