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Area of the region ????

goku900

New member
Nov 25, 2013
8
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
Feb 5, 2012
1,621
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. :)

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