What is the area of a region with r = 2 + cos(theta)?

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In summary, the conversation discussed finding the area of a region defined by the equation r = 2 + cos(theta). After setting the equation to 0 and solving for theta, the integration was set up for the interval [0, 2pi] and simplified to 1/4 integral cos2theta + 4costheta + 5. However, the resulting answer did not match the expected area of 18.64, with the correct answer being 14.14. The discrepancy can be verified by graphing the region and finding a rectangle with a smaller area that encloses it.
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Vanrichten
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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 ?
 
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  • #2
goku900 said:
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]
 

FAQ: What is the area of a region with r = 2 + cos(theta)?

What is the equation for the area of a region with a polar equation of r=2+cos(θ)?

The equation for the area of a region with a polar equation of r=2+cos(θ) is:A = ∫(1/2)r^2dθwhere r is the polar equation and θ is the angle of rotation.

How do you find the limits of integration when calculating the area of a polar region with r=2+cos(θ)?

The limits of integration for a polar region with r=2+cos(θ) can be found by setting the polar equation equal to 0 and solving for θ. In this case, the limits would be θ= 0 and θ= 2π.

Can the area of a polar region with r=2+cos(θ) be calculated using rectangular coordinates?

Yes, the area of a polar region with r=2+cos(θ) can also be calculated using rectangular coordinates. The equation for the area in rectangular coordinates would be:A = ∫(1/2)(x^2+y^2)dxwhere x= rcos(θ) and y= rsin(θ).

How does the value of the constant in the polar equation r=2+cos(θ) affect the shape of the region?

The value of the constant in the polar equation r=2+cos(θ) affects the size of the region, with larger values resulting in a larger region. However, it does not affect the overall shape of the region, which is determined by the cosine function.

Can the area of a region with a polar equation of r=2+cos(θ) be negative?

No, the area of a region cannot be negative. The integral used to calculate the area only takes positive values, so the resulting area will always be positive or 0.

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