# What is the area bounded by the the inside of polar curve #1+cos theta# and outside the polar curve #r(1+cos theta)=1#?

# "Area" = (3pi)/4+ 4/3 #

If we plot the polar curve, and shadethe area sought:

We observe that the points of intersection are:

#theta = -pi/2# and#theta=pi/2#

We calculate area in polar coordinates using :

# A = 1/2 \ int_alpha^beta \ r^2 \ d theta #

Thus, the enclosed area is:

# A = 1/2 \ int_(-pi/2)^(pi/2) \ (1+cos theta)^2 - (1/(1+cos theta))^2 \ d theta #

# :. 2A = int_(-pi/2)^(pi/2) \ (1+cos theta)^2 \ d theta - int_(-pi/2)^(pi/2) (1/(1+cos theta))^2 \ d theta #

Consider the first integral;:

# I_1= int_(-pi/2)^(pi/2) \ (1+cos theta)^2 \ d theta #

# \ \ \ = int_(-pi/2)^(pi/2) \ 1+ 2cos theta + cos^2 theta \ d theta #

# \ \ \ = int_(-pi/2)^(pi/2) \ 1+ 2cos theta + (1+cos 2theta)/2 \ d theta #

# \ \ \ = int_(-pi/2)^(pi/2) \ 3/2+ 2cos theta + (cos 2theta)/2 \ d theta #

# \ \ \ = [(3theta)/2+ 2sin theta + (sin 2theta)/4]_(-pi/2)^(pi/2) \ #

# \ \ \ = ((3(pi/2))/2+ 2sin (pi/2)+ (sin (2pi))/4) - ((3(-pi/2))/2+ 2sin (-pi/2)+ (sin (-2pi))/4)#

# \ \ \ = ((3pi)/4+ 2+ 0) - (-(3pi)/4-2-0)#

# \ \ \ = (3pi)/2+ 4#

And, now the second integral:

# I_2 = int_(-pi/2)^(pi/2) (1/(1+cos theta))^2 \ d theta #

For which, we perform, a tangent half-angle substitution:

# u =tan(theta/2) => (du)/(d theta) = 1/2sec^2(theta/2) #

When:

# theta = { (pi/2), (-pi/2) :} => u = { (1), (-1) :} #

And we can manipulate the integral and perform the substitution, to get:

# I_2 = int_(-pi/2)^(pi/2) 1/( (1-tan^2(theta/2))/(1+tan^2(theta/2))+1)^2 \ d theta #

# \ \ \ = int_(-pi/2)^(pi/2) 1/( (1-tan^2(theta/2))/(sec^2(theta/2))+1)^2 \ d theta #

# \ \ \ = int_(-pi/2)^(pi/2) 1/ ((1-tan^2(theta/2)+1+tan^2(theta/2))/(1+tan^2(theta/2)))^2 \ d theta #

# \ \ \ = int_(-pi/2)^(pi/2) 1/ ((2)/(sec^2(theta/2)))^2 \ d theta #

# \ \ \ = int_(-pi/2)^(pi/2) 1/ ((2)/(sec^2(theta/2)))^2 \ d theta #

# \ \ \ = 1/4 \ int_(-pi/2)^(pi/2) sec^2 (theta/2) sec^2 (theta/2) \ d theta #

# \ \ \ = 1/2 \ int_(-pi/2)^(pi/2) (1+tan^2 (theta/2)) \ 1/2sec^2 (theta/2) \ d theta #

# \ \ \ = 1/2 \ int_(-1)^(1) 1+u^2 \ du #

# \ \ \ = 1/2 [u+u^3/3]_(-1)^(1) #

# \ \ \ = 1/2 {(1+1/3) - (-1-1/3)} #

# \ \ \ = 4/3 #

Combining both results, we get:

# 2A = I_1 + I_2 #

# \ \ \ \ \ = (3pi)/2+ 4 + 4/3 #

# \ \ \ \ \ = (3pi)/2+ 16/3 #

Hence:

# A = (3pi)/4+ 4/3 #

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The area bounded by the inside of the polar curve (r = 1 + \cos(\theta)) and outside the polar curve (r(1 + \cos(\theta)) = 1) can be found by computing the definite integral of the difference between the two curves from the limits where they intersect.

First, find the points of intersection by equating the two curves:

[1 + \cos(\theta) = \frac{1}{r}]

[r = \frac{1}{1 + \cos(\theta)}]

To find the limits of integration, set (1 + \cos(\theta) = \frac{1}{r}) equal to (0) to find where the curves intersect:

[1 + \cos(\theta) = 0]

[\cos(\theta) = -1]

This gives us (\theta = \pi).

Then, the limits of integration for (\theta) are from (0) to (\pi).

The area (A) is given by the integral:

[A = \frac{1}{2} \int_{0}^{\pi} \left((1 + \cos(\theta))^2 - \left(\frac{1}{1 + \cos(\theta)}\right)^2\right) d\theta]

Solving this integral will give the area bounded by the two curves.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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