Find the area enclosed by the curve r=1+cosθ and r=1 the radius vectors at θ=0 to θ=pi/2.

1. Break Down the Problem

To find the area enclosed by the curves $r = 1 + \cos \theta$ and $r = 1$ from $\theta = 0$ to $\theta = \frac{\pi}{2}$, we will:

1. Identify the intersection points of the curves.
2. Set up the integral to calculate the area between the two curves.

2. Relevant Concepts

The area $A$ enclosed by polar curves can be calculated using the formula: $A = \frac{1}{2} \int_{\alpha}^{\beta} (r_{\text{outer}}^2 - r_{\text{inner}}^2) \, d\theta$ where $r_{\text{outer}}$ is the curve that is further from the origin and $r_{\text{inner}}$ is the curve that is closer.

3. Analysis and Detail

1. Intersection: Set the equations equal to find the intersection points: $1 + \cos \theta = 1$ This simplifies to: $\cos \theta = 0 \Rightarrow \theta = \frac{\pi}{2}$ Thus, the curves intersect at $\theta = \frac{\pi}{2}$.

2. Determine the Area: In the interval $[0, \frac{\pi}{2}]$:

   • For $\theta = 0$, $r = 1 + \cos(0) = 2$
   • For $\theta = \frac{\pi}{2}$, $r = 1 + \cos\left(\frac{\pi}{2}\right) = 1$

   The area needs to be calculated for the interval, where:

   • $r_{\text{outer}} = 1 + \cos \theta$
   • $r_{\text{inner}} = 1$

3. Set Up the Integral: $A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} ((1 + \cos \theta)^2 - 1^2) \, d\theta$

4. Expanding the Integral: $A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} ((1 + 2\cos \theta + \cos^2 \theta) - 1) \, d\theta$ $= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (2\cos \theta + \cos^2 \theta) \, d\theta$

4. Verify and Summarize

Calculating the area:

1. Integral of $2\cos \theta$: $\int 2\cos \theta \, d\theta = 2\sin \theta$

2. Integral of $\cos^2 \theta$ using the identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$: $\int \cos^2 \theta \, d\theta = \frac{1}{2} \left( \theta + \frac{\sin(2\theta)}{2} \right)$

3. Combine and evaluate from $0$ to $\frac{\pi}{2}$: $A = \frac{1}{2} \left[ \left( 2\sin \theta + \frac{1}{2} \left( \theta + \frac{\sin(2\theta)}{2} \right) \right) \bigg|_0^{\frac{\pi}{2}} \right]$ Evaluating:

   • At $\frac{\pi}{2}$: $2\sin \left(\frac{\pi}{2}\right) = 2$, the contribution from $\cos^2$ is $\frac{\pi}{4}$.
   • At $0$: both contributions are $0$.

Final area becomes: $A = \frac{1}{2} \left( 2 + \frac{\pi}{4} \right) = 1 + \frac{\pi}{8}$

Final Answer

The area enclosed by the curves from $\theta = 0$ to $\theta = \frac{\pi}{2}$ is: $\boxed{1 + \frac{\pi}{8}}$