given the graphs of polar equations ( r = 3+cos(\theta) ) and ( r = 3-cos(\theta) )\n(a) set up the integral…

Explanation:

Step1: Use the formula for the area in polar coordinates

The formula for the area (A) of a polar region is (A=\frac{1}{2}\int_{\alpha}^{\beta}[r_{1}^{2}(\theta)-r_{2}^{2}(\theta)]d\theta). Due to symmetry, we can find the area of the upper - half of the shaded region and then double it. For (r = 3+\cos\theta) and (r = 3 - \cos\theta), and considering the symmetry about the (y) - axis ((\theta=-\frac{\pi}{2}) to (\theta=\frac{\pi}{2})), the integral for the area of the shaded region is (A = 2\times\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[(3 + \cos\theta)^{2}-(3-\cos\theta)^{2}]d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[(3 + \cos\theta)^{2}-(3-\cos\theta)^{2}]d\theta). Expand ((a + b)^{2}-(a - b)^{2}=4ab). Here (a = 3) and (b=\cos\theta), so ((3 + \cos\theta)^{2}-(3-\cos\theta)^{2}=4\times3\times\cos\theta = 12\cos\theta). The integral becomes (A=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}12\cos\theta d\theta).

Step2: Evaluate the integral

Recall that (\int\cos\theta d\theta=\sin\theta+C). Using the fundamental theorem of calculus (\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}12\cos\theta d\theta=12[\sin\theta]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}). Substitute the upper and lower limits: (12\left(\sin\frac{\pi}{2}-\sin(-\frac{\pi}{2})\right)). Since (\sin\frac{\pi}{2}=1) and (\sin(-\frac{\pi}{2})=- 1), we have (12(1-(-1))).

Answer:

The integral for the area is (A=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[(3 + \cos\theta)^{2}-(3-\cos\theta)^{2}]d\theta) and the value of the area is (24).