Let's break down the problem: ### **Problem Recap** - A circular corral has a radius of 11. - A goat is tied to the **fence** with a rope of length \( a \), where \( 0 \leq a \leq 22 \). - The **goat is tied at the edge** of the circle (the fence), and the rope can reach either inside or along the edge. - We are asked for the area **inside the corral** that the goat can graze. - We need to set up the **integral(s)** for this area, using the provided choices. --- ### **Analysis** Let’s: - Place the origin at the point where the goat is tied, and the center of the corral at (0, 11). - The circle's equation (centered at (0, 11), radius 11): \[ x^2 + (y-11)^2 = 121 \] - The region the goat can graze is the area inside the **corral** and also within the **circle of radius \( a \)** centered at the origin. #### **Limits** - The goat can graze in the overlap of: - The circle of radius \( a \) centered at the origin - The corral (of radius 11 centered at (0,11)) We use **polar coordinates** (\( r, \theta \)), with the origin at the tie point (the edge of the corral). #### **Angle Limits** - The goat can graze from \( \theta = -\sin^{-1}(a/22) \) to \( \theta = \sin^{-1}(a/22) \). - For \( |\theta| < \sin^{-1}(a/22) \), the goat's reach is limited by the rope (\( r = a \)). - For \( \sin^{-1}(a/22) < |\theta| < \pi/2 \), the goat's reach is limited by the corral's fence. #### **Integral Setup** The area is the sum of two regions: 1. For \( 0 \leq \theta \leq \sin^{-1}(a/22) \), the goat can move out to distance \( a \). 2. For \( \sin^{-1}(a/22) \leq \theta \leq \pi/2 \), the goat can move out to the corral's edge (which is at distance \( 22 \cos\theta \) from the tie point, but since the radius is 11, this must be handled carefully). But the answer choices are asking for the setup, not the explicit computation. --- ### **Matching With Choices** - **Choice C:** \[ \int_0^{\sin^{-1}(a/22)} \left(\boxed{\frac{1}{2}a^2}\right) d\theta + \int_{\sin^{-1}(a/22)}^{\pi/2} \left(\boxed{\frac{1}{2}(22\cos\theta)^2}\right) d\theta \] But since the radius of the corral is 11, the maximum distance from the tie point to the opposite side of the corral is \( 22 \) along the diameter. - The correct setup should be: - For \( 0 \leq \theta \leq \sin^{-1}(a/22) \): area out to radius \( a \) (the rope) - For \( \sin^{-1}(a/22) \leq \theta \leq \pi/2 \): area out to the corral's edge (\( 22 \cos\theta \)) So, **Choice C** is correct, assuming the blank is filled with the appropriate radius squared (\( a^2 \) or \( (22 \cos\theta)^2 \)). --- ### **Final Answer** The answer is **Choice C**. The filled-in integrals would be: \[ \int_{0}^{\sin^{-1}(a/22)} \frac{1}{2} a^2 d\theta + \int_{\sin^{-1}(a/22)}^{\pi/2} \frac{1}{2} (22\cos\theta)^2 d\theta \] #### **Explanation** - The first term is the area swept by the rope for angles where it doesn't exceed the corral. - The second term is the area swept to the corral's edge for larger angles. - This matches the geometry of the problem and the required integral setup. --- ### **Boxed Final Answer** **C.** \[ \int_{0}^{\sin^{-1}(a/22)} \frac{1}{2} a^2 d\theta + \int_{\sin^{-1}(a/22)}^{\pi/2} \frac{1}{2} (22\cos\theta)^2 d\theta \]