1. **Problem Statement:** (a) Find the area common to the interiors of the cardioids $r = 1 + \cos \theta$ and $r = 1 - \cos \theta$. (b) Find the volume of the wedge cut from the cylinder $x^2 + y^2 = 1$ by the planes $z = -y$ and $z = 0$. 2. **Formulas and Important Rules:** - Area enclosed by a polar curve $r(\theta)$ between $\alpha$ and $\beta$ is given by: $$A = \frac{1}{2} \int_{\alpha}^{\beta} r^2(\theta) d\theta$$ - Volume under a surface $z = f(x,y)$ over a region $R$ in the $xy$-plane is: $$V = \iint_R f(x,y) \, dA$$ - For the wedge, volume can be found by integrating $z$ between the planes over the circular base. 3. **Part (a) Area of Common Region of Cardioids:** - The cardioids are $r_1 = 1 + \cos \theta$ and $r_2 = 1 - \cos \theta$. - They intersect where $r_1 = r_2 \Rightarrow 1 + \cos \theta = 1 - \cos \theta \Rightarrow \cos \theta = 0$. - So intersections at $\theta = \frac{\pi}{2}$ and $\theta = \frac{3\pi}{2}$. - The common region is symmetric about the vertical axis. - Area common = $2 \times$ area between $\theta = 0$ to $\frac{\pi}{2}$ where $r_1 \leq r_2$. - Area = $2 \times \frac{1}{2} \int_0^{\pi/2} \min(r_1^2, r_2^2) d\theta = \int_0^{\pi/2} (1 - \cos \theta)^2 d\theta$ because $r_2 \leq r_1$ in this interval. - Expand: $$(1 - \cos \theta)^2 = 1 - 2 \cos \theta + \cos^2 \theta$$ - Integral: $$\int_0^{\pi/2} (1 - 2 \cos \theta + \cos^2 \theta) d\theta = \int_0^{\pi/2} 1 d\theta - 2 \int_0^{\pi/2} \cos \theta d\theta + \int_0^{\pi/2} \cos^2 \theta d\theta$$ - Evaluate each: 1. $\int_0^{\pi/2} 1 d\theta = \frac{\pi}{2}$ 2. $\int_0^{\pi/2} \cos \theta d\theta = \sin \theta \big|_0^{\pi/2} = 1$ 3. Use identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$: $$\int_0^{\pi/2} \cos^2 \theta d\theta = \int_0^{\pi/2} \frac{1 + \cos 2\theta}{2} d\theta = \frac{1}{2} \left[ \theta + \frac{\sin 2\theta}{2} \right]_0^{\pi/2} = \frac{1}{2} \left( \frac{\pi}{2} + 0 \right) = \frac{\pi}{4}$$ - Substitute back: $$\frac{\pi}{2} - 2(1) + \frac{\pi}{4} = \frac{\pi}{2} - 2 + \frac{\pi}{4} = \frac{3\pi}{4} - 2$$ - So area common = $\frac{3\pi}{4} - 2$. 4. **Part (b) Volume of Wedge:** - Cylinder: $x^2 + y^2 = 1$. - Planes: $z = -y$ and $z = 0$. - Volume between $z=0$ (top) and $z=-y$ (bottom) over the unit disk. - Since $z=0$ is above $z=-y$ when $y \leq 0$, and below when $y \geq 0$, we consider volume where $z$ is positive. - Volume: $$V = \iint_{x^2 + y^2 \leq 1} (0 - (-y)) dA = \iint_{x^2 + y^2 \leq 1} y \, dA$$ - Because $y$ is odd in $y$ and region symmetric about $x$-axis, integral over full disk is zero. - But wedge is between $z=0$ and $z=-y$, so volume is positive where $-y \geq 0 \Rightarrow y \leq 0$. - So integrate over lower half disk $y \leq 0$: $$V = \iint_{x^2 + y^2 \leq 1, y \leq 0} (-y) dA$$ - Use polar coordinates: $x = r \cos \theta$, $y = r \sin \theta$, $r \in [0,1]$, $\theta \in [\pi, 2\pi]$ (lower half circle). - Then: $$V = \int_{\pi}^{2\pi} \int_0^1 (-r \sin \theta) r dr d\theta = \int_{\pi}^{2\pi} (-\sin \theta) \int_0^1 r^2 dr d\theta = \int_{\pi}^{2\pi} (-\sin \theta) \frac{1}{3} d\theta = \frac{1}{3} \int_{\pi}^{2\pi} -\sin \theta d\theta$$ - Evaluate: $$\int_{\pi}^{2\pi} -\sin \theta d\theta = [\cos \theta]_{\pi}^{2\pi} = \cos(2\pi) - \cos(\pi) = 1 - (-1) = 2$$ - So volume: $$V = \frac{1}{3} \times 2 = \frac{2}{3}$$ **Final Answers:** - (a) Area common to cardioids = $\boxed{\frac{3\pi}{4} - 2}$ - (b) Volume of wedge = $\boxed{\frac{2}{3}}$