Let's break down and solve the problem step by step: **Given:** - The cross-section of the cylinder is defined by: \( x^2 + y^2 \leq 4x \) - Bounded below by \( z = 0 \) - Bounded above by \( z + x = 8 \) (so \( z = 8 - x \)) We are to **find the volume** of this solid. --- ## **Step 1: Rewrite the Cross-Section** The cross-section equation is: \[ x^2 + y^2 \leq 4x \] Rewrite it by completing the square: \[ x^2 - 4x + y^2 \leq 0 \implies (x-2)^2 + y^2 \leq 4 \] This is a **circle** of radius 2 centered at \((2,0)\) in the \(xy\)-plane. --- ## **Step 2: Set Up the Volume Integral** The volume is given by: \[ V = \iint_{R} \left[ \text{Upper z-bound} - \text{Lower z-bound} \right]\, dA \] where \( R \) is the region \( (x-2)^2 + y^2 \leq 4 \). The upper \(z\)-bound is \( z = 8 - x \), the lower is \( z = 0 \): \[ V = \iint_{R} (8 - x) \, dA \] --- ## **Step 3: Change to Polar Coordinates** Let: - \( x = r\cos\theta + 2 \) - \( y = r\sin\theta \) - \( (x-2)^2 + y^2 = r^2 \leq 4 \implies r \in [0, 2] \) - \( \theta \in [0, 2\pi] \) - \( dA = r\, dr\, d\theta \) Now, \( x = r\cos\theta + 2 \), so \( 8-x = 8 - (r\cos\theta + 2) = 6 - r\cos\theta \). So: \[ V = \int_0^{2\pi} \int_0^2 (6 - r\cos\theta) r\, dr\, d\theta \] --- ## **Step 4: Evaluate the Integral** Expand the integrand: \[ (6 - r\cos\theta) r = 6r - r^2\cos\theta \] So, \[ V = \int_0^{2\pi} \int_0^2 [6r - r^2\cos\theta] dr\, d\theta \] First, integrate with respect to \( r \): - \( \int_0^2 6r\, dr = 6 \cdot \frac{1}{2} r^2 \Big|_0^2 = 3 \cdot 4 = 12 \) - \( \int_0^2 r^2\, dr = \frac{1}{3} r^3 \Big|_0^2 = \frac{8}{3} \) So, \[ V = \int_0^{2\pi} [12 - \cos\theta \cdot \frac{8}{3} ] d\theta \] --- ## **Step 5: Integrate with Respect to \( \theta \)** \[ V = \int_0^{2\pi} 12\, d\theta - \frac{8}{3} \int_0^{2\pi} \cos\theta\, d\theta \] - \( \int_0^{2\pi} 12\, d\theta = 12 \cdot 2\pi = 24\pi \) - \( \int_0^{2\pi} \cos\theta\, d\theta = 0 \) (since cosine is periodic and symmetric about \(0\) over a full period) --- ## **Step 6: Write the Final Answer** \[ \boxed{V = 24\pi} \] --- ### **Final Answer** \[ \boxed{24\pi} \] This is the volume of the cylinder bounded as described.