Let's solve the problem step by step: ## **Step 1: Sketch the Region** Given curves: - \( y = 5\sqrt{36-x^2} \) - \( y = 0 \) (x-axis) - \( x = 3 \) - \( x = 5 \) This is the region under the curve \( y = 5\sqrt{36-x^2} \) from \( x=3 \) to \( x=5 \), above the x-axis. --- ## **Step 2: Set Up the Volume Integral (Disk Method)** When rotating about the x-axis, the radius of the disk at a point \( x \) is \( y = 5\sqrt{36-x^2} \). The thickness is \( dx \). **Volume of a typical disk:** \[ dV = \pi [\text{radius}]^2 dx = \pi \left(5\sqrt{36-x^2}\right)^2 dx \] **Total Volume:** \[ V = \int_{x=3}^{x=5} \pi \left(5\sqrt{36-x^2}\right)^2 dx \] --- ## **Step 3: Simplify the Integral** \[ V = \pi \int_{3}^{5} 25(36-x^2) dx \] \[ = 25\pi \int_{3}^{5} (36-x^2) dx \] --- ## **Step 4: Integrate** \[ \int (36-x^2) dx = 36x - \frac{1}{3}x^3 \] Apply the limits: \[ V = 25\pi \left[36x - \frac{1}{3}x^3\right]_{3}^{5} \] Plug in the values: For \( x = 5 \): \[ 36(5) - \frac{1}{3}(5^3) = 180 - \frac{125}{3} \] For \( x = 3 \): \[ 36(3) - \frac{1}{3}(3^3) = 108 - \frac{27}{3} = 108 - 9 = 99 \] Subtract: \[ \left(180 - \frac{125}{3}\right) - 99 = 180 - \frac{125}{3} - 99 = (180 - 99) - \frac{125}{3} = 81 - \frac{125}{3} \] --- ## **Step 5: Final Simplification** \[ 81 - \frac{125}{3} = \frac{243}{3} - \frac{125}{3} = \frac{243 - 125}{3} = \frac{118}{3} \] So, \[ V = 25\pi \left(\frac{118}{3}\right) = \frac{2950}{3} \pi \] --- ### **Final Answer** \[ \boxed{V = \frac{2950}{3} \pi} \] --- ## **Sketch** - The region is under \( y = 5\sqrt{36-x^2} \) from \( x = 3 \) to \( x = 5 \), above x-axis. - A typical disk at \( x \) has radius \( 5\sqrt{36-x^2} \). **[You can visualize this as a thick "slice" of a circle rotated around the x-axis, forming a solid with a curved outer surface.]** If you want a sketch, let me know!

Certainly! Here's a step-by-step description to help you visualize and sketch the solid: --- ### **Step 1: Sketch the Region in the xy-Plane** - Draw the coordinate axes. - The curve \( y = 5\sqrt{36 - x^2} \) represents the upper half of a circle with radius 6 centered at the origin, scaled vertically by a factor of 5. - Plot the semicircle: - When \( x=0 \), \( y=5 \times 6=30 \). - When \( x= \pm 6 \), \( y=0 \). - Shade the region between \( x=3 \) and \( x=5 \), above the x-axis, under the curve \( y=5\sqrt{36 - x^2} \). --- ### **Step 2: Visualize the Rotation** - The region from \( x=3 \) to \( x=5 \) is being rotated about the **x-axis**. - Each vertical slice at position \( x \) becomes a disk when rotated: - **Radius of the disk:** \( y = 5\sqrt{36 - x^2} \). - The disk's cross-section looks like a circle with that radius. --- ### **Step 3: Sketch the 3D Solid** - Imagine revolving that shaded region around the x-axis: - The result is a **solid of revolution** with a curved outer surface. - It resembles a "washer" or "bulb" shape with the outer edge following the curve \( y=5\sqrt{36 - x^2} \) from \( x=3 \) to \( x=5 \). --- ### **Step 4: Typical Disk or Washer** - At a typical \( x \) between 3 and 5: - Draw a circle in cross-section with radius \( y=5\sqrt{36 - x^2} \). - This circle is perpendicular to the x-axis. - As \( x \) varies, the radius changes, creating the curved surface of the solid. --- ### **Summary Sketch** - **2D view:** A segment of a semicircular arc between \( x=3 \) and \( x=5 \); this forms the boundary of the region. - **3D view:** Rotating this segment around the x-axis produces a solid with a smoothly curved outer surface, tapering towards the ends. --- If you'd like, I can also provide a simple hand-drawn style sketch to help visualize this!

Certainly! Here's a step-by-step description of how to draw a hand-drawn style sketch of the region and the resulting solid: 1. **Draw the coordinate axes:** - Sketch horizontal (x-axis) and vertical (y-axis) lines. 2. **Plot the semicircular boundary:** - The curve \( y = 5 \sqrt{36 - x^2} \) is the upper half of a circle \( x^2 + (y/5)^2 = 36 \). - Since \( y = 5 \sqrt{36 - x^2} \), the circle has radius 6 (since \( x^2 + (y/5)^2 = 36 \)). - Draw a circle centered at the origin with radius 6. - Sketch only the upper semicircle. 3. **Identify the region:** - Mark the vertical lines \( x=3 \) and \( x=5 \). - Shade the region under the semicircle between these two vertical lines. - This is a "slice" of the semicircle from \( x=3 \) to \( x=5 \). 4. **Visualize the solid of revolution:** - Imagine rotating this shaded region about the x-axis. - For each \( x \) between 3 and 5, draw a horizontal line from the x-axis up to the curve (radius \( y \)). - Draw a circle at each \( x \) with radius equal to \( y=5 \sqrt{36 - x^2} \). These circles represent the cross-sections of the solid. 5. **Sketch the 3D shape:** - Connect the circles smoothly to form a curved, bulbous shape. - The shape will look like a "washer" with curved outer surface, tapering towards the ends at \( x=3 \) and \( x=5 \). 6. **Add labels:** - Label the axes, the boundary points \( x=3, 5 \), and the curve. - Indicate the radius at a typical \( x \) to show the disk being rotated. --- **Tip:** Use light, curved lines to suggest the 3D nature of the solid, and dashed lines for the parts behind the surfaces if you want to add depth. --- Would you like me to generate a simple sketch image based on this description?