Let's break down each part of the problem step by step: --- ### Given: - **Region R**: Bounded by \( y = 9 - x^2 \) (right/top), \( y \)-axis (left), and \( x \)-axis (bottom). - **Solid**: Generated by revolving \( R \) about the \( y \)-axis. - **Hole**: Cylinder of radius 1 along the \( y \)-axis is bored out. - **Density**: Constant, \( \rho \). --- ## (a) **Volume by Horizontal Slices** ### **Step 1: Describe the Region** - \( x \) ranges from \( 0 \) (y-axis) to \( \sqrt{9-y} \) (solve \( y = 9-x^2 \) for \( x \)). - \( y \) ranges from \( 0 \) to \( 9 \) (where parabola meets x-axis). ### **Step 2: Horizontal Slice** - At height \( y \), \( x \) ranges \( 0 \leq x \leq \sqrt{9-y} \). - When revolved, generates a disk with a hole (if \( \sqrt{9-y} > 1 \)), i.e., an annulus for \( y \leq 8 \), and a full disk for \( 8 < y \leq 9 \) (since \( \sqrt{9-y} < 1 \)). But in this case, since the hole is bored through the entire solid, for all \( y \), if \( \sqrt{9-y} > 1 \), the inner radius is 1, else the disk disappears. So, for \( y \) from 0 to 8: - Outer radius: \( R_o = \sqrt{9-y} \) - Inner radius: \( R_i = 1 \) For \( y \) from 8 to 9: - \( \sqrt{9-y} < 1 \), so there is no solid left in this region since the hole's radius is larger than the solid. ### **Step 3: Volume of a Horizontal Slice** - Area at height \( y \): \( A(y) = \pi [(\sqrt{9-y})^2 - 1^2] = \pi [(9-y) - 1] = \pi (8-y) \), but only for \( y \) where \( \sqrt{9-y} \geq 1 \), i.e., \( y \leq 8 \). - For \( y > 8 \), \( A(y) = 0 \). ### **Step 4: Integrate to Find Volume** \[ V = \int_{y=0}^{8} \pi (8-y) \, dy \] \[ V = \pi \int_{0}^{8} (8-y) \, dy \] \[ = \pi \left[ 8y - \frac{y^2}{2} \right]_{0}^{8} \] \[ = \pi \left[ 8(8) - \frac{8^2}{2} \right] \] \[ = \pi [64 - 32] = \pi \cdot 32 \] **Final answer for (a):** \[ \boxed{32\pi} \] --- ## (b) **Volume by Annular Cylinders (Shell Method)** ### **Step 1: Set Up Shell Method** - For a thin shell at radius \( x \), height \( y = 9-x^2 \), thickness \( dx \). - \( x \) ranges from \( x = 1 \) (hole) to \( x = 3 \) (where \( 9 - x^2 = 0 \implies x = 3 \)). ### **Step 2: Volume Element** - Circumference: \( 2\pi x \) - Height: \( 9-x^2 \) - Thickness: \( dx \) - Volume: \( dV = 2\pi x (9-x^2) dx \) ### **Step 3: Integrate** \[ V = \int_{x=1}^{3} 2\pi x (9-x^2) dx \] \[ = 2\pi \int_{1}^{3} (9x - x^3) dx \] \[ = 2\pi \left[ \frac{9x^2}{2} - \frac{x^4}{4} \right]_{1}^{3} \] Calculate at \( x=3 \): \[ \frac{9 \cdot 9}{2} - \frac{81}{4} = \frac{81}{2} - \frac{81}{4} = \frac{162-81}{4} = \frac{81}{4} \] At \( x=1 \): \[ \frac{9 \cdot 1}{2} - \frac{1}{4} = \frac{9}{2} - \frac{1}{4} = \frac{18-1}{4} = \frac{17}{4} \] Difference: \[ \frac{81}{4} - \frac{17}{4} = \frac{64}{4} = 16 \] Multiply by \( 2\pi \): \[ V = 2\pi \cdot 16 = 32\pi \] **Final answer for (b):** \[ \boxed{32\pi} \] --- ## (c) **\( y \)-coordinate of Center of Mass (\( \bar{y} \))** \[ \bar{y} = \frac{1}{V} \int y \, dV \] For horizontal slices: \[ dV = \pi (8-y) dy \] \[ \int y \, dV = \int_{0}^{8} y \cdot \pi (8-y) dy = \pi \int_{0}^{8} (8y - y^2) dy \] Calculate: \[ \pi \left[ 8 \cdot \frac{y^2}{2} - \frac{y^3}{3} \right]_{0}^{8} \] \[ = \pi \left[ 4y^2 - \frac{y^3}{3} \right]_{0}^{8} \] At \( y=8 \): \[ 4 \cdot 64 - \frac{512}{3} = 256 - \frac{512}{3} \] So: \[ \int y \, dV = \pi \left(256 - \frac{512}{3}\right) = \pi \left(\frac{768 - 512}{3}\right) = \pi \left(\frac{256}{3}\right) \] Recall \( V = 32\pi \): \[ \bar{y} = \frac{\pi \frac{256}{3}}{32\pi} = \frac{256}{3 \times 32} = \frac{256}{96} = \frac{8}{3} \] **Final answer for (c):** \[ \boxed{\bar{y} = \frac{8}{3}} \] --- ## (d) **Moment of Inertia about \( y \)-axis** For a shell at radius \( x \), \( dI = r^2\,dm \), and \( dm = \rho \, dV \). - \( dV = 2\pi x (9-x^2) dx \) - \( dI = x^2 \cdot \rho \cdot 2\pi x (9-x^2) dx = 2\pi \rho x^3 (9-x^2) dx \) Integrate: \[ I_y = 2\pi \rho \int_{x=1}^{3} x^3 (9-x^2) dx \] \[ = 2\pi \rho \int_{1}^{3} (9x^3 - x^5) dx \] \[ = 2\pi \rho \left[ \frac{9x^4}{4} - \frac{x^6}{6} \right]_{1}^{3} \] At \( x=3 \): \[ \frac{9 \cdot 81}{4} - \frac{729}{6} = \frac{729}{4} - \frac{729}{6} \] Common denominator 12: \[ = \frac{2187}{12} - \frac{1458}{12} = \frac{729}{12} \] At \( x=1 \): \[ \frac{9}{4} - \frac{1}{6} = \frac{27-2}{12} = \frac{25}{12} \] Difference: \[ \frac{729}{12} - \frac{25}{12} = \frac{704}{12} = \frac{176}{3} \] Multiply by \( 2\pi\rho \): \[ I_y = 2\pi\rho \cdot \frac{176}{3} = \frac{352}{3}\pi\rho \] **Final answer for (d):** \[ \boxed{I_y = \frac{352}{3}\pi\rho} \] --- ### **Summary Table** | Part | Answer | |------|-------------------------------| | (a) | \( 32\pi \) | | (b) | \( 32\pi \) | | (c) | \( \frac{8}{3} \) | | (d) | \( \frac{352}{3}\pi\rho \) | If you need any part explained in more detail, let me know!

Certainly! Here are the final answers: **(a)** \( \boxed{32\pi} \) **(b)** \( \boxed{32\pi} \) **(c)** \( \boxed{\dfrac{8}{3}} \) **(d)** \( \boxed{\dfrac{352}{3}\pi \rho} \)