Let's solve **Problem 1** step by step: ### 1) Shell Method, Region bounded by \( x = \frac{y^2}{2} \), \( x = 0 \), \( y = 2 \), revolved about the **x-axis**. --- #### **Step 1: Sketch and Setup** - The curve: \( x = \frac{y^2}{2} \) - Vertical line: \( x = 0 \) - Upper bound: \( y = 2 \) - Lower bound: Since \( x = 0 \Rightarrow y = 0 \) **Region**: From \( y = 0 \) to \( y = 2 \), between \( x = 0 \) and \( x = \frac{y^2}{2} \). --- #### **Step 2: Shell Method** When rotating around the **x-axis**, use **horizontal shells** (integrate with respect to \( y \)): - **Radius** (distance from shell to x-axis): \( y \) - **Height** (horizontal distance): \( \frac{y^2}{2} - 0 = \frac{y^2}{2} \) **Volume Element**: \( dV = 2\pi (\text{radius})(\text{height})\,dy = 2\pi y \cdot \frac{y^2}{2} dy = \pi y^3 dy \) --- #### **Step 3: Set up the Integral** Integrate from \( y = 0 \) to \( y = 2 \): \[ V = \int_{0}^{2} \pi y^3 \, dy \] --- #### **Step 4: Compute the Integral** \[ V = \pi \int_{0}^{2} y^3 \, dy = \pi \left[ \frac{y^4}{4} \right]_{0}^{2} = \pi \left( \frac{2^4}{4} - \frac{0^4}{4} \right) = \pi \left( \frac{16}{4} \right) = 4\pi \] --- ### **Final Answer** \[ \boxed{4\pi} \] --- Would you like to move to the next problem?

Let's proceed with **Problem 2**: --- ### 2) Volume of the solid generated by revolving the region bounded by \( y = (x - 3)^2 - 1 \), \( x=0 \), \( y=7 \) about the **y-axis**. --- #### **Step 1: Understand the region** - The curve: \( y = (x - 3)^2 - 1 \) - \( x=0 \), \( y=7 \) - The region is bounded above by \( y=7 \), and below by the parabola \( y = (x-3)^2 - 1 \). Find the intersection points of the parabola and \( y=7 \): \[ 7 = (x - 3)^2 - 1 \Rightarrow (x - 3)^2 = 8 \Rightarrow x - 3 = \pm \sqrt{8} = \pm 2\sqrt{2} \] Thus: \[ x = 3 \pm 2 \sqrt{2} \] The region in the \( xy \)-plane between \( x = 3 - 2\sqrt{2} \) and \( x=3 + 2\sqrt{2} \), from the parabola up to \( y=7 \). --- ### **Method 1: Disk/Washer Method (about y-axis)** - Since revolving around the y-axis, use **vertical slices** (integrate with respect to \( x \)). - The **radius** of a typical shell at \( x \): \( x \) - The **height** of the shell: \( y_{\text{top}} - y_{\text{bottom}} = 7 - [(x-3)^2 - 1] = 8 - (x-3)^2 \) - Shell volume element: \[ dV = 2\pi (\text{radius})(\text{height})\, dx = 2\pi x [8 - (x-3)^2] dx \] - Integration bounds: \( x \) from \( 3 - 2 \sqrt{2} \) to \( 3 + 2 \sqrt{2} \). --- ### **Step 2: Set up the integral** \[ V = \int_{x=3 - 2\sqrt{2}}^{3 + 2\sqrt{2}} 2\pi x [8 - (x - 3)^2] dx \] --- ### **Step 3: Simplify the integrand** Let \( u = x - 3 \), then: - When \( x = 3 - 2\sqrt{2} \), \( u = -2 \sqrt{2} \) - When \( x = 3 + 2 \sqrt{2} \), \( u = 2 \sqrt{2} \) And: \[ x = u + 3 \] Express the integrand: \[ 2 \pi (u + 3) [8 - u^2] \] The integral becomes: \[ V = 2 \pi \int_{-2 \sqrt{2}}^{2 \sqrt{2}} (u + 3)(8 - u^2) du \] --- ### **Step 4: Expand the integrand** \[ (u + 3)(8 - u^2) = u \cdot 8 - u \cdot u^2 + 3 \cdot 8 - 3 u^2 = 8 u - u^3 + 24 - 3 u^2 \] Thus: \[ V = 2 \pi \int_{-2 \sqrt{2}}^{2 \sqrt{2}} (8 u - u^3 + 24 - 3 u^2) du \] --- ### **Step 5: Integrate term-by-term** Integral: \[ V = 2 \pi \left[ \int 8 u du - \int u^3 du + \int 24 du - \int 3 u^2 du \right]_{-2 \sqrt{2}}^{2 \sqrt{2}} \] Compute each: - \( \int 8 u du = 4 u^2 \) - \( \int u^3 du = \frac{u^4}{4} \) - \( \int 24 du = 24 u \) - \( \int 3 u^2 du = u^3 \) Putting it all together: \[ V = 2 \pi \left[ 4 u^2 - \frac{u^4}{4} + 24 u - u^3 \right]_{-2 \sqrt{2}}^{2 \sqrt{2}} \] --- ### **Step 6: Evaluate at bounds** At \( u = 2\sqrt{2} \): \[ 4 (2\sqrt{2})^2 - \frac{(2\sqrt{2})^4}{4} + 24 (2\sqrt{2}) - (2\sqrt{2})^3 \] Calculate each: - \( (2 \sqrt{2})^2 = 4 \times 2 = 8 \) - \( (2 \sqrt{2})^4 = ( (2 \sqrt{2})^2 )^2 = 8^2 = 64 \) - \( (2 \sqrt{2})^3 = (2 \sqrt{2}) \times (2 \sqrt{2})^2 = 2 \sqrt{2} \times 8 = 16 \sqrt{2} \) Now substitute: \[ 4 \times 8 - \frac{64}{4} + 24 \times 2 \sqrt{2} - 16 \sqrt{2} = 32 - 16 + 48 \sqrt{2} - 16 \sqrt{2} \] Simplify: \[ (32 - 16) + (48 \sqrt{2} - 16 \sqrt{2}) = 16 + 32 \sqrt{2} \] At \( u = - 2 \sqrt{2} \): - \( u^2 = 8 \) - \( u^4 = 64 \) - \( u^3 = -16 \sqrt{2} \) Substitute: \[ 4 \times 8 - \frac{64}{4} + 24 \times (-2 \sqrt{2}) - (-16 \sqrt{2}) = 32 - 16 - 48 \sqrt{2} + 16 \sqrt{2} \] Simplify: \[ (32 - 16) + (-48 \sqrt{2} + 16 \sqrt{2}) = 16 - 32 \sqrt{2} \] --- ### **Step 7: Compute the definite integral** Subtract: \[ \left[ 16 + 32 \sqrt{2} \right] - \left[ 16 - 32 \sqrt{2} \right] = (16 - 16) + (32 \sqrt{2} + 32 \sqrt{2}) = 0 + 64 \sqrt{2} \] Multiply by \( 2 \pi \): \[ V = 2 \pi \times 64 \sqrt{2} = 128 \pi \sqrt{2} \] --- ### **Final answer:** \[ \boxed{128 \pi \sqrt{2}} \] --- **Summary:** - **Method used:** Shell method about the y-axis - **Region bounds:** \( x = 3 \pm 2 \sqrt{2} \) - **Volume:** \(\boxed{128 \pi \sqrt{2}}\) Would you like to proceed with the next problem?