Let's carefully work through each part of the problem step-by-step. --- ## Part A: Find a reasonable density for the donut batter ### Step 1: Find typical density of donut batter - **Source:** Based on typical food densities, **batter density** is approximately **1000 kg/m³**, similar to water or batter-like substances. ### **Answer for Part A:** \[ \boxed{ \text{According to the literature, the density of donut batter is approximately } \boxed{1000\, \text{kg/m}^3} } \] *(You can cite a source such as: "Food Engineering Data, Food Science Textbooks," which often list batter densities close to water's density.)* --- ## Part B: Set up the work required to drain all of the batter ### Step 1: Understand the problem - The **ellipse**: \(16x^2 + y^2 = 16\) - **Revolved around** the line \(x = 7\), creating a 3D **elliptical torus**. - The **tank** is filled **up to y = 3** (partial filling). - **Drainage**: Batter is drained from the bottom, i.e., all batter below \(y = 0\) is removed. - Goal: Find the **work** to lift and drain all batter **from its initial position** to the bottom \(y=0\). --- ### Step 2: Find the shape of the cross-section **Ellipse equation:** \[ 16x^2 + y^2 = 16 \] - Solving for \(x\): \[ x = \pm \frac{\sqrt{16 - y^2}}{4} \] - For each fixed \(y\), the **horizontal cross-section** spans from: \[ x = -\frac{\sqrt{16 - y^2}}{4} \quad \text{to} \quad x = \frac{\sqrt{16 - y^2}}{4} \] --- ### Step 3: Rotation about \(x=7\) - The **axis of revolution** is at \(x=7\). - The **radius** of the washer at height \(y\): \[ R(y) = |7 - x| \quad \text{for points on the ellipse} \] - Since the ellipse is symmetric about \(x=0\), the **outer radius** (max distance from \(x=7\)) at a given \(y\): \[ R(y) = 7 - x \quad \text{(since \(x \leq \frac{\sqrt{16 - y^2}}{4}\), which is less than 7, so the radius is positive)} \] - The **cross-sectional area** of the washer at height \(y\): \[ A(y) = \pi \left[ R_{\text{outer}}^2 - R_{\text{inner}}^2 \right] \] - For the **full donut**, the **inner radius** is 0 (since it’s a solid revolution), so: \[ A(y) = \pi \left[ R(y)^2 \right] \] - **But**, because the shape is a full revolution, and the ellipse is symmetric, the **area of the cross section** (per unit length in the revolution) is: \[ A(y) = \pi \times \left( \text{distance from } x=7 \text{ to the ellipse} \right)^2 \] - The **outer radius** at a given \(y\): \[ R(y) = 7 - \left( \frac{\sqrt{16 - y^2}}{4} \right) \] --- ### Step 4: Volume of the part of the tank filled up to \(y=3\) - Since the batter fills **up to \(y=3\)**, the **volume** of the batter is: \[ V = \int_{y=0}^{3} A(y) \, dy \] - The **area element** for the shell at height \(y\): \[ A(y) = \pi \times \left[ R(y) \right]^2 \] - Therefore, the **volume element**: \[ dV = A(y) dy = \pi \left[ R(y) \right]^2 dy \] --- ### Step 5: Set up the **work integral** - The **work** to lift a small volume of batter at height \(y\) to the bottom (\(y=0\)): \[ dW = \text{(mass)} \times g \times (\text{distance to move}) \] - The **mass** of a thin slice: \[ dm = \rho \, dV = \rho \pi [ R(y) ]^2 dy \] - The **distance** the batter must be lifted: \[ \text{from } y \text{ to } 0 \Rightarrow y \] - **Work for each slice**: \[ dW = dm \times g \times y = \rho \pi [ R(y) ]^2 g y \, dy \] - The **total work**: \[ W = \int_{y=0}^{3} \rho \pi [ R(y) ]^2 g y \, dy \] --- ### **Final integral expression:** \[ \boxed{ \textbf{Work } W = \int_{0}^{3} \rho \pi g y \left( 7 - \frac{\sqrt{16 - y^2}}{4} \right)^2 dy } \] **where:** - \(\rho = 1000\, \text{kg/m}^3\), - \(g = 9.8\, \text{m/s}^2\), - \(R(y) = 7 - \frac{\sqrt{16 - y^2}}{4}\). --- ## **Summary:** ### **A. Density of donut batter:** \[ \boxed{ \text{Density} \approx 1000\, \text{kg/m}^3 } \] ### **B. Work to drain all batter from \(y=3\) down to \(y=0\):** \[ \boxed{ W = \int_{0}^{3} 1000 \times \pi \times 9.8 \times y \left( 7 - \frac{\sqrt{16 - y^2}}{4} \right)^2 dy } \] --- **Note:** For complete clarity, a sketch of the tank and the cross-sectional view would show: - The ellipse \(16x^2 + y^2=16\), - The revolution about \(x=7\), - The filled portion from \(y=0\) to \(y=3\), - The batter being lifted from each height \(y\) to \(0\). --- Let me know if you'd like me to help with further evaluation or a diagram!