Let's solve the problem step by step: --- ### **a. Describe the curve in \(\mathbb{R}^3\) in which these two surfaces intersect. Where is this curve located?** The surfaces intersect where \[ z = 16 - x^2 - y^2 = 3x^2 + 3y^2 \] Setting the two expressions for \(z\) equal: \[ 16 - x^2 - y^2 = 3x^2 + 3y^2 \implies 16 = 4x^2 + 4y^2 \implies x^2 + y^2 = 4 \] The intersection is when \(x^2 + y^2 = 4\). Plug back into either equation to get \(z\): Using \(z = 3x^2 + 3y^2\): \[ z = 3(x^2 + y^2) = 3 \times 4 = 12 \] **So the curve is the circle \(x^2 + y^2 = 4\) at height \(z = 12\) in \(\mathbb{R}^3\).** --- ### **b. In terms of \(x\) and \(y\), write an equation describing the projection of this curve onto the \(xy\)-plane.** The projection onto the \(xy\)-plane ignores the \(z\)-coordinate, so: \[ x^2 + y^2 = 4 \] This is a circle of radius 2 centered at the origin. --- ### **c. What coordinate system do you think is most natural for an iterated integral that gives the volume of the solid?** Since the boundary is a circle (\(x^2 + y^2 = 4\)), **polar coordinates** are the most natural. --- ### **d. Set up, but do not evaluate, an integral expression whose value is the average \(z\)-value of points inside the solid region \(S\).** The average value of \(z\) over the solid is: \[ \text{Average } z = \frac{1}{\text{Volume}} \int\int\int_S z \, dV \] For this solid, the region in the \(xy\)-plane is the disk \(x^2 + y^2 \leq 4\), and for each \((x, y)\), \(z\) goes from \(z_\text{lower} = 3x^2 + 3y^2\) to \(z_\text{upper} = 16 - x^2 - y^2\). In **polar coordinates** (\(x = r\cos\theta, y = r\sin\theta\)), the setup: - \(0 \leq r \leq 2\) - \(0 \leq \theta \leq 2\pi\) - \(z\) goes from \(3r^2\) to \(16 - r^2\) **Volume:** \[ V = \int_0^{2\pi} \int_0^2 \left[ (16 - r^2) - (3r^2) \right] r\, dr\, d\theta = \int_0^{2\pi} \int_0^2 (16 - 4r^2) r\, dr\, d\theta \] **Average z:** \[ \text{Average } z = \frac{1}{V} \int_0^{2\pi} \int_0^2 \int_{z=3r^2}^{z=16 - r^2} z\, dz \, r\, dr\, d\theta \] --- ### **e. Use GeoGebra to plot the two surfaces. Locate the curve from part a. Using the GeoGebra graph, estimate the average value of \(z\) on the solid.** - In GeoGebra, input: - \(z_1 = 16 - x^2 - y^2\) - \(z_2 = 3x^2 + 3y^2\) - The intersection curve is the circle \(x^2 + y^2 = 4\) at \(z = 12\). - The solid lies between these two surfaces above the disk \(x^2 + y^2 \leq 4\). **Estimate:** Since the top surface ranges from \(z = 16\) at the center (\(r=0\)) down to \(z = 12\) at \(r=2\), and the bottom surface goes from \(z = 0\) at the center up to \(z = 12\) at \(r=2\), the average \(z\) will be somewhat above the halfway point between the two surfaces at each \((x, y)\). By inspection (or geometric intuition), the average \(z\) should be **slightly above \(z=8\)**, possibly in the range **\(z = 9\) to \(10\)**. --- ### **Summary of Final Answers** **a.** The intersection curve is \(x^2 + y^2 = 4\), \(z=12\) in \(\mathbb{R}^3\). **b.** The projection is \(x^2 + y^2 = 4\). **c.** Polar coordinates. **d.** \[ \text{Average } z = \frac{1}{V} \int_0^{2\pi} \int_0^2 \int_{z=3r^2}^{z=16 - r^2} z\, dz\, r\, dr\, d\theta \] where \(V = \int_0^{2\pi} \int_0^2 (16 - 4r^2) r\, dr\, d\theta\). **e.** From the graph, estimate average \(z \approx 9\) to \(10\). (Use GeoGebra to refine this estimate.)