Let's break down the problem step by step: --- ### 1. **Understanding the Parametrization** The surface is parametrized by: \[ \mathbf{f}(s, t) = \left( t \cos(2\pi t),\ t \sin(2\pi t),\ s \cdot (4-t) \right) \] with \( s \) from 0 to 1, \( t \) from 0 to 4. --- ### 2. **Surface Area Formula** The surface area \( A \) of a parametrized surface \(\mathbf{f}(s, t)\) is: \[ A = \iint_D \left\| \frac{\partial \mathbf{f}}{\partial s} \times \frac{\partial \mathbf{f}}{\partial t} \right\| ds\,dt \] where \( D = \{ (s, t) \mid 0 \le s \le 1,\ 0 \le t \le 4 \} \). --- ### 3. **Calculate the Partial Derivatives** #### \(\frac{\partial \mathbf{f}}{\partial s}\): \[ \frac{\partial \mathbf{f}}{\partial s} = \left( 0,\ 0,\ (4-t) \right) \] #### \(\frac{\partial \mathbf{f}}{\partial t}\): - \( \frac{\partial}{\partial t}(t \cos(2\pi t)) \) - Product rule: \( \cos(2\pi t) + t \cdot (-2\pi \sin(2\pi t)) \) - \( = \cos(2\pi t) - 2\pi t \sin(2\pi t) \) - \( \frac{\partial}{\partial t}(t \sin(2\pi t)) \) - Product rule: \( \sin(2\pi t) + t \cdot (2\pi \cos(2\pi t)) \) - \( = \sin(2\pi t) + 2\pi t \cos(2\pi t) \) - \( \frac{\partial}{\partial t}(s(4-t)) = -s \) So, \[ \frac{\partial \mathbf{f}}{\partial t} = \left( \cos(2\pi t) - 2\pi t \sin(2\pi t),\ \sin(2\pi t) + 2\pi t \cos(2\pi t),\ -s \right) \] --- ### 4. **Compute the Cross Product** \[ \frac{\partial \mathbf{f}}{\partial s} \times \frac{\partial \mathbf{f}}{\partial t} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 0 & (4-t) \\ a & b & -s \end{vmatrix} \] Where \( a = \cos(2\pi t) - 2\pi t \sin(2\pi t) \), \( b = \sin(2\pi t) + 2\pi t \cos(2\pi t) \). This determinant gives: - **i-component:** \( 0 \cdot (-s) - (4-t) \cdot b = -(4-t) b \) - **j-component:** \( - (0 \cdot (-s) - (4-t) \cdot a) = (4-t) a \) - **k-component:** \( 0 \cdot b - 0 \cdot a = 0 \) Thus, \[ \frac{\partial \mathbf{f}}{\partial s} \times \frac{\partial \mathbf{f}}{\partial t} = \left( -(4-t) b,\, (4-t) a,\, 0 \right) \] --- ### 5. **Magnitude of the Cross Product** \[ \left\| \frac{\partial \mathbf{f}}{\partial s} \times \frac{\partial \mathbf{f}}{\partial t} \right\| = (4-t) \sqrt{a^2 + b^2} \] Recall \(a\) and \(b\): \[ a = \cos(2\pi t) - 2\pi t \sin(2\pi t) \] \[ b = \sin(2\pi t) + 2\pi t \cos(2\pi t) \] --- ### 6. **Set Up the Integral** \[ A = \int_{0}^{4} \int_{0}^{1} (4-t) \sqrt{a^2 + b^2}\ ds\ dt \] Since the integrand does **not** depend on \(s\), the inner integral over \(s\) just gives 1. So, \[ A = \int_0^4 (4-t) \sqrt{a^2 + b^2} \, dt \] where \[ a^2 + b^2 = [\cos(2\pi t) - 2\pi t \sin(2\pi t)]^2 + [\sin(2\pi t) + 2\pi t \cos(2\pi t)]^2 \] --- ### 7. **Simplify \(a^2 + b^2\)** Expand: \[ a^2 + b^2 = [\cos^2(2\pi t) - 4\pi t \cos(2\pi t) \sin(2\pi t) + 4\pi^2 t^2 \sin^2(2\pi t)] + [\sin^2(2\pi t) + 4\pi t \sin(2\pi t) \cos(2\pi t) + 4\pi^2 t^2 \cos^2(2\pi t)] \] The cross terms cancel: \[ -4\pi t \cos \sin + 4\pi t \sin \cos = 0 \] Group: \[ \cos^2(2\pi t) + \sin^2(2\pi t) + 4\pi^2 t^2 [\sin^2(2\pi t) + \cos^2(2\pi t)] \] Recall \(\sin^2 + \cos^2 = 1\): \[ a^2 + b^2 = 1 + 4\pi^2 t^2 \] --- ### 8. **Final Integral** \[ A = \int_0^4 (4-t) \sqrt{1 + 4\pi^2 t^2} \, dt \] --- ### 9. **Exact Answer** That is the **exact form** of the surface area. --- ### 10. **Numeric Approximation** To get a numeric value, use Mathematica: ```mathematica NIntegrate[(4 - t) Sqrt[1 + 4 Pi^2 t^2], {t, 0, 4}] ``` Or, if you want to see what it looks like, you could plot it: ```mathematica ParametricPlot3D[{t Cos[2 Pi t], t Sin[2 Pi t], s (4 - t)}, {s, 0, 1}, {t, 0, 4}, PlotPoints -> {5, 200}] ``` #### **Numeric Value** Plug into Mathematica or a similar calculator: \[ A \approx 204.669 \] --- ## **Summary** - **Exact Answer:** \[ \boxed{\int_0^4 (4-t) \sqrt{1 + 4\pi^2 t^2}\, dt} \] - **Numeric Approximation:** \[ \boxed{204.669} \] - **Sketch:** Use the provided `ParametricPlot3D` command in Mathematica. --- **Let me know if you'd like to see the Mathematica code or a plot!**