Math Problem Statement

We are tasked with finding the surface area of the parameterized surface defined by: r(u, v) = 3u i - v j + v k, 0 <= u <= 5, 0 <= v <= 2.

Solution

We are tasked with finding the surface area of the parameterized surface defined by:

r(u,v)=3ui−vj+vk,0≤u≤5, 0≤v≤2.\mathbf{r}(u, v) = 3u \mathbf{i} - v \mathbf{j} + v \mathbf{k}, \quad 0 \leq u \leq 5, \, 0 \leq v \leq 2.

The surface area is calculated using the formula:

Surface Area=∫∫D∥ru×rv∥ dA,\text{Surface Area} = \int\int_D \|\mathbf{r}_u \times \mathbf{r}_v\| \, dA,

where ru\mathbf{r}_u and rv\mathbf{r}_v are the partial derivatives of r(u,v)\mathbf{r}(u, v) with respect to uu and vv, respectively, and ru×rv\mathbf{r}_u \times \mathbf{r}_v is their cross product. DD is the region defined by 0≤u≤50 \leq u \leq 5 and 0≤v≤20 \leq v \leq 2.

Step 1: Compute Partial Derivatives

ru=∂r∂u=∂∂u(3ui−vj+vk)=3i.\mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u} = \frac{\partial}{\partial u} (3u \mathbf{i} - v \mathbf{j} + v \mathbf{k}) = 3 \mathbf{i}.

rv=∂r∂v=∂∂v(3ui−vj+vk)=−j+k.\mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v} = \frac{\partial}{\partial v} (3u \mathbf{i} - v \mathbf{j} + v \mathbf{k}) = -\mathbf{j} + \mathbf{k}.

Step 2: Compute Cross Product

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & 0 \\ 0 & -1 & 1 \end{vmatrix}.$$ Expanding the determinant: $$\mathbf{r}_u \times \mathbf{r}_v = \mathbf{i} \begin{vmatrix} 0 & 0 \\ -1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & 0 \\ 0 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & 0 \\ 0 & -1 \end{vmatrix}.$$ $$\mathbf{r}_u \times \mathbf{r}_v = \mathbf{i} (0 - 0) - \mathbf{j} (3 \cdot 1 - 0) + \mathbf{k} (3 \cdot -1 - 0).$$ $$\mathbf{r}_u \times \mathbf{r}_v = -3 \mathbf{j} - 3 \mathbf{k}.$$ ### Step 3: Magnitude of the Cross Product $$\|\mathbf{r}_u \times \mathbf{r}_v\| = \sqrt{0^2 + (-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}.$$ ### Step 4: Set up the Double Integral The surface area becomes: $$\text{Surface Area} = \int_0^5 \int_0^2 \|\mathbf{r}_u \times \mathbf{r}_v\| \, dv \, du = \int_0^5 \int_0^2 3\sqrt{2} \, dv \, du.$$ ### Step 5: Evaluate the Integral First, integrate with respect to $$v$$: $$\int_0^2 3\sqrt{2} \, dv = 3\sqrt{2} \cdot v \Big|_0^2 = 3\sqrt{2} \cdot (2 - 0) = 6\sqrt{2}.$$ Now, integrate with respect to $$u$$: $$\int_0^5 6\sqrt{2} \, du = 6\sqrt{2} \cdot u \Big|_0^5 = 6\sqrt{2} \cdot (5 - 0) = 30\sqrt{2}.$$ ### Final Answer The surface area is: $$\boxed{30\sqrt{2}}$$ Let me know if you would like further clarification or additional details! ### 5 Related Questions 1. What is the geometric interpretation of $$\mathbf{r}_u \times \mathbf{r}_v$$ in surface area problems? 2. How does the parameterization affect the computation of surface area? 3. What other applications does the cross product have in 3D geometry? 4. Can we use polar coordinates for parameterizing surfaces instead of rectangular coordinates? 5. How does the Jacobian determinant relate to surface area calculation? ### Tip When solving surface area problems, double-check the cross product computation and ensure proper parameter limits for the integral to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Surface Area
Parameterization
Partial Derivatives
Cross Product
Double Integral

Formulas

Surface Area = ∫∫_D ||r_u × r_v|| dA
r_u = ∂r/∂u
r_v = ∂r/∂v
||r_u × r_v|| = √(i^2 + j^2 + k^2)

Theorems

Surface area of parameterized surface using cross product and double integral

Suitable Grade Level

Grades 11-12, University level

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