The surface area of a parametric surface is calculated using the formula:

$A = \int\int_D \|\mathbf{r}_u \times \mathbf{r}_v\| \, dA,$ where $\mathbf{r}_u$ and $\mathbf{r}_v$ are the partial derivatives of the position vector $\mathbf{r}(u, v)$ with respect to $u$ and $v$, respectively, and $D$ is the domain of the parameters $u$ and $v$.

Parametric Equations

$\mathbf{r}(u, v) = \langle u^2, uv, \frac{1}{2}v^2 \rangle$

Step 1: Compute Partial Derivatives

1. Partial derivative with respect to $u$: $\mathbf{r}_u = \frac{\partial}{\partial u} \langle u^2, uv, \frac{1}{2}v^2 \rangle = \langle 2u, v, 0 \rangle$

2. Partial derivative with respect to $v$: $\mathbf{r}_v = \frac{\partial}{\partial v} \langle u^2, uv, \frac{1}{2}v^2 \rangle = \langle 0, u, v \rangle$

Step 2: Compute the Cross Product $\mathbf{r}_u \times \mathbf{r}_v$

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2u & v & 0 \\ 0 & u & v \end{vmatrix} = \mathbf{i} \begin{vmatrix} v & 0 \\ u & v \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2u & 0 \\ 0 & v \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2u & v \\ 0 & u \end{vmatrix}$$ Simplify each determinant: 1. For $$\mathbf{i}$$: $$\begin{vmatrix} v & 0 \\ u & v \end{vmatrix} = v \cdot v - 0 \cdot u = v^2$$ 2. For $$\mathbf{j}$$: $$\begin{vmatrix} 2u & 0 \\ 0 & v \end{vmatrix} = 2u \cdot v - 0 \cdot 0 = 2uv$$ 3. For $$\mathbf{k}$$: $$\begin{vmatrix} 2u & v \\ 0 & u \end{vmatrix} = 2u \cdot u - 0 \cdot v = 2u^2$$ Thus: $$\mathbf{r}_u \times \mathbf{r}_v = \langle v^2, -2uv, 2u^2 \rangle$$ ### Step 3: Compute the Magnitude $$\|\mathbf{r}_u \times \mathbf{r}_v\| = \sqrt{(v^2)^2 + (-2uv)^2 + (2u^2)^2}$$ $$= \sqrt{v^4 + 4u^2v^2 + 4u^4}$$ Factorize: $$\|\mathbf{r}_u \times \mathbf{r}_v\| = \sqrt{(v^2 + 2u^2)^2} = v^2 + 2u^2$$ ### Step 4: Set Up the Integral $$A = \int_0^3 \int_0^4 (v^2 + 2u^2) \, du \, dv$$ ### Step 5: Compute the Integral 1. Inner integral with respect to $$u$$: $$\int_0^4 (v^2 + 2u^2) \, du = \int_0^4 v^2 \, du + \int_0^4 2u^2 \, du$$ $$= v^2 \int_0^4 1 \, du + 2 \int_0^4 u^2 \, du$$ \[ = v^2 [u]_0^4 + 2 \left[\frac{u^3}{3}\right]_0^4 \] $$= v^2 (4 - 0) + 2 \cdot \frac{4^3}{3}$$ $$= 4v^2 + \frac{128}{3}$$ 2. Outer integral with respect to $$v$$: $$A = \int_0^3 \left(4v^2 + \frac{128}{3}\right) \, dv$$ $$= \int_0^3 4v^2 \, dv + \int_0^3 \frac{128}{3} \, dv$$ $$= 4 \int_0^3 v^2 \, dv + \frac{128}{3} \int_0^3 1 \, dv$$ \[ = 4 \left[\frac{v^3}{3}\right]_0^3 + \frac{128}{3} [v]_0^3 \] $$= 4 \cdot \frac{3^3}{3} + \frac{128}{3} \cdot 3$$ $$= 4 \cdot 9 + 128$$ $$= 36 + 128 = 164$$ ### Final Answer: The area of the surface is: $$\boxed{164 \text{ square units}}$$ Would you like to discuss any step in more detail? ### Related Questions: 1. What is the formula for the area of parametric surfaces in general? 2. How do you calculate cross products in vector calculus? 3. How does factorization simplify the magnitude calculation? 4. What are other methods to calculate surface area besides parametric representation? 5. How can this process be applied to cylindrical or spherical coordinates? ### Tip: Always verify your parameter domain boundaries carefully to avoid integration errors.