Evaluate the surface integral $\iint_{S} \mathbf{F} \cdot d\mathbf{S}$ for the given vector field $\mathbf{F}$ and the oriented surface $S$. In other words, find the flux of $\mathbf{F}$ across $S$. For closed surfaces, use the positive (outward) orientation. $\mathbf{F}(x, y, z) = xy \mathbf{i} + 12x^2 \mathbf{j} + yz \mathbf{k}$ $S$ is the surface $z = xe^y$, $0 \leq x \leq 1$, $0 \leq y \leq 4$, with upward orientation.

Neetesh Kumar | November 13, 2024

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This is the solution to Math 1D
Assignment: 16.7 Question Number 14
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Step-by-step solution:

To evaluate the flux of the vector field $\mathbf{F}$ across the surface $S$, we use the surface integral $\iint_{S} \mathbf{F} \cdot d\mathbf{S}$.

Since $S$ is not a closed surface but has an upward orientation, we’ll calculate $d\mathbf{S}$ in terms of the surface parameters.

Step 1: Parametrize the Surface

The surface $S$ is given by $z = xe^y$. We can parametrize it as:

$\mathbf{r}(x, y) = x \mathbf{i} + y \mathbf{j} + xe^y \mathbf{k}$

where $0 \leq x \leq 1$ and $0 \leq y \leq 4$.

Step 2: Calculate $d\mathbf{S}$

To find $d\mathbf{S}$, we need $\mathbf{r}_x \times \mathbf{r}_y \, dx \, dy$.

1. Compute $\mathbf{r}_x = \frac{\partial \mathbf{r}}{\partial x} = \mathbf{i} + e^y \mathbf{k}$
2. Compute $\mathbf{r}_y = \frac{\partial \mathbf{r}}{\partial y} = \mathbf{j} + x e^y \mathbf{k}$

Now, find the cross product $\mathbf{r}_x \times \mathbf{r}_y$:

$\mathbf{r}_x \times \mathbf{r}_y = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & e^y \\ 0 & 1 & x e^y \end{vmatrix}$

Expanding this determinant:

$\mathbf{r}_x \times \mathbf{r}_y = (0 - e^y) \mathbf{i} - (1 \cdot x e^y - 0) \mathbf{j} + (1 \cdot 1 - 0) \mathbf{k}$

Simplifying, we get:

$\mathbf{r}_x \times \mathbf{r}_y = -e^y \mathbf{i} - x e^y \mathbf{j} + \mathbf{k}$

Thus, $d\mathbf{S} = \mathbf{r}_x \times \mathbf{r}_y \, dx \, dy = (-e^y \mathbf{i} - x e^y \mathbf{j} + \mathbf{k}) \, dx \, dy$

Step 3: Compute $\mathbf{F} \cdot d\mathbf{S}$

The vector field $\mathbf{F}(x, y, z) = xy \mathbf{i} + 12x^2 \mathbf{j} + yz \mathbf{k}$. On the surface $S$, we substitute $z = xe^y$, so $\mathbf{F}$ on $S$ is:

$\mathbf{F}(x, y, xe^y) = xy \mathbf{i} + 12x^2 \mathbf{j} + yxe^y \mathbf{k}$

Now, calculate the dot product $\mathbf{F} \cdot d\mathbf{S}$:

$\mathbf{F} \cdot d\mathbf{S} = (xy \mathbf{i} + 12x^2 \mathbf{j} + yxe^y \mathbf{k}) \cdot (-e^y \mathbf{i} - x e^y \mathbf{j} + \mathbf{k})$

Expanding this dot product:

$\mathbf{F} \cdot d\mathbf{S} = xy(-e^y) + 12x^2(-x e^y) + yxe^y(1)$

Simplify each term:

1. $xy(-e^y) = -xye^y$
2. $12x^2(-x e^y) = -12x^3 e^y$
3. $yxe^y(1) = yxe^y$

Combining these, we get:

$\mathbf{F} \cdot d\mathbf{S} = -xye^y - 12x^3 e^y + yxe^y$

Simplify further:

$\mathbf{F} \cdot d\mathbf{S} = (-xye^y + yxe^y) - 12x^3 e^y$

$\mathbf{F} \cdot d\mathbf{S} = -12x^3 e^y$

Step 4: Set Up and Evaluate the Integral

Now we integrate over the region $0 \leq x \leq 1$ and $0 \leq y \leq 4$:

$\iint_{S} \mathbf{F} \cdot d\mathbf{S} = \int_{0}^{1} \int_{0}^{4} -12x^3 e^y \, dy \, dx$

Separate the integrals:

$= -12 \int_{0}^{1} x^3 \, dx \int_{0}^{4} e^y \, dy$

Evaluate each integral separately.

1. $\int_{0}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{4}$
2. $\int_{0}^{4} e^y \, dy = \left[ e^y \right]_{0}^{4} = e^4 - 1$

Combine results:

$\iint_{S} \mathbf{F} \cdot d\mathbf{S} = -12 \cdot \frac{1}{4} \cdot (e^4 - 1)$

$\iint_{S} \mathbf{F} \cdot d\mathbf{S} = 3(1 - e^4)$

Final Answer

The flux of $\mathbf{F}$ across $S$ is $\boxed{3(1 - e^4)}$.

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