To solve this problem using Stokes' Theorem, we need to proceed with the following steps:

1. Understanding Stokes' Theorem:

Stokes' Theorem relates a surface integral of a curl over a surface $S$ to a line integral over the boundary curve $C$. Mathematically, it is written as:

$\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$

Where:

• $\mathbf{F}$ is the vector field (in our case $\mathbf{F} = xyi + (x^2 + y^2)j$),
• $C$ is the boundary curve of the surface $S$,
• $d\mathbf{r}$ is the differential along the curve $C$,
• $\nabla \times \mathbf{F}$ is the curl of the vector field $\mathbf{F}$,
• $d\mathbf{S}$ is the differential area element of the surface $S$.

In this problem, $C$ is a curve along the parabola $y = x^2 - 4$ in the xy-plane, going from $(2, 0)$ to $(4, 12)$.

2. Calculate the Curl of $\mathbf{F}$:

The vector field given is $\mathbf{F} = xyi + (x^2 + y^2)j$. Let's compute its curl:

$\nabla \times \mathbf{F} = \left( \frac{\partial F_j}{\partial x} - \frac{\partial F_i}{\partial y} \right) \hat{k}$

Where:

• $F_i = xy$,
• $F_j = x^2 + y^2$.

First, calculate the partial derivatives: $\frac{\partial F_j}{\partial x} = \frac{\partial (x^2 + y^2)}{\partial x} = 2x$ $\frac{\partial F_i}{\partial y} = \frac{\partial (xy)}{\partial y} = x$

Thus, the curl is: $\nabla \times \mathbf{F} = (2x - x) \hat{k} = x \hat{k}$

3. Surface Integral over $S$:

The surface $S$ lies in the xy-plane, so the differential surface element is $d\mathbf{S} = \hat{k} \, dx \, dy$. Therefore, the surface integral becomes:

$\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_S x \, dx \, dy$

Now we need to find the bounds of integration.

4. Set the Bounds of Integration:

The curve $C$ is along the parabola $y = x^2 - 4$, which gives us the bounds for $y$ in terms of $x$. The limits for $x$ are from $2$ to $4$ (from the points $(2, 0)$ to $(4, 12)$).

Thus, the double integral becomes:

$\iint_S x \, dx \, dy = \int_2^4 \int_{x^2 - 4}^{12} x \, dy \, dx$

5. Evaluate the Double Integral:

First, perform the integration with respect to $y$:

$\int_{x^2 - 4}^{12} x \, dy = x \left( 12 - (x^2 - 4) \right) = x \left( 16 - x^2 \right)$

Now, integrate with respect to $x$:

$\int_2^4 x(16 - x^2) \, dx = \int_2^4 (16x - x^3) \, dx$

Evaluating the integral:

$\int_2^4 16x \, dx = 8x^2 \Big|_2^4 = 8(16) - 8(4) = 128 - 32 = 96$

$\int_2^4 x^3 \, dx = \frac{x^4}{4} \Big|_2^4 = \frac{256}{4} - \frac{16}{4} = 64 - 4 = 60$

Thus, the total integral is:

$96 - 60 = 36$

Final Answer:

The value of the line integral using Stokes' Theorem is 36.

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Do you want more details on any step or have any questions?

Here are 5 related questions you might find interesting:

1. How is the curl of a vector field computed?
2. Can Stokes' theorem be applied in 3D?
3. How would this problem change if the curve $C$ was closed?
4. How is the surface integral related to the line integral in vector calculus?
5. What are some applications of Stokes' Theorem in physics?

Tip: In vector calculus, always check the orientation of your surface and curve when applying Stokes' theorem!