integrate xyi + (x^2 + y^2)j using stokes theoram for y=x^2-4 in xy plane from (2,0) to (4,12)

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**.

---

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!

integrate xyi + (x^2 + y^2)j using Stokes theorem for y=x^2-4 in xy plane from (2,0) to (4,12)

Learn how to apply Stokes' Theorem to solve the integral of xyi + (x^2 + y^2)j for the curve y = x^2 - 4 in the xy plane. Follow the step-by-step calculation of the curl, surface integral, and boundary curve for this advanced calculus problem. Suitable for university-level students studying vector calculus.

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