Calculating Integral J With Stokes' Theorem Step-by-Step Guide

Hey guys! Ever get stuck on a tricky integral and feel like you're going in circles? Today, we're diving deep into how to solve the integral $J=\int_{C} y^{2}z{2}dx + x^{2}z{2}dy + x^{2}y{2}dz$ using the powerful Stokes' Theorem. If you're wrestling with line integrals, surface integrals, and vector calculus, you're in the right place. Let’s break this down step by step and make sure you've got a solid understanding. Our mission is to transform complex calculations into a walk in the park!

Understanding the Problem

Before we jump into the solution, let’s make sure we're all on the same page. The integral we're tackling is a line integral, specifically $J=\int_{C} y^{2}z{2}dx + x^{2}z{2}dy + x^{2}y{2}dz$. This integral is defined along a curve C, which in this case is defined parametrically. We have:

$$C:\begin{cases}x=a\cos(t)\\y=b\sin(t)\\z=c\end{cases}$$

where $0 \leq t \leq 2\pi$. Notice that the curve C is an ellipse in the xy-plane at a constant height z = c. This is a crucial piece of information, and understanding the geometry of the curve will help us visualize the problem better.

Now, why are we using Stokes’ Theorem? Well, Stokes' Theorem provides a neat way to convert a line integral over a closed curve into a surface integral over a surface bounded by that curve. This can often simplify the calculation, especially when the line integral is complex. In essence, Stokes' Theorem states:

$$\oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{S} (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$$

Here, $\mathbf{F}$ is a vector field, C is the closed curve, S is the surface bounded by C, and $\nabla \times \mathbf{F}$ is the curl of the vector field. This theorem is a cornerstone in vector calculus and a powerful tool for solving problems like this one. The beauty of Stokes’ Theorem lies in its ability to transform a potentially difficult line integral into a surface integral, often making the problem more manageable. So, let’s see how we can apply this to our specific integral.

Breaking Down the Components

To apply Stokes' Theorem effectively, let’s first identify the components of our problem. The vector field $\mathbf{F}$ can be extracted from the integral as:

$$\mathbf{F} = (y^{2}z^{2}, x^{2}z^{2}, x^{2}y^{2})$$

This vector field is what we will be working with, and its curl will play a significant role in our calculations. We also need to define the surface S bounded by the curve C. Since C is an ellipse in the xy-plane at z = c, a natural choice for S is the elliptical disk in the plane z = c with the same boundary. This simplifies our surface integral significantly.

Now, let's think about the orientation. The curve C is traversed counterclockwise when viewed from above (positive z-direction), so we'll choose the upward-pointing normal vector for the surface S. This ensures that the orientations of C and S are consistent, which is a requirement for Stokes' Theorem to hold. The normal vector $\mathbf{n}$ to the surface S (which lies in the plane z = c) is simply $\mathbf{k} = (0, 0, 1)$. This choice of normal vector aligns with the counterclockwise direction of the curve C, making our calculations smoother and more intuitive.

Applying Stokes' Theorem

Okay, guys, now for the exciting part – actually using Stokes' Theorem! We've got our vector field $\mathbf{F}$ and our surface S. The first step is to compute the curl of $\mathbf{F}$, which is given by:

$$\nabla \times \mathbf{F} = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right) \times (y^{2}z^{2}, x^{2}z^{2}, x^{2}y^{2})$$

This might look a bit intimidating, but let's break it down. The curl is a vector operation that measures the rotation of a vector field. Calculating the curl involves taking partial derivatives, so let’s go through it carefully:

$$\nabla \times \mathbf{F} = \left(\frac{\partial}{\partial y}(x^{2}y^{2}) - \frac{\partial}{\partial z}(x^{2}z^{2}), \frac{\partial}{\partial z}(y^{2}z^{2}) - \frac{\partial}{\partial x}(x^{2}y^{2}), \frac{\partial}{\partial x}(x^{2}z^{2}) - \frac{\partial}{\partial y}(y^{2}z^{2})\right)$$

Taking the partial derivatives, we get:

$$\nabla \times \mathbf{F} = (2x^{2}y - 2y^{2}z, 2y^{2}z - 2x^{2}y, 2xz^{2} - 2yz^{2})$$

Now that we have the curl, we need to compute the surface integral of the curl over the surface S:

$$\iint_{S} (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_{S} (\nabla \times \mathbf{F}) \cdot \mathbf{n} dS$$

Since our surface S lies in the plane z = c, the normal vector $\mathbf{n}$ is $\mathbf{k} = (0, 0, 1)$. Therefore, the dot product simplifies to:

$$(\nabla \times \mathbf{F}) \cdot \mathbf{n} = (2x^{2}y - 2y^{2}z, 2y^{2}z - 2x^{2}y, 2xz^{2} - 2yz^{2}) \cdot (0, 0, 1) = 2xz^{2} - 2yz^{2}$$

So, our surface integral becomes:

$$\iint_{S} (2xz^{2} - 2yz^{2}) dS$$

Setting Up the Surface Integral

Alright, let's get this surface integral rolling! We're integrating over the elliptical disk in the plane z = c. To make our lives easier, we can parameterize the surface S using the same parameters as the curve C, but now treating them as parameters for a surface:

$$\mathbf{r}(x, y) = (x, y, c)$$

where (x, y) lie within the ellipse $\frac{x^{2}}{a{2}} + \frac{y^{2}}{b{2}} \leq 1$. This parameterization is super handy because it directly connects back to the original curve C, making the transition from the line integral to the surface integral much smoother.

However, a more convenient parameterization involves using elliptical coordinates:

$$x = a r \cos(\theta)$$

$$y = b r \sin(\theta)$$

$$z = c$$

where $0 \leq r \leq 1$ and $0 \leq \theta \leq 2\pi$. This transformation simplifies the integral because it aligns with the elliptical geometry of our surface. The Jacobian for this transformation is:

$$J = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} a \cos(\theta) & -a r \sin(\theta) \\ b \sin(\theta) & b r \cos(\theta) \end{vmatrix} = abr$$

Now, we substitute the parameterization into our integrand:

$$2xz^{2} - 2yz^{2} = 2(a r \cos(\theta))c^{2} - 2(b r \sin(\theta))c^{2} = 2c^{2}r(a \cos(\theta) - b \sin(\theta))$$

So, the integral becomes:

$$\iint_{S} (2xz^{2} - 2yz^{2}) dS = \int_{0}^{2\pi} \int_{0}^{1} 2c^{2}r(a \cos(\theta) - b \sin(\theta)) |J| dr d\theta$$

Evaluating the Integral

Time to roll up our sleeves and evaluate this double integral! We have:

$$\int_{0}^{2\pi} \int_{0}^{1} 2c^{2}r(a \cos(\theta) - b \sin(\theta)) abr dr d\theta = 2abc^{2} \int_{0}^{2\pi} \int_{0}^{1} r^{2}(a \cos(\theta) - b \sin(\theta)) dr d\theta$$

First, let’s handle the inner integral with respect to r:

$$\int_{0}^{1} r^{2} dr = \left[\frac{1}{3}r^{3}\right]_{0}^{1} = \frac{1}{3}$$

Now, our integral simplifies to:

$$2abc^{2} \int_{0}^{2\pi} \frac{1}{3}(a \cos(\theta) - b \sin(\theta)) d\theta = \frac{2abc^{2}}{3} \int_{0}^{2\pi} (a \cos(\theta) - b \sin(\theta)) d\theta$$

Next, we integrate with respect to $\theta$:

$$\int_{0}^{2\pi} (a \cos(\theta) - b \sin(\theta)) d\theta = \left[a \sin(\theta) + b \cos(\theta)\right]_{0}^{2\pi} = (a \sin(2\pi) + b \cos(2\pi)) - (a \sin(0) + b \cos(0)) = (0 + b) - (0 + b) = 0$$

So, the final result is:

$$\frac{2abc^{2}}{3} \cdot 0 = 0$$

Conclusion: The Final Answer

Drumroll, please! After all that work, we've found that the integral J is equal to 0. Yes, you heard it right – zero! This result might seem surprising, but it's a testament to the power and elegance of Stokes' Theorem. By converting the line integral into a surface integral, we were able to simplify the problem and arrive at a clean, concise answer.

In summary, we started with a complex line integral, applied Stokes' Theorem to transform it into a surface integral, carefully calculated the curl of the vector field, parameterized the surface, and evaluated the resulting double integral. Each step was crucial, and understanding the underlying concepts made the process much smoother. Remember, guys, practice makes perfect, so keep tackling those integrals! By understanding the core principles and applying them methodically, you’ll become a master of integral calculus in no time. Whether it's line integrals, surface integrals, or anything in between, you've got this!