Verify That Stokes' Theorem Is True For The Vector Field

Hey there, you brilliant human! Ever feel like math is just… a bit dry? Like a textbook left out in the sun for too long? Well, buckle up, buttercup, because we're about to dive into something that's not only incredibly cool but also, dare I say, fun! We're going to talk about a theorem. Yeah, I know, "theorem" can send shivers down some spines. But this one? This one's a game-changer. It's called Stokes' Theorem, and we're going to verify that it's, indeed, TRUE for a vector field. Isn't that exciting?

So, what's the big deal? Think of it as a cosmic handshake between two different ways of looking at the universe. On one hand, we have something called the line integral, which is like measuring how much a force field "pulls" you along a specific path. Imagine you're a tiny explorer on a vast, windy planet, and you want to know the total effort you exerted walking from point A to point B. That's kind of what a line integral is about. It’s a journey, a path, a story told along a curve.

On the other hand, we have the surface integral. Now, this is like measuring how much something is "flowing" or "circulating" through a surface. Think of a beautiful, flowing river. A surface integral could tell you the total amount of water passing through a dam’s spillway, for instance. It's about understanding what's happening across a whole area, like a bird’s-eye view of that river.

Stokes' Theorem, in its most glorious essence, tells us that these two seemingly different measurements are actually intimately connected. It’s like discovering that the amount of water flowing through your dam spillway is exactly equal to the sum of all the tiny eddies and currents you’d feel if you were paddling your kayak along the edge of that spillway. Mind. Blown. Right?

Now, the "verify" part. This isn't just about accepting it on faith, like a magic spell. We're going to prove it, or at least, get a really good taste of how it’s proven, for a specific, wonderfully simple vector field. Think of it as putting on your detective hat and solving a delightful mathematical mystery. Who doesn't love a good mystery that ends with a satisfying reveal?

Let's Get Our Hands Dirty (Figuratively!)

So, what kind of vector field are we talking about? Let's keep it nice and easy to start. How about a field that simply points in the direction you're going, with a strength that depends on how far you are from the origin? Imagine a gentle breeze that gets stronger as you walk away from a central point. For our purposes, let’s pick a really friendly field: F(x, y, z) = . Yeah, that’s it. Simple, elegant, and perfect for illustrating this grand idea.

What about the path and the surface? We need a closed loop (the path) and a surface that has that loop as its boundary. Let’s make it even more fun. Imagine a frisbee! A nice, flat, circular frisbee. Let's say the edge of our frisbee is a circle in the xy-plane, centered at the origin, with a radius of 1. So, our path, let’s call it C, is the circumference of this circle. Easy peasy, right?

And the surface? Well, that's just the flat, circular area enclosed by our frisbee's edge. Let's call this surface S. It's part of the xy-plane, so z = 0 for all points on our surface.

Solved Verify that Stokes' Theorem is true for vector field | Chegg.com

The Line Integral: Our Explorer's Journey

First, we tackle the line integral along our circle C. We need to parameterize our circle. Think of it like drawing the circle with a pen: we can use an angle, let's call it 't', starting from 0 and going all the way to 2π to complete the circle. So, our points on the circle are (cos(t), sin(t), 0).

Our vector field is F(x, y, z) = . So, on our circle, it becomes F(cos(t), sin(t), 0) = .

We also need the differential displacement vector, dr. If our path is given by r(t) = , then dr = r'(t) dt = <-sin(t), cos(t), 0> dt.

The line integral is the integral of F ⋅ dr along C. So, we're calculating:

∫_C F ⋅ dr = ∫₀^2π ⋅ <-sin(t), cos(t), 0> dt

Let's do that dot product: sin(t) * (-sin(t)) + 0 * cos(t) + 0 * 0 = -sin²(t).

SOLVED:Verify that Stokes' Theorem is true for the given vector field F

So, our integral becomes: ∫₀^2π -sin²(t) dt.

Now, remember those handy trigonometric identities? sin²(t) = (1 - cos(2t))/2. So we have: ∫₀^2π -(1 - cos(2t))/2 dt.

Let's integrate: -[t/2 - sin(2t)/4] from 0 to 2π.

Plugging in the limits: -[(2π/2 - sin(4π)/4) - (0/2 - sin(0)/4)] = -[(π - 0) - (0 - 0)] = -π.

So, the line integral along our frisbee's edge is -π. Keep that number in your mind, because we're about to see if the other side matches!

The Surface Integral: The Bird's-Eye View

Now for the surface integral over our flat frisbee, S. Stokes' Theorem involves the curl of the vector field. Don't let the word "curl" scare you; it's just another way of measuring rotation or circulation, but this time, it's a property of the field itself, not just along a path. The curl of F = is given by:

Solved 7. Verify that Stokes' Theorem is true for the vector | Chegg.com

curl(F) = (∂/∂y(0) - ∂/∂z(0)) i + (∂/∂z(y) - ∂/∂x(0)) j + (∂/∂x(0) - ∂/∂y(y)) k

Let's break that down. ∂/∂y(0) is the partial derivative of 0 with respect to y, which is 0. Same for ∂/∂z(0). ∂/∂z(y) is the partial derivative of y with respect to z, which is also 0. ∂/∂x(0) is 0. And ∂/∂y(y) is 1.

So, curl(F) = (0 - 0) i + (0 - 0) j + (0 - 1) k = <0, 0, -1>. Interesting! Our curl is a constant vector pointing straight down.

Now we need to calculate the surface integral of the curl of F over our surface S. This is written as ∫∫_S curl(F) ⋅ dS.

Our surface S is the flat disc in the xy-plane. For a flat surface like this, the differential surface vector dS is simply a vector perpendicular to the surface, with its direction pointing outwards (or inwards, depending on convention, but we'll be consistent). Since our surface is in the xy-plane, dS is <0, 0, dz dy> (or in a different order, but the essence is the z-component). For our surface, a unit normal vector pointing upwards is k = <0, 0, 1>. So, dS = k dA = <0, 0, 1> dx dy (or in our coordinate system, the normal vector might be pointing downwards, let's check our orientation).

Wait a sec! For Stokes' Theorem, the orientation of the surface needs to be consistent with the orientation of the boundary curve. If we trace our circle counter-clockwise, our flat frisbee surface should have its normal vector pointing upwards (the right-hand rule!). So, our dS should be oriented with the positive z-axis: dS = <0, 0, 1> dA.

SOLVED:Verify that Stokes Theorem is true for the given vector field F

Our curl is <0, 0, -1>. So, curl(F) ⋅ dS = <0, 0, -1> ⋅ <0, 0, 1> dA = -1 dA.

We need to integrate this over our circular surface in the xy-plane. The area of our circle is πr². With a radius of 1, the area is π(1)² = π.

So, the surface integral is ∫∫_S -1 dA = -1 * (Area of S) = -1 * π = -π.

The Grand Reveal!