Green's theorem example 2 Another example applying Green's Theorem
Green's theorem example 2
- Let's say I have a path in the xy plane that's essentially
- the unit circle.
- So this is my y-axis, this is my x-axis, and our path is
- going to be the unit circle.
- And we're going to traverse it just like that.
- We're going to traverse it clockwise.
- I think you get the idea.
- And so its equation is the units circle.
- So the equation of this is x squared plus y squared is
- equal to 1; has a radius of 1 unit circle.
- And what we're concerned with is the line integral
- over this curve c.
- It's a closed curve c.
- It's actually going in that direction of 2y dx minus 3x dy.
- So, we are probably tempted to use Green's
- theorem and why not?
- So let's try.
- So this is our path.
- So Green's theorem tells us that the integral of some curve
- f dot dr over some path where f is equal to-- let me write
- it a little nit neater.
- Where f of x,y is equal to P of x, y i plus Q of x, y j.
- That this integral is equal to the double integral over the
- region-- this would be the region under question
- in this example.
- Over the region of the partial of Q with respect to x
- minus the partial of P with respect to y.
- All of that dA, the differential of area.
- And of course, the region is what I just showed you.
- Now, you may or may not remember-- well, there's a
- slight, subtle thing in this, which would give
- you the wrong answer.
- In the last video we said that Green's theorem applies when
- we're going counterclockwise.
- Notice, even on this little thing on the integral I made
- it go counterclockwise.
- In our example, the curve goes clockwise.
- The region is to our right.
- Green's theorem-- this applies when the region is to our left.
- So in this situation when the region is to our right and
- we're going-- so this is counterclockwise.
- So in our example, where we're going clockwise, the region is
- to our right, Green's theorem is going to be the
- negative of this.
- So in our example, we're going to have the integral of c and
- we're going to go in the clockwise direction.
- So maybe I'll draw it like that of f dot dr. This is going
- to be equal to the double integral over the region.
- You could just swap these two-- the partial of P with respect
- to y minus the partial of Q with respect to x da.
- So let's do that.
- So this is going to be equal to, in this example, the
- integral over the region-- let's just keep it
- abstract for now.
- We could start setting the boundaries, but let's just
- keep the region abstract.
- And what is the partial of P with respect-- let's remember,
- this right here is our-- I think we could recognize right
- now that if we take f dot dr we're going to get this.
- The dr contributes those components.
- The f contributes these two components.
- So this is P of x,y.
- And then this is Q of x,y.
- And we've seen it.
- I don't want to go into the whole dot dr and take the dot
- product over and over again.
- I think you can see that this is the dot product
- of two vectors.
- This is the x component of f, y component of f.
- This is the x component of dr, y component of dr. So let's
- take the partial of P with respect to y.
- You take the derivative of this with respect to y, you get 2.
- Derivative of 2y is just 2.
- So you get 2, and then, minus the derivative of
- Q with respect to x.
- Derivative of this with respect to x is minus 3.
- So we're going to get minus 3, and then all of that da.
- And this is equal to the integral over the region.
- What's this, it's 2 minus minus 3?
- That's the same thing as 2 plus 3.
- So it's the integral over the region of 5 dA.
- 5 is just a constant, so we can take it out of the integral.
- So this is going to turn out to be quite a simple problem.
- So this is going to be equal to 5 times the double integral
- over the region R dA.
- Now what is this thing?
- What is this thing right here?
- It looks very abstract, but we can solve this.
- This is just the area of the region.
- That's what that double integral represents.
- You just sum up all the little dA's.
- That's a dA, that's a dA.
- You sum up the infinite sums of those little
- dA's over the region.
- Well, what's the area of this unit circle?
- Here we just break out a little bit of ninth grade-- actually,
- even earlier than that-- pre-algebra or middle
- school geometry.
- Area is equal to pi r squared.
- What's our radius?
- So unit circle, our radius is 1.
- Length is 1.
- So the area here is pi.
- So this thing over here, that whole thing is
- just equal to pi.
- So the answer to our line integral is just 5 pi, which
- is pretty straightforward.
- I mean, we could have taken the trouble of setting up a double
- integral where we take the antiderivative with respect to
- y first and write y is equal to the negative square root of 1
- minus x squared y is equal to the positive square root.
- x goes from minus 1 to 1.
- But that would have been super hairy and a huge pain.
- And we just have to realize, no, this is just the area.
- And the other interesting thing is I challenge you to solve
- the same integral without using Green's theorem.
- You know, after generating a parameterization for this
- curve, going in that direction, taking the derivatives
- of x of t and y of t.
- Multiplying by the appropriate thing and then taking the
- antiderivative-- way hairier than what we just did using
- Green's theorem to get 5 pi.
- And remember, the reason why it wasn't minus 5 pi here
- is because we're going in a clockwise direction.
- If we were going in a counterclockwise direction we
- could have applied the straight up Green's theorem, and we
- would have gotten minus 5 pi.
- Anyway, hopefully you found that useful.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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