# Greens' Theorem (finding area) -

In summary, the conversation discusses the use of Green's theorem to compute the area inside a given curve. The method involves using the equation \frac{1}{2} \oint_c x\,dy - y\,dx and integrating with respect to t. The conversation includes calculations and a correction to a miscalculation, which results in the final answer of 4/3.
Ok, so I am obviously doing something wrong... but I don't know what it is. Here is the problem:

Use Green's theorem to compute the area inside the curve.
$$\vec r(t)=(2\sin(t)\cos(t),\sin(t))\,\,\,\,\,\,0\leq t\leq \pi$$

Greens Theorem
$$\frac{1}{2} \oint_c x\,dy - y\,dx$$

$$x = 2\sin(t)\cos(t)$$
$$y = \sin(t)$$

I'm guessing this is where I'm making my error, since I'm making an assumption about the notation.
$$dy = \frac{dy}{dt} = \cos(t)$$

$$dx = \frac{dx}{dt} = 4(\cos(t))^2 - 2$$

$$A = \frac{1}{2}\oint_0^\pi [(2\sin(t)\cos(t))(\cos(t)]\,dt-[(\sin(t))(4(\cos(t))^2 - 2)]\,dt$$

Letting my trusty TI-89 eat up the integral and I get:

$$2\pi-\frac{4}{3}$$

Well the online homework doesn't like this answer, so it must be wrong. I'm just unsure where I'm making the mistake. The notation is a little off the wall for me, so I'm not 100% sure with what I'm doing. Any help is appreciated. Thank you.

$$x = 2sin(t)cos(t) = sin(2t), dx = 2cos(2t)$$

$$y = sin(t), dy = cos(t)$$

$$\frac{1}{2} \oint_c x\,dy - y\,dx$$

$$\frac{1}{2} \oint_c(sin(2t)cos(t) - 2sin(t)cos(2t))dt$$

Maple says that equals 4/3

I think you miscalculated 'dx'.

$$x = 2sin(t)cos(t)$$

$$dx = 2(cos(t)cos(t) - sin(t)sin(t))$$

$$dx = 2(cos^2(t)-sin^2(t) = 2cos(2t) \mbox{ by a double angle identity.}$$

Thanks man, the help was very much appreciated

## What is Greens' Theorem and how does it relate to finding area?

Greens' Theorem is a mathematical theorem that relates the line integral around a closed curve to the double integral over the region enclosed by the curve. This allows us to use techniques from multivariable calculus to find the area of a two-dimensional region.

## Why is Greens' Theorem useful for finding area?

Greens' Theorem allows us to use simpler techniques from multivariable calculus, such as integrating over a single variable, to find the area of a two-dimensional region. This can save time and effort compared to other methods.

## What are the conditions for applying Greens' Theorem to find area?

To apply Greens' Theorem, the region must be bounded by a simple, closed curve and the curve must be smooth and oriented counterclockwise. Additionally, the region must be simply connected, meaning it does not contain any holes.

## Can Greens' Theorem be used for irregular or non-convex regions?

Yes, Greens' Theorem can be used for irregular or non-convex regions as long as the conditions for applying the theorem are met. The region can be broken down into smaller, simpler regions and the area can be calculated for each region separately.

## Are there any limitations to using Greens' Theorem for finding area?

Greens' Theorem can only be applied to two-dimensional regions, not three-dimensional volumes. Additionally, the curve that bounds the region must be smooth, meaning it has a continuous tangent vector, and the region must be simply connected.

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