# Double integral bounded by closed parametric curve

1. Jul 10, 2009

### okkvlt

question:

how do i find the area under f[x,y] bounded by a closed parametric curve x[t],y[t]? it doesnt look like i can use a change of variables. it seems as though double integrals only with functions where the curve is given explicitly such as y[x] or x[y].

2. Jul 10, 2009

### tiny-tim

Hi okkvlt!
I don't understand … what double integral?

The area is a single integral … ∫y dx, or ∫y[t] x'[t] dt.

3. Jul 11, 2009

### okkvlt

thats a typo i meant

how do i find the area under f[x,y] bounded by a closed parametric curve x[t],y[t]? it doesnt look like i can use a change of variables. it seems as though double integrals only work with functions where the curve is given explicitly such as y[x] or x[y].

what about when the boundary is given by a parametric curve?

4. Jul 11, 2009

### HallsofIvy

Use Green's theorem:
$$\int P(x,y)dx+ Q(x,y)dy= \int\int \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}dxdy$$
Taking Q(x,y)= y so that $\partial Q/\partial y= 1$ and P(x,y)= 0 so that $\partial P/\partial x= 0$ and
$$Area= \int\int dxdy= \int y dx= \int y(t) (dx/dt)dt$$
just as tiny-tim said! But notice this is a path integral, not an integral over an area.

For example, if x= cos(t) and y= sin(t), with t going from 0 to $2\pi$ (this is a circle of radius 1 with center at (0,0)), (dx/dt) dt= -sin(t) dt and ydx= -sin^2(t) dt. The area is
$$\int_{t=0}^{2\pi} -sin^2(t)dt= -\frac{1}{2}\int_0^{2\pi}(1- cos(2t))dt$$
$$= (1/2)(2\pi)= \pi$$