Double integral bounded by closed parametric curve

[okkvlt]

question:

how do i find the area under f[x,y] bounded by a closed parametric curve x[t],y[t]? it doesn't 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].

 

[tiny-tim]

Hi okkvlt!

how do i find the area under f[x,y] bounded by a closed parametric curve x[t],y[t]? it doesn't 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].

I don't understand … what double integral?

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

 

[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 doesn't 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?

 

[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