Use green's theorem to find the area of the ellipse given by
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This is a standard application, a way to use Green's Theorem to compute areas by doing line integrals.
Let DD be the ellipse, and CC its boundary x2a2+y2b2=1x2a2+y2b2=1. The area you are trying to compute is
∫∫D1dA.∫∫D1dA.
According to Green's Theorem, if you write 1=∂Q∂x−∂P∂y1=∂Q∂x−∂P∂y, then this integral equals
∮C(Pdx+Qdy).∮C(Pdx+Qdy).
There are many possibilities for PP and QQ. Pick one. Then use the parametrization of the ellipse
xy=acost=bsintx=acosty=bsint
to compute the line integral.
As you can probably see, the idea of finding PP and QQ with ∂Q∂x−∂P∂y=1∂Q∂x−∂P∂y=1 can be used to compute the area of any region enclosed by a simple closed curve. Of course, the line integral may be more complicated than the area computation, but that's another kettle of fish.
Let DD be the ellipse, and CC its boundary x2a2+y2b2=1x2a2+y2b2=1. The area you are trying to compute is
∫∫D1dA.∫∫D1dA.
According to Green's Theorem, if you write 1=∂Q∂x−∂P∂y1=∂Q∂x−∂P∂y, then this integral equals
∮C(Pdx+Qdy).∮C(Pdx+Qdy).
There are many possibilities for PP and QQ. Pick one. Then use the parametrization of the ellipse
xy=acost=bsintx=acosty=bsint
to compute the line integral.
As you can probably see, the idea of finding PP and QQ with ∂Q∂x−∂P∂y=1∂Q∂x−∂P∂y=1 can be used to compute the area of any region enclosed by a simple closed curve. Of course, the line integral may be more complicated than the area computation, but that's another kettle of fish.
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