FP3 Polar Area question

Submitted by Anonymous (not verified) on

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r=2(cosθ-sinθ)
r=2((x-y)/r)
r²=2x-2y
...
(x-1)²+(y+1)²=2
so r=√2

So I have two forms to work out the area.

A=πr²
=2π

A=½∫r^2 dθ between 2π and 0
=½∫4(cos²θ-2cosθsinθ+sin²θ) dθ
=2∫(1-sin2θ) dθ
=2[θ+½cos2θ] between 2π and 0
=2[(2π+½)-(½)]
=4π

The answer is 2π so the error is in the polar bit probably something stupid

Wrong limits

Submitted by Rayman (not verified) on

The curve doesn't exist in the full range for $0 \leq \theta \leq 2\pi$.
Need to find when $ r \geq 0$ and the valid range(s) for $\theta$, then integrate using the limits.

Thanks :)

Submitted by O________O (not verified) on

I didn't realise r>0 and neither did my teacher :/ ill probs look at the proof again :)

ty

Submitted by O________O (not verified) on

this does seem better than the ones I found ~ v helpful

Polar coordinates

Submitted by stcs on

There does seem to be an ambiguity about the possible sign of $r$ in polar coordinates, which I only came across when looking at A-level syllabuses.

To my mind, $r$ is the distance from the point to the origin and so is never negative.

I can see that it might be useful in some cases to have a negative $r$, but it is never necessary and (again to my mind) just complicates things.

Stephen