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## Homework Statement

[itex]\int_{C}|y|ds[/itex] where C is the curve [itex](x^{2}+y^{2})^{2}=2^{2}(x^{2}-y^{2})[/itex]

## Homework Equations

## The Attempt at a Solution

i used polar coordinates [itex]x = r cos \theta[/itex] and [itex]y = r sin \theta[/itex]

then substituted into the equation to get [itex]r = 2\sqrt{cos 2\theta}[/itex]

since [itex]r\geq0[/itex] gives [itex]-\frac{\pi}{4}\leq \theta\leq \frac{\pi}{4}[/itex]

substituting back gives [itex]x = 2\sqrt{cos 2\theta}cos \theta[/itex] and [itex]y = 2\sqrt{cos 2\theta}sin \theta[/itex]

then i calculated [itex]\frac{dx}{d\theta}=-\frac{2sin 2\theta cos \theta}{\sqrt{cos 2\theta}}-\frac{2sin \theta}{\sqrt{cos 2\theta}}[/itex] and [itex]\frac{dy}{d\theta}=-\frac{2sin 2\theta sin \theta}{\sqrt{cos 2\theta}}+\frac{2cos \theta}{\sqrt{cos 2\theta}}[/itex]

and found [itex]ds=\sqrt{(\frac{dx}{d\theta})^{2}+(\frac{dy}{d\theta})^{2}}d\theta=(4tan 2\theta sec 2\theta+4 cos 2\theta)d\theta[/itex]

now substituting back to the integral gives [itex]8\int^{-\frac{\pi}{4}}_{\frac{\pi}{4}}\sqrt{cos 2\theta}|sin \theta|(tan 2\theta sec 2\theta+ cos 2\theta)d\theta[/itex]

which looks terribly difficult, so i inferred that i did something wrong somewhere, can someone please shed a bit of light? thanks

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