7. Unusual Domain

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Question statement

Consider the rose $f(\theta)=\var{k}\cos\var{n}\theta$ .

{domain()}

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			Name	Type	Generated Value

k	integer	8
n	integer	4
gcd	number	8

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The limits of integration for $r$ are simply $0$ and $f$.

Looking at the boundaries of the rightmost leaf, the upper side reaches $r=0$ when $\var{n}\theta=\frac{\pi}{2}$, and the lower side when $\var{n}\theta=-\frac{\pi}{2}$.

The limits of integration for $\theta$ are therefore $\pm\pi/\var{2*n}$.

Because we just want the area, we integrate over $r$.

\[
A=\int_{-\pi/\var{2*n}}^{\pi/\var{2*n}}\int_0^{\var{k}\cos\var{n}\theta} r\, \mathrm{d}r\,\mathrm{d}\theta
=\int_{-\pi/\var{2*n}}^{\pi/\var{2*n}} \frac{1}{2}r^2 \, \bigg|_0^{\var{k}\cos\var{n}\theta} \,\mathrm{d}\theta
=\int_{-\pi/\var{2*n}}^{\pi/\var{2*n}} \simplify[fractionNumbers]{{k^2/2}}\cos^2\var{n}\theta \,\mathrm{d}\theta
\]

We know $\cos^2\theta=\cos2\theta+1$ from the double angle formulae, so we deduce that $\cos^2\var{n}\theta=\cos\var{2*n}\theta+1$.

\[
A=\simplify[fractionNumbers]{{k^2/2}} \int_{-\pi/\var{2*n}}^{\pi/\var{2*n}} \cos\var{2*n}\theta+1 \,\mathrm{d}\theta
=\simplify[fractionNumbers]{{k^2/2}}\bigg(\frac{1}{\var{2*n}}\sin\var{2*n}\theta+\theta\bigg) \bigg|_{-\pi/\var{2*n}}^{\pi/\var{2*n}}
=\simplify[fractionNumbers]{{k^2/(2*n)}}\pi
\]

Over an interval of $2\pi$, $f$ produces an equal number of positive and negative leaves. That is to say the leaves have positive and negative radii, and so they all cancel out with each other and therefore $A=0$.

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7. Unusual Domain

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Nicholas Barker 7 years, 1 month ago

Nicholas Barker 7 years, 1 month ago

Nicholas Barker 7 years, 1 month ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Mathematical expression

Advanced settings

Checking accuracy

Variable value generators

Pattern restriction

Variables

String restrictions

Generated value: `integer`