Calculus

Integrals

Since the area of a curve in polar coordinates $\rho \left(\theta \right)$ between the angles $\alpha$ and $\beta$ is

[11] $$A = \frac{1}{2}{\int }_{\alpha }^{\beta }{\rho }^{2}d\theta$$

it is possible to demonstrate that the area of the siluroid between two generic angles is

[12] $$A ={\overline{)n^2\left(2\theta +\sin \left(4\theta \right)+2\sin \left(2\theta \right)-\frac{2}{3}{\sin }^3\left(2\theta \right)\right)}}_{ \alpha }^{ \beta }$$

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Total area

The total area of the siluroid is $2\pi n^2$

Main lobe

The area of the main lobe is

[13] $$\left(\pi + \frac{8}{3}\right)n^2$$

Secondary lobe

The area of a secondary lobe is

[14] $$\frac{1}{2}\left(\pi - \frac{8}{3}\right)n^2$$

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Derivatives

Let us use the equation in polar coordinates to calculate the derivatives of a siluroid. The first derivative is

[15] $$\dot{\rho } = -4n \sin \left(\theta \right)\left(3\cos \left(2\theta \right) + 2\right)$$

whereas the second derivative is

[16] $$\ddot{\rho } = -4n \cos \left(\theta \right)\left(9\cos \left(2\theta \right) - 4\right)$$

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First derivative

The curve corresponding to the first derivative of the mother equation is

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Second derivative

The curve corresponding to the second derivative of the mother equation is

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