# 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] $$A={\overline{){n}^{2}\left(2\theta +\mathrm{sin}\left(4\theta \right)+2\mathrm{sin}\left(2\theta \right)-\frac{2}{3}{\mathrm{sin}}^{3}\left(2\theta \right)\right)}}_{\alpha}^{\beta}$$### 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] $$\frac{1}{2}\left(\pi -\frac{8}{3}\right){n}^{2}$$## Derivatives

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

[15] $$\dot{\rho}=-4n\u200a\mathrm{sin}\left(\theta \right)\left(3\mathrm{cos}\left(2\theta \right)+2\right)$$whereas the second derivative is

[16] $$\ddot{\rho}=-4n\u200a\mathrm{cos}\left(\theta \right)\left(9\mathrm{cos}\left(2\theta \right)-4\right)$$### First derivative

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

### Second derivative

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