# Problems from Areas of enclosures in the plane

Calculate the area of the enclosure delimited by the ellipse of equation:

$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$$See development and solution ### Development: We identify the domain $$D$$: $$D= \big\{ (x,y)\in\mathbb{R}^2 \ | \ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 \big\}$$$

In this case it will be useful to change to angular coordinates since the inside of an ellipse can be parametrized as follows:

$$\left\{ \begin{array}{l} x(r,\theta)=a\cdot r\cdot \cos(\theta) \\ y(r,\theta) =b\cdot r\cdot \sin(\theta) \end{array} \right. \quad \text{ with } \quad r\in[0,1] \ \text{ and } \ \theta\in[0,2\pi]$$$We have to calculate the determinant of the Jacobian matrix of the change: $$\begin{array}{ll} x_r=a\cos(\theta) & x_\theta=-a\cdot r\cdot \sin(\theta) \\ y_r=b\cdot\sin(\theta) & y_\theta=b\cdot r\cdot\cos(\theta) \end{array}$$ $$\Rightarrow \det(J)=a\cdot b\cdot r\cdot\cos^2(\theta)+ a\cdot b\cdot r\cdot\sin^2(\theta)=a\cdot b\cdot r$$ Thus we have: $$\begin{array}{rl} \text{Area}(D)=& \int_D 1 \ dx \ dy=\int_0^1\Big( \int_0^{2\pi} a\cdot b\cdot r \ d\theta \Big) \ dr \\ =& \int_0^1 \Big( a\cdot b\cdot r \int_0^{2\pi}\ d\theta \Big) \ dr= a\cdot b\cdot\int_0^1 2\pi r \ dr \\ =& 2\pi \cdot a\cdot b\cdot \Big[ \dfrac{r^2}{2} \Big]_0^1 = \pi\cdot a\cdot b \end{array}$$$

### Solution:

The area is $$\pi a b$$.

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