Question

Definite Integrals Calculus. Can someone help me on where to get started with this. Prove the area of this ellipse

Answer 1

If you already know some multivariable calculus, you can simply compute the double integral

$\displaystyle\iint_E\mathrm dx\,mathrm dy$

where $E$ denotes the region bounded by the ellipse with equation

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

We can solve for $y$:

$y=\pm\sqrt{b^2-\dfrac{b^2x^2}{a^2}}=\pm\dfrac ba\sqrt{a^2-x^2}$

then the integral becomes

$\displaystyle\int_{x=-a}^{x=a}\int_{y=-\frac ba\sqrt{a^2-x^2}}^{y=\frac ba\sqrt{a^2-x^2}}\mathrm dy\,\mathrm dx$

We also could have solve for $x$ instead and swapped the order of integration, so that the area is

$\displaystyle\int_{y=-b}^{y=b}\int_{x=-\frac ab\sqrt{b^2-y^2}}^{x=\frac ab\sqrt{b^2-y^2}}\mathrm dx\,\mathrm dy$

If you don't know about double integrals yet, these basically reduce to either of the single-variate integrals,

$\displaystyle\frac{2b}a\int_{x=-a}^{x=a}\sqrt{a^2-x^2}\,\mathrm dx=\frac{4b}a\int_{x=0}^{x=a}\sqrt{a^2-x^2}\,\mathrm dx$

(making use of the fact that $\sqrt{a^2-x^2}$ is symmetric about 0) or

$\displaystyle\frac{2a}b\int_{y=-b}^{y=b}\sqrt{b^2-y^2}\,\mathrm dx=\frac{4a}b\int_{y=0}^{y=b}\sqrt{b^2-y^2}\,\mathrm dy$

either of which can be evaluated with a trigonometric substitution. For instance, taking $x=a\sin t$, gives $\mathrm dx=a\cos t\,\mathrm dt$, and the integral becomes

$\displaystyle\frac{4b}a\int_{a\sin t=0}^{a\sin t=a}\sqrt{a^2-(a\sin t)^2}\,a\cos t\,\mathrm dt=4ab\int_{t=0}^{t=\pi/2}\sqrt{1-\sin^2t}\cos t\,\mathrm dt$

$=\displaystyle4ab\int_{t=0}^{t=\pi/2}\cos^2t\,\mathrm dt$

$=\displaystyle2ab\int_{t=0}^{t=\pi/2}(1+\cos2t)\,\mathrm dt$

$=2ab\left(t+\dfrac12\sin2t\right)\bigg|_{t=0}^{t=\pi/2}$

$=2ab\left(\dfrac\pi2\right)=\pi ab$

The integral with respect to $y$ can be resolved in a similar way.

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We also could have converted to polar coordinates first, parameterizing the region $E$ by

$\begin{cases}x=ar\cos t\\y=br\sin t\\0\le r\le1\\0\le t\le2\pi\end{cases}$

The Jacobian matrix for this transformation is

$\mathbf J=\begin{bmatrix}\dfrac{\partial x}{\partial r}&\dfrac{\partial x}{\partial t}\\\\\dfrac{\partial y}{\partial r}&\dfrac{\partial y}{\partial t}\end{bmatrix}=\begin{bmatrix}a\cos t&-ar\sin t\\b\sin t&br\cos t\end{bmatrix}$

and its determinant gives $|\det\mathbf J|=abr$. So the integral reduces to

$\displaystyle\iint_E\mathrm dx\,\mathrm dy=\iint_E|\det\mathbf J|\,\mathrm dr\,\mathrm dt=ab\int_{t=0}^{t=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm dt$

$=\displaystyle\frac{ab}2\int_{t=0}^{t=2\pi}\mathrm dt$

$=\dfrac{ab}2(2\pi)=\pi ab$