If you already know some multivariable calculus, you can simply compute the double integral
[tex]\displaystyle\iint_E\mathrm dx\,mathrm dy[/tex]
where [tex]E[/tex] denotes the region bounded by the ellipse with equation
[tex]\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1[/tex]
We can solve for [tex]y[/tex]:
[tex]y=\pm\sqrt{b^2-\dfrac{b^2x^2}{a^2}}=\pm\dfrac ba\sqrt{a^2-x^2}[/tex]
then the integral becomes
[tex]\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[/tex]
We also could have solve for [tex]x[/tex] instead and swapped the order of integration, so that the area is
[tex]\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[/tex]
If you don't know about double integrals yet, these basically reduce to either of the single-variate integrals,
[tex]\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[/tex]
(making use of the fact that [tex]\sqrt{a^2-x^2}[/tex] is symmetric about 0) or
[tex]\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[/tex]
either of which can be evaluated with a trigonometric substitution. For instance, taking [tex]x=a\sin t[/tex], gives [tex]\mathrm dx=a\cos t\,\mathrm dt[/tex], and the integral becomes
[tex]\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[/tex]
[tex]=\displaystyle4ab\int_{t=0}^{t=\pi/2}\cos^2t\,\mathrm dt[/tex]
[tex]=\displaystyle2ab\int_{t=0}^{t=\pi/2}(1+\cos2t)\,\mathrm dt[/tex]
[tex]=2ab\left(t+\dfrac12\sin2t\right)\bigg|_{t=0}^{t=\pi/2}[/tex]
[tex]=2ab\left(\dfrac\pi2\right)=\pi ab[/tex]
The integral with respect to [tex]y[/tex] can be resolved in a similar way.
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We also could have converted to polar coordinates first, parameterizing the region [tex]E[/tex] by
[tex]\begin{cases}x=ar\cos t\\y=br\sin t\\0\le r\le1\\0\le t\le2\pi\end{cases}[/tex]
The Jacobian matrix for this transformation is
[tex]\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}[/tex]
and its determinant gives [tex]|\det\mathbf J|=abr[/tex]. So the integral reduces to
[tex]\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[/tex]
[tex]=\displaystyle\frac{ab}2\int_{t=0}^{t=2\pi}\mathrm dt[/tex]
[tex]=\dfrac{ab}2(2\pi)=\pi ab[/tex]
