Q.Find the area of the region in the first quadrant bounded by the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$, using integration.

Integrate $y=\sqrt{1-x^2/2}$ over the first quadrant and multiply by 4 for the full ellipse.

The ellipse $\dfrac{x^2}{2}+\dfrac{y^2}{1}=1$ has $a=\sqrt2$, $b=1$, so $y=\sqrt{1-\dfrac{x^2}{2}}$ in the first quadrant.

By symmetry, total area $=4\displaystyle\int_0^{\sqrt2}\sqrt{1-\dfrac{x^2}{2}}\,dx$.

Let $x=\sqrt2\sin\theta$, $dx=\sqrt2\cos\theta\,d\theta$; when $x=0,\theta=0$, when $x=\sqrt2,\theta=\pi/2$. Also $\sqrt{1-x^2/2}=\cos\theta$.

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