Area of an Ellipse (Standard Form)

The standard equation of an ellipse centred at the origin, with semi-axes $a$ (along $x$) and $b$ (along $y$), is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$. Solving for the upper half,

Notice that this is exactly the circle's semicircle expression $\sqrt{a^2-x^2}$ (a circle of radius $a$), multiplied by the constant factor $\dfrac{b}{a}$ -- an ellipse is, in this precise sense, a circle of radius $a$ compressed (or stretched) vertically by the factor $b/a$. This observation is the key that lets the circle-area result of Section 3 be reused here almost without extra work.

Setting up the quarter-ellipse integral. By the same symmetry argument as for the circle, the area of the whole ellipse is $4$ times the area of its first-quadrant portion, which is bounded by the arc, the $x$-axis, and the ordinates $x=0$ and $x=a$:

Reusing the circle result. The remaining integral, $\int_0^a \sqrt{a^2-x^2}\,dx$, is precisely the quarter-circle integral of Section 3 with $r$ replaced by $a$, which was already evaluated there as $\dfrac{\pi a^2}{4}$. So

Scaling up to the full ellipse. Multiplying by $4$, exactly as for the circle,

the standard result, now derived rather than quoted -- and it correctly reduces to the circle formula $\pi r^2$ when $a=b=r$ (an ellipse with equal semi-axes is simply a circle), a useful check on the formula's correctness.

Getting a quarter or other partial ellipse. As with the circle, only the limits and the multiplying factor change for a partial region:

• Quarter ellipse (one quadrant only): the result $A_1 = \dfrac{\pi ab}{4}$ found above, with no further scaling.
• Half ellipse (bounded by the arc and the major axis, i.e. the $x$-axis): twice the quarter-ellipse result, $\dfrac{\pi ab}{2}$.
• Full ellipse: four times the quarter-ellipse result, $\pi ab$, as derived above. …