Evaluating $\int_{-a}^{a}\sqrt{a^2-x^2}dx$

Question

Question:

How to evaluate

$$\int_{-a}^{a} \sqrt{a^2 - x^2} dx$$

This came up while trying to prove that the area of an ellipse is give by $\pi a b$ where $a$ and $b $are the major and minor axes respectively.

I'm curious as to how one could evaluate this. It should be $$\frac{\pi a^2}{2}$$ but I have no idea how to get that result.

Could someone show me?

Answer 1

Hint: Use this substitution. $$x = a\sin(\theta)$$

Answer 2

Hint: Let $x = a \sin t $ then $dx = a \cos t\, dt$ and $$\int_{-\pi/2}^{\pi/2} \sqrt{a^2 - a^2 \sin^2 t}\,\, a\cos t \, dt =\int_{- \pi/2}^{ \pi/2} a^2\cos^2 t \, dt$$

Answer 3

Write $I = \int \sqrt{a^2 - x^2} \ dx$. Then integrating by parts,

$$I = x\sqrt{a^2 - x^2} + \int \frac{x^2}{\sqrt{a^2-x^2}} \ dx = x\sqrt{a^2 - x^2} - I + \int \frac{a^2}{\sqrt{a^2 - x^2}} \ dx $$

This last integral should look standard. Thus

$$I = \frac{1}{2} \left( x\sqrt{a^2 - x^2} + a^2 \arcsin(x/a) \right) $$

Hence

$$\int_{-a}^a \sqrt{a^2 - x^2} dx = \ \frac{a^2}{2} \left[ \arcsin(x/a) \right]_{-a}^a = \frac{a^2}{2} ( \pi/2 - (-\pi/2)) = \frac{\pi a^2}{2}$$

Answer 4

If $y=\sqrt{a^2-x^2}$ then $y \geq 0$ and $$x^2+y^2=a^2$$

This shows that your integral represents the area between the $x$-axis and the upper half circle $x^2+y^2=a^2$.

The area of this half disk is $\frac{\pi a^2}{2}$.