# Finding the area of the top half of a circle

Alright, I'm trying to calculate the area of the top half of a circle of radius $a$. Here's what I did so far:

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

So I wrote $x$ as $a \cdot \sin \theta$:

$$\int_{-a}^a \sqrt{(a^2 - a^2\sin^2 \theta )}$$

$$\int_{-a}^a a \sqrt{( 1 - \sin \theta^2)}$$

$$\int_{-a}^a [a \cdot \cos \theta]$$

$$2 \sin(a) a$$

The problem is that my textbook states that the area is actually:

$$\frac{\pi a^2}{2}$$

I've done this calculation over and over and I'm sure there are no mistakes, so what is going on here?

• You haven't done the substitution correctly - remember the $dx$ needs to change at some point. Make sure to keep the $dx$ at the end of integrals to prevent this type of mistake! – B. Mehta Dec 23 '17 at 1:05
• Both answers seem to me to be circular in their core. While they appear to prove that the area of a circle is $\pi r^2$, I believe both rely on the assumption that an arc length is $\theta r$ and a a "full revolution" is $2\pi$. In other words, from assumed knowledge $\int\sqrt{a^2 - x^2}dx$ provides no proof, rather a strange way of writing a familiar relationship. – Jared Dec 23 '17 at 4:49
• if $x=a\sin\theta$ then $dx=a\cos\theta d\theta$ – janmarqz Dec 24 '17 at 16:52

$$x=a\sin t\implies dx=a\cos t\,dt$$

and from here

$$\int_{-a}^a\sqrt{a^2-x^2}\,dx=a\int_{-\frac\pi2}^\frac\pi2\sqrt{1-\sin^2 t}\,a\cos t\,dt=a^2\int_{-\frac\pi2}^\frac\pi2\cos^2t=$$

$$=\left.\frac{a^2}2(t+\cos t\sin t)\right|_{-\frac\pi2}^{\frac\pi2}=\frac{a^2\pi}2$$

• How do you know that $a = \frac{\pi}{2}$? – Trey Dec 23 '17 at 1:13
• @Trey Because of the substitution. If $\;x=a\sin t\;$ , then when $\;x=a\;$ we get $\;a=a\sin t\implies \sin t=1\implies t=\pi/2\;$ , and likewise with the lower limit. – DonAntonio Dec 23 '17 at 1:15

Use double integral:-

$r:0\to a$

$\theta: 0 \to \pi$

$dxdy=rdrd\theta$.

$A=\int_{\theta=0}^{\pi}\int_{r=0}^{a} rdrd\theta=\int_{\theta=0}^{\pi}d\theta\int_{r=0}^{a} rdr=\frac{\pi}{2}a^2.$