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Could someone help me out with calculating this integral.

$$\int_{-a}^a \sqrt{a^2-t^2}dt$$

Where $a>0$.

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In other words, half the area of a circle of radius $a$... – anon Dec 13 '12 at 16:30
Set $t=a\cos\theta$ (if you can't appeal to geometric methods). – David Mitra Dec 13 '12 at 16:31
Why the area of a circle ? – Kasper Dec 13 '12 at 16:31
(The graph of the upper half of the circle (of radius $a$) is given by $y=\sqrt{a^2-x^2}$ from $x=-a$ to $x=a$, since the implicit equation of the full circle is $x^2+y^2=a^2$. Integrating will give you the area. However this should just serve to tell you the answer so you can think of a method to compute the integral - the answer involving $\pi$ indicates trigonometric substitution may be prudent, as per David's comment.) – anon Dec 13 '12 at 16:34
up vote 3 down vote accepted

$$ \int_{-a}^a dt \sqrt{a^2-t^2} = -a \int_{\pi}^0 d\theta \ \sin \theta \sqrt{a^2-\left(a \cos \theta\right)^2} = a^2 \int_{0}^{\pi} d\theta \ \sin^2 \theta $$ You can finish it. $\sin^2 \theta = \left[1 - \cos\left(2\theta\right)\right]/2$ should help.

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