Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Could someone help me out with calculating this integral.

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

Where $a>0$.

share|improve this question
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

1 Answer 1

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.