# Integral of a quarter-circle (1 Viewer)

#### Heresy

##### Member
Find the area bounded by the curve y = (4-x^2)^1/2, the x-axis and the y-axis in the first quadrant.

I know what the integral is, but whenever I try to solve it I end up with 16/3 - when the answer should be pi units^2

Any help would be appreciated. Thanks.

Last edited:

Cheers!

#### fan96

##### 617 pages
something that should be explicitly noted is that

$\bg_white \sqrt{1-\sin^2 \theta} = |\cos \theta|$

(and in general, $\bg_white \sqrt{x^2} = |x|$)

usually the bounds of the integral will mean that $\bg_white |\cos \theta| = \cos \theta$ but one should still be careful when dealing with situations like this.

#### HeroWise

##### Active Member
Yes definetly, i have been in situations where i got that wrong lol. aka Weird vector questions which has both multiple positive cases and the negative cases which i missed because of that silly thing

#### Heresy

##### Member
something that should be explicitly noted is that

$\bg_white \sqrt{1-\sin^2 \theta} = |\cos \theta|$

(and in general, $\bg_white \sqrt{x^2} = |x|$)

usually the bounds of the integral will mean that $\bg_white |\cos \theta| = \cos \theta$ but one should still be careful when dealing with situations like this.
Thank you.

On a completely unrelated note - how do you type in that font with the equations - I'm curious lol - it will make it 10 times easier next time I inevitably ask for help