# Math Help - average height of function

1. ## average height of function

find the average height of y=(2/3)*(sqrt (9 - x^2) ) over the interval [-3,3]
i used the average height formula but i get the answer as 0 while the correct answer is supposed to be Pi/2
can someone help me out?

2. $\frac{1}{3-(-3)} \int_{-3}^3 \frac{2}{3} \sqrt{9 - x^2} \, dx
$

$\frac{1}{9} \int_{-3}^3 \sqrt{9 - x^2} \, dx$

$\frac{1}{9} \cdot \frac{\pi \cdot 3^2}{2}$

3. how did you integrate to get to the last step?
i think thats where im erring

4. you don't need calculus to integrate, only geometry ... the graph of $y = \sqrt{9 - x^2}$ from -3 to 3 is a semicircle of radius 3.

5. oh i see
but is it possible to use calculus and find the area because thats what the question wants i think

6. possible, but a bit more involved.

use the trig substitution $x = 3\sin{\theta}$