Use areas to evaluate the integral:

[sqrt(9-x^2)]dx on interval [-3,3]

I am trying to figure out what would be the easiest way to do this. I know that you have to use trigonometry in some way.

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- Jan 23rd 2013, 08:05 AMMikusboiUsing known areas to find integrals
Use areas to evaluate the integral:

[sqrt(9-x^2)]dx on interval [-3,3]

I am trying to figure out what would be the easiest way to do this. I know that you have to use trigonometry in some way. - Jan 23rd 2013, 08:17 AMibduttRe: Using known areas to find integrals
Use the formula ∫▒〖√(a^2- x^2 ) dx= 1/2 〗 √(a^2- x^2 )+ 1/2 a^2 sin^(-1)〖x/a+C〗 from -3 to 3

- Jan 23rd 2013, 08:49 AMBobPRe: Using known areas to find integrals
The integral, when evaluated, will be the area 'under the curve' $\displaystyle y=\sqrt{9-x^{2}}$ between -3 and +3.

The equation is that of a circle, so the result will be the area of the semi-circle. - Jan 23rd 2013, 08:52 AMPlatoRe: Using known areas to find integrals
- Jan 23rd 2013, 09:02 AMMikusboiRe: Using known areas to find integrals
9pi/2?

- Jan 23rd 2013, 09:06 AMMikusboiRe: Using known areas to find integrals
wow that is very simple. I completely forgot about the way circles were represented in equation form.

- Jan 23rd 2013, 06:44 PMhollywoodRe: Using known areas to find integrals
If there's a really nice shortcut like that, of course you have to take it.

There's a more generally applicable method that you should know about, though, which is trig substitution. It should be in your calculus book (and probably explained there 1000 times better). Here's the quick overview:

If you have $\displaystyle \sqrt{x^2+a^2}$, substitute $\displaystyle x=a\tan\theta$

If you have $\displaystyle \sqrt{x^2-a^2}$, substitute $\displaystyle x=a\sec\theta$

If you have $\displaystyle \sqrt{a^2-x^2}$, substitute $\displaystyle x=a\sin\theta$

Then you can use a trigonometric identity to make the square root go away. It's kind of a reverse substitution - instead of u = some function of x, you have x = some function of theta. Here's how it works on your example:

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

Substitute $\displaystyle x=3\sin\theta$, so $\displaystyle dx=3\cos\theta\,d\theta$. By the trig identity $\displaystyle 1-\sin^2\theta=\cos^2\theta$, $\displaystyle \sqrt{9-x^2}=3\cos\theta$. You also have to change the limits: $\displaystyle x=-3$ corresponds to $\displaystyle \theta=-\frac{\pi}{2}$, while $\displaystyle x=3$ corresponds to $\displaystyle \theta=\frac{\pi}{2}$. So the integral becomes:

$\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(3\cos\theta)( 3\cos\theta)\,d\theta = 9\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2\theta\,d\theta$

And substituting $\displaystyle \cos^2\theta=\frac{1}{2}(1+\cos\2\theta)$ gives

$\displaystyle 9\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2}\,(1+\cos{2\theta})\,d\theta= $

$\displaystyle \left[ \frac{9}{2}\,\theta\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} + \frac{9}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos{2\theta}\,d \theta = $

$\displaystyle \frac{9\pi}{2} + \left[\frac{9}{4}\sin{2\theta}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{9\pi}{2}$

- Hollywood