Using known areas to find integrals

**Mikusboi** · January 23rd 2013, 08:05 AM

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.

**ibdutt** · January 23rd 2013, 08:17 AM

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

**BobP** · January 23rd 2013, 08:49 AM

The integral, when evaluated, will be the area 'under the curve' $y=\sqrt{9-x^{2}}$ between -3 and +3.
The equation is that of a circle, so the result will be the area of the semi-circle.

**Plato** · January 23rd 2013, 08:52 AM

Originally Posted by Mikusboi

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.

It is absolutely simple.
$\sqrt{9-x^2}$ is a semicircle center at $(0,0)$ with radius $3$ . What is its area?

**Mikusboi** · January 23rd 2013, 09:02 AM

9pi/2?

**Mikusboi** · January 23rd 2013, 09:06 AM

wow that is very simple. I completely forgot about the way circles were represented in equation form.

**hollywood** · January 23rd 2013, 06:44 PM

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

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:

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

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

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

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

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

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

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

- Hollywood

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