# Thread: Centre of Mass by integration

1. You're not pulling sine and cosine out of the air, you're using a substitution. We're saying that x will be:

$x = r\sin{\theta}$

Which automatically means that:

$\frac{dx}{d\theta} = r\cos{\theta}$

And by the properties of differentials we can conclude that:

$dx = r\cos{\theta} ~d\theta$

The rest is the result of the substitution.

2. [QUOTE=skeeter;461233]other than recognizing that the area of a semicircle is $\frac{\pi r^2}{2}$ , no.

So if I am allowed to recognize that the area of the semicircle is as above....
how could I integrate y=sqr r^2-x^2
without using substitution?

3. It would be much harder to interpret the answer without using some form of trigonometric substitution. This substitution is useful because of the form of the function in the integral and because of the circular shape.

4. Ok, but why does skeeter evaluate r=pi/2?
where did the value of r come from?

5. Originally Posted by wolfhound
Ok, but why does skeeter evaluate r=pi/2?
where did the value of r come from?
in using substitution, one must reset the limits of integration when evaluating a definite integral.

note that the substitution $x = r\sin{\theta}$ was used.

the upper limit w/r to $x$ was $r$ , we need to change that to a value of $\theta$ when integrating w/r to $\theta$

substituting $r$ for $x$ ...

$r = r\sin{\theta}$

$1 = \sin{\theta}$

$\theta = \frac{\pi}{2}$

6. ## Thank you

I see now thanks for the help
I would share my apple pie with you all but all this maths made me hungry and I ate it

7. Originally Posted by skeeter
other than recognizing that the area of a semicircle is $\frac{\pi r^2}{2}$ , no.

...
Here is a slightly different approach:

$\int \sqrt{r^2-x^2} dx = \int\left(\frac{r^2}{\sqrt{r^2-x^2}} - x \cdot \frac{x}{\sqrt{r^2-x^2}}\right)dx$

The integral of the first summand is an arcsin-function and the integral of the second summand is an irrational function (you have to use partial integration).

But of course this way isn't very easy at all.

8. ## pumpkin pi

$2r^2\int_0^{\frac{\pi}{2}} \cos^2{\theta} \, d\theta$

$2r^2\int_0^{\frac{\pi}{2}} \frac{1 - \cos(2\theta)}{2} \, d\theta$
sorry, but how does cos^2 change into 1-cos (2theta) below?

9. Originally Posted by wolfhound
$2r^2\int_0^{\frac{\pi}{2}} \cos^2{\theta} \, d\theta$

$2r^2\int_0^{\frac{\pi}{2}} \frac{1 - \cos(2\theta)}{2} \, d\theta$
sorry, but how does cos^2 change into 1-cos (2theta) below?
using a double angle identity which you should already be familiar ...

$\cos(2x) = 2\cos^2{x} - 1$

solve for $\cos^2{x}$

10. I'm sorry but I am not familiar because I have only started integration last thursday,I done 2 hours of class on definite and indefinite integrals , nothing at all on this
If you dont want to help thats fine thanks

11. Originally Posted by wolfhound
I'm sorry but I am not familiar because I have only started integration last thursday,I done 2 hours of class on definite and indefinite integrals , nothing at all on this
If you dont want to help thats fine thanks

Calculus makes frequent use of trigonometric identities, which SHOULD (but aren't always) be covered prior to taking the course. If you are not familiar with them then here is a site for them:

Trig Identities

12. but is says on that site cos^2 u = 1 + cos(2u)/2
$2r^2\int_0^{\frac{\pi}{2}} \cos^2{\theta} \, d\theta$

$2r^2\int_0^{\frac{\pi}{2}} \frac{1 - \cos(2\theta)}{2} \, d\theta$
but here it is 1 - cos(2theta)/2
why is it - here?

13. Originally Posted by wolfhound
but is says on that site cos^2 u = 1 + cos(2u)/2
$2r^2\int_0^{\frac{\pi}{2}} \cos^2{\theta} \, d\theta$

$2r^2\int_0^{\frac{\pi}{2}} \frac{1 - \cos(2\theta)}{2} \, d\theta$
but here it is 1 - cos(2theta)/2
why is it - here?
my mistake ... it should be $\frac{1+\cos(2\theta)}{2}
$

14. wolfound, you will benefit greatly by strengthening your trig skills. that substitution, $\cos^2(\theta)=\frac{1+\cos(2\theta)}{2}$, can be derived from the manifest $cos^2\theta +sin^2\theta =1$identity and the angle sum formulas as derived, and proven geometrically here:
The Sine of the Sum of Angles

this website even has a drill section with solutions, so you can test your skills once you study their modules.

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# com of a semicircular disv using integral

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