1. ## Integral

Hello everyone, I have this procedure correct?

I have used trigonometric substitution:

$\displaystyle\int_{0}^{2\pi} \displaystyle\int_{0}^{a} r^2 cos ( \phi ) \sqrt[2 ]{a^2 -r^2} dr d\phi$

$r = asen(\theta), dr = acos\theta.d\theta$

And replacing the integral I is:

$cos(\phi). a^2.sen^2(\theta) \sqrt[ ]{a^2 - (asen.\theta)^2} .acos\theta.d\theta$

Factoring $a^2$ and taking square root:

$cos(\phi). a^2.sen^2(\theta).a. \sqrt[ ]{1 - sen^2\theta} .acos\theta.d\theta$

$= cos(\phi).a^4.sen^2(\theta).cos^2(\theta).d\theta$

A greeting and thank you very much!

2. Originally Posted by Dogod11
Hello everyone, I have this procedure correct?

I have used trigonometric substitution:

$\displaystyle\int_{0}^{2\pi} \displaystyle\int_{0}^{a} r^2 cos ( \phi ) \sqrt[2 ]{a^2 -r^2} dr d\phi$

$r = asen(\theta), dr = acos\theta.d\theta$

And replacing the integral I is:

$cos(\phi). a^2.sen^2(\theta) \sqrt[ ]{a^2 - (asen.\theta)^2} .acos\theta.d\theta$

Factoring $a^2$ and taking square root:

$cos(\phi). a^2.sen^2(\theta).a. \sqrt[ ]{1 - sen^2\theta} .acos\theta.d\theta$

$= cos(\phi).a^4.sen^2(\theta).cos^2(\theta).d\theta$

A greeting and thank you very much!
If I remember correctly sin(x)=sen(x) if so

Use the identity

$\sin^2(\theta)=1-\cos^2(\theta)$

Then use $\cos^2(\theta)=\frac{1}{2}(1+\cos(2\theta))$

Three times to expand out

$(cos^2(\theta))^2$

I hope this helps

3. If I remember correctly sin(x)=sen(x) if so
Hello, you are right, sorry, my natural language is not English.

But I've been very helpful, also confirmed that it was rather the beginning,

Thanks again,

Greetings